Seaholme

Hey, I'm trying to create a page which will list all the users currently online, but all I can find via google are scripts to say how many are online, not who they are. I thought I would do it so that, when people log in, their 'online/offline' column in the database is changed from 0 to 1. However, I don't know how to make it so that when the session times out, this column will revert back to 0. Can anybody help me work out how I might be able to do this? Thanks!

Thanks so much for mentioning MySQL joins, I hadn't heard of them before. I hope you guys don't mind, but I'm finding it quite difficult to stick all this stuff together -- this is my SQL join, it's giving me the following error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'trainingentrants, entrantsWHERE trainingentrants.eventid = training.idGROUP BY t' at line 1 ... and I'm not sure why, as I'm not familiar with this function at all! I followed a tutorial I found to piece it together. // Make a MySQL Connection // Construct our join query $query = "SELECT trainingentrants.entrantscore, trainingentrants.entrantid, trainingentrants.eventid, training.gold, training.silver, training.run AS entrantstuff". "FROM trainingentrants, entrants". "WHERE trainingentrants.eventid = training.id". "GROUP BY traininentrants.entrantscore"; $result = mysql_query($query) or die(mysql_error()); // Print out the contents of each row into a table while($row = mysql_fetch_array($result)){ echo $row['entrantscore']. " - ". $row['entrantid']; echo "<br />";} Can anybody help me sort this out? :[

Thanks! I'm trying to use it, but I'm still quite new to PHP and am finding this function extremely counterintuitive. This is what I have so far: <?php // Make a MySQL Connection ..... $date = date("d/m/y"); $result = mysql_query("SELECT * FROM training WHERE rundate='$date') or die(mysql_error()); while($row = mysql_fetch_array( $result )) { foreach ($row as $value) { $entrants = mysql_query("SELECT * FROM trainingentrants WHERE eventid='$row[\'id\']' ORDER BY entrantscore ASC") or die(mysql_error()); while($entrantinfo = mysql_fetch_array( $entrants )) { $first = $row['gold']; $entrantid = $entrantinfo['entrantid']; mysql_query("UPDATE dogs SET TFA = TFA + '10' WHERE id = '$entrantid'"); } } ?> I can't work out what $value should be! Which part should I call "value"? :/

Hey all, This isn't so much a problem with code I already have as a request for some help as to what code to use in the first place! What I have are two tables, one with information about Events -- the ID of the event and the prize you get if you win it -- and another with information about the individuals entered into this event -- their ID, the ID of the event they were entered into, and their score. What I want to do is create a script which will allow me to take every single row of the table with the Events information in it and do the following: 1. gather all the information relating to a Event 2. gather the information about the individuals entered in an Event 3. list the individuals entered in the event by their score 4. award prizes to those with the 1st, 2nd and 3rd highest scores 5. delete the event My problem is not so much how to complete stages 1-5, but more how to make a script which will run through the database of Events and complete stages 1-5 on each event individually. The only way I know how to do it would be for me to manually enter all the event IDs and do it for each event separately, but I'm sure there must be some sort of PHP which will complete these events for each row in turn. Can anybody help me work out how to make this happen for every line of the database? I don't just want to gather information about each line, but to have functions 1-4 happen on it! Thanks in advance for any ideas!

Hey, Thanks very much for the suggestion; I did what you said and moved the opening of the form outside the loop. Unfortunately it doesn't seem to have fixed the problem! The submit button still doesn't seem to work independently for further options after the first one :/ This is what it's been changed to: if(isset($_POST['submit'])) { $type = $_POST['type']; $level = $_POST['level']; $entrantid = $_POST['dogid']; // Get all the data from the dogs table $result2 = mysql_query("SELECT * FROM training WHERE type='$type' AND level='$level'") or die(mysql_error()); if($result2 == ''){ echo "There are no events matching these details.";} else{ echo "<form action=arena.php method=post>"; echo "<table border='0'>"; echo "<tr><th>Event Name</th> <th>Level</th> <th>Type</th> <th>Entry Fee</th> <th>Enter Dog</th> </tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result2 )) { $eventno = $row['id']; $creator = $row['runby']; $fee = $row['fee']; $submit = ($eventno + '1'); // Print out the contents of each row into a table echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['level']; echo "</td><td>"; echo $row['type']; echo "</td><td>"; echo $row['fee']; echo "</td><td>"; echo " <input type=hidden name=eventno value=$eventno> <input type=hidden name=entrantid value=$entrantid>"; echo "<input type=submit name=submit2 value='Enter Dog!'>"; echo "</td></tr>"; } echo "</table></form>"; }} if(isset($_POST['submit2'])) { $eventid = $_POST['eventno']; $entrantid = $_POST['entrantid']; $entrantscore = '2'; // Get all the data from the dogs table $result3 = mysql_query("SELECT * FROM trainingentrants WHERE eventid='$eventid' AND entrantid='$entrantid'") or die(mysql_error()); if($result3 != ''){ echo "Oops, this dog is already participating in this event!";} else { echo "Success! Your Dog has been entered into this Event."; $enterDog = @mysql_query("INSERT INTO trainingentrants (eventid, entrantid, entrantscore) VALUES ('$eventid', '$entrantid', '$entrantscore')") or die("Error: ".mysql_error());

You should be able to use this: http://php.net/manual/en/function.preg-replace.php Example 4 looks much like what you want to do, if you just put your own variables in I'm not claiming this is the best way or that I'm especially experienced at this! I just reckon that would be a solution. Also, what you just used is the hash key/a sharp sign. £ is a pound sign!

Hey, I'm trying to get a form to appear in a loop, but I'm having some issues with how to make it so that each individual submit button works for its individual form -- I can't seem to come up with a solution as to how to make the whole thing work :/ This is the relevant section of PHP: // Get all the data from the dogs table $result2 = mysql_query("SELECT * FROM training WHERE type='$type' AND level='$level'") or die(mysql_error()); if($result2 == ''){ echo "There are no events matching these details.";} else{ echo "<table border='0'>"; echo "<tr><th>Event Name</th> <th>Level</th> <th>Type</th> <th>Entry Fee</th> <th>Enter Dog</th> </tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result2 )) { $eventno = $row['id']; $creator = $row['runby']; $fee = $row['fee']; $submit = ($eventno + '1'); // Print out the contents of each row into a table echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['level']; echo "</td><td>"; echo $row['type']; echo "</td><td>"; echo $row['fee']; echo "</td><td>"; echo "<form action=arena.php method=post> <input type=hidden name=eventno value=$eventno> <input type=hidden name=entrantid value=$entrantid>"; echo "<input type=submit name=submit2 value='Enter Dog!'>"; echo "</td></tr>"; } echo "</table></form>"; }} if(isset($_POST['submit2'])) { $eventid = $_POST['eventno']; $entrantid = $_POST['entrantid']; $entrantscore = '2'; // Get all the data from the dogs table $result3 = mysql_query("SELECT * FROM trainingentrants WHERE eventid='$eventid' AND entrantid='$entrantid'") or die(mysql_error()); if($result3 != ''){ echo "Oops, this dog is already participating in this event!";} else { echo "Success! Your Dog has been entered into this Event."; $enterDog = @mysql_query("INSERT INTO trainingentrants (eventid, entrantid, entrantscore) VALUES ('$eventid', '$entrantid', '$entrantscore')") or die("Error: ".mysql_error()); // give entry fee to creator $sql4=mysql_query("UPDATE players SET onhand = onhand + '$fee' WHERE id='$creator'"); $result4=mysql_query($sql4); Basically the bit where the formaction=arena.php I want it so that every single Event (just to make this clear, the idea is that you are presented with a list of "Events" from the database to enter your dogs into!) has a submit button which will enter the dog into the event. I'm not sure whether the error is in the submit button or what, but basically I really need help working out how to make the form function in the loop as well as just showing up in it! :S All and any advice would be appreciated! Thanks!

Bizarre xD It sounds like a made-up word. Got that one wrong Anyway thanks a bunch for the help, it fixed it

Hey, I put "pagination" in little bunny ear things because it's a stupid made-up word, however it was the name of the tutorial I used to try and get my forum to automatically generate further rows in new pages after its counted a certain number of rows. Basically I can get the "Next" and "Last" buttons to appear, and it does show 10 rows... only those rows are completely blank (it's not filling in any of the data from the database) and perceptibly the "next" and "last" buttons change the URL but don't seem to be changing the number of blank rows displayed from 10 -- and there aren't 20 posts total on the board, so that leads me to conclude that they don't work to show the remaining posts on the next page :/ I'm hesitant to ask here as so little of this actually -IS- working, so it seems to me like a lot of work will be required to fix it, but if people could point out any glaring errors to help me at least get some of it functioning, that would be really appreciated. I've included pretty much all of the PHP from that page below. I guess it should be quite clear which bits are from the tutorial and which bits are mine, on account of it talking in the third person! // fetch line id from URL $id=$_GET['id']; // find all posts on this board $sql="SELECT * FROM $tbl_name WHERE forumid=$id ORDER BY lastreply DESC"; // ORDER BY id descending $result=mysql_query($sql); ?> <table width="70%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> <tr> <td width="50%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td> <td width="12%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td> <td width="12%" align="center" bgcolor="#E6E6E6"><strong>Author</strong></td> <td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td> <td width="13%" align="center" bgcolor="#E6E6E6"><strong>Last Reply</strong></td> </tr> <?php if (!(isset($pagenum))) { $pagenum = 1; } //Here we count the number of results //Edit $data to be your query $data = mysql_query("SELECT * FROM $tbl_name WHERE forumid=$id ORDER BY lastreply DESC") or die(mysql_error()); $rows = mysql_num_rows($data); //This is the number of results displayed per page $page_rows = 10; //This tells us the page number of our last page $last = ceil($rows/$page_rows); //this makes sure the page number isn't below one, or more than our maximum pages if ($pagenum < 1) { $pagenum = 1; } elseif ($pagenum > $last) { $pagenum = $last; } //This sets the range to display in our query $max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; $data_p = mysql_query("SELECT * FROM $tbl_name $max") or die(mysql_error()); //This is where you display your query results while($info = mysql_fetch_array( $data_p )) { ?> <tr> <td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td> <td align="center" bgcolor="#FFFFFF"><a href=profile.php?id=<? echo $rows['name']; ?>><? echo $rows['displayname']; ?> (#<? echo $rows['name']; ?>)</a></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td> <td align="center" bgcolor="#FFFFFF"><? echo $rows['lastreply']; ?> by (#<? echo $rows['lastreplyauthor']; ?>)</td> </tr> <?php echo "<br>"; } echo "<p>"; // This shows the user what page they are on, and the total number of pages echo " --Page $pagenum of $last-- <p>"; // First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page. if ($pagenum == 1) { } else { echo " <a href='{$_SERVER['PHP_SELF']}?id=$id&pagenum=1'> <<-First</a> "; echo " "; $previous = $pagenum-1; echo " <a href='{$_SERVER['PHP_SELF']}?id=$id&pagenum=$previous'> <-Previous</a> "; } //just a spacer echo " ---- "; //This does the same as above, only checking if we are on the last page, and then generating the Next and Last links if ($pagenum == $last) { } else { $next = $pagenum+1; echo " <a href='{$_SERVER['PHP_SELF']}?id=$id&pagenum=$next'>Next -></a> "; echo " "; echo " <a href='{$_SERVER['PHP_SELF']}?id=$id&pagenum=$last'>Last ->></a> "; } // Start looping table row while($rows=mysql_fetch_array($result)){ ?> <?php // Exit looping and close connection } mysql_close(); ?> <tr> <td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php?id=<? echo $id; ?>"> <strong>Create New Topic</strong></a></td> </tr> </table> <?php //close max/min if statement Thanks!

Wow, thanks so much! All sorted. Also, you don't have to answer this as technically my problem is solved... but can you be a little more detailed as to what XSS injection might be? I've been paying zero attention to security with my site, really. Is it people inputting PHP onto the page somehow? And if so, do you suggest something along the lines of that function should accompany every userinput? I'm quite new to the world of coding so sorry if these are pretty basic things!

It's not quite when they're displayed -- they display okay, it's just when it comes to the form which updates them. I'll give you an example, it's really hard to explain! xD The error occurs in this HTML here: <table width="700" border="1" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="4">Update your user information: </td> </tr> <tr> <td align="center">Username</td> <td align="center">Camp Name</td> <td align="center">Password</td> <td align="center">Email</td> <td align="center"> </td> </tr> <tr> <td><input name="displayname" type="text" id="displayname" value="<? echo $rows['displayname']; ?>"></td> <td><input name="campname" type="text" id="campname" value="<? echo $rows['campname']; ?>"></td> <td><input name="password" type="password" id="password" value=""></td> <td><input name="email" type="text" id="email" value="<? echo $rows['email']; ?>"></td> </tr> <tr> <td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> </tr> </table> </td> </form> </tr> </table> (Sorry it's kinda not formatted that beautifully). Basically if somebody enters in a displayname that contains " or ', then the part of the form which is meant to echo the displayname shows up as a textarea with anything after the " or ' showing up outside the textarea. To demonstrate this a bit more clearly, if you imagine that [ and ] define the beginning and end of the textarea, a name originally entered like this: [<b>Name</b">] Ends up like this: [<b>Name</b ] "> Does that help explain it a bit better?

Heya, I have a little issue which is that whenever people update their displaynames using " and ', it interferes with the update field by effectively cutting off anything that goes after it. I tried this: $displaynamex = htmlspecialchars($displayname, ENT_QUOTES); However, I don't want it to take away ALL the characters -- just ' and ", I'd really like it if HTML tags like > and < stayed put. And although I thought that this would technically work, I've had no luck with it, it still cuts out... $displaynamex = str_replace('"', '\"' ,$displayname); Can anybody suggest any solutions for somehow making " and ' safe to enter in? Thanks!

Hi, I don't know if this is exactly what you're looking for (I'm a little confused as to how you would do this without a database), but I use this function in order to add on (or subtract) time from a current one. It's extremely simple -- even I don't get confused by it!! And seriously, that's a saying a lot, unfortunately! xD http://php.net/manual/en/function.strtotime.php Hope it helps.

Thanks so much guys. Unfortunately I've put this variable in multiple places already so I might have to go with the ugly solution, or go back and change tonnes of stuff!

Hi there, I've no idea how this broke, but somehow I've managed to generate an error in this and I have no idea what's wrong to go about fixing it This is the error I get Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INT, TBY) VALUES ('Name', 'Gender', 'Breed', 'Region', '07/12/10'' at line 1 // Finally, create the record. $newDog = @mysql_query("INSERT INTO dogs (name, gender, breed, region, date, owner, trait1, trait2, agree, STR, SPD, END, HRD, TFA, AGI, SND, FIT, INT, TBY) VALUES ('$name', '$gender', '$breed', '$region', '$date', $owner, '$trait1', '$trait2', '$agreeableness', '$high', '$verylow', '$high2', '$veryhigh', '$low', '$TFA', '$low2', '$SND', '$FIT', '$medium')") or die("Error: ".mysql_error()); I've looked particularly in the area it's complaining about, but no success :/ Does anybody have any idea why it might not be working? Just in case this helps, I had no errors in this until I added in the following variables: $verylow = rand(10, 13); $low = rand(13, 16); $low2 = rand(13, 16); $medium = rand(16, 21); $high = rand(21, 25); $high2 = rand(21, 25); $veryhigh = rand(25, 30); $TFA = rand(15, 70); $SND = rand(60, 90); $FIT = '50'; Thanks!

Thankyou thankyou thankyou thankyou! Working now

Hey all (again) I'm back to ask for more assistance with the exact same dropdown list. Yep. I keep discovering new and unsolvable (by myself, anyway!) problems with just one thing, and for some reason nothing else is going wrong for me! I promise I'll stop plaguing this forum with questions at some point! Anyway: the problem is that I have a dropdown list which fetches a list from the database and shows all the entries fitting a particular criterion in that list... however, no matter what entry is selected from the dropdown list, the only one ever sent by the form is always the last entry. So effectively there's no point in having a dropdown list, just the final entry. This is the PHP // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table $dogid = $row['id']; } echo '<bR><br>Select a dog for a basic checkup<br><form action="basiccheckup.php" method=post> <select name=dogid>'; // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo "<option value=\"$dogid\">{$row['name']}</option>"; } echo '</select>'; Any ideas as to how to solve it? I assume the problem must lie in the form somehow, but I'm not 100%.

Urgh, why did I do that?? *headbash* Thanks so much. Clearly I don't pay enough attention!

Oof. Okay I still haven't solved this, so here's the full code for each of the pages (bar, of course, the connecting to the server bits)... // Connect to server and select database. mysql_connect("$dbhost", "$dbuser", "$dbpass")or die("cannot connect"); mysql_select_db("$dbname")or die("cannot select DB"); echo "Welcome to the Vets (n.b. not yet working)<br> Here the trained veterinarians will check your dogs over and give them certificates of clean health,<br> which will need to be renewed regularly in order to compete in training and races."; // Make a MySQL Connection mysql_connect("", "", "") or die(mysql_error()); mysql_select_db("") or die(mysql_error()); // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table $dogid = $row['id']; } echo '<bR><br>Select a dog for a basic checkup<br><form action="basiccheckup.php" method=post> <select name=\"$dogid\">'; // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo '<option value=\"$dogid\">'; echo $row['name']; echo '</option>'; } echo '</select>'; // A simple example of a form. echo ' <input type=submit name=submit value="Basic Health Check? $100"> </form>'; // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); echo '<bR>Select a dog for a special pre-race checkup<br><form action=preracecheckup.php method=post> <select name=name>'; while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo '<option value="dog2">'; echo $row['name']; echo '</option>'; } echo '</select>'; // A simple example of a form. echo ' <input type=submit name=submit value="Pre-race Health Check? $150"> </form>'; include('footer.php'); and then the second page... // Connect to server and select database. mysql_connect("$dbhost", "$dbuser", "$dbpass")or die("cannot connect"); mysql_select_db("$dbname")or die("cannot select DB"); // fetch info from form $dogname = $_POST['$dogid']; echo $dogname; // test to see if person has enough money to complete the checkup $result = mysql_query("SELECT onhand FROM players WHERE id=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { $testmoney = $row['onhand'];} if(($testmoney - '100') < '0') { echo "Oops, you don't have enough money for this!"; } else { // set buyer as new owner $sql2=("UPDATE dogs SET vetstatus = 'Vet Certificate Obtained' WHERE id='$dogname'"); $result2=mysql_query($sql2); // subtract money from buyer $sql3=mysql_query("UPDATE players SET onhand = onhand - '100' WHERE id=".$_SESSION['id']); $result3=mysql_query($sql3); // if successfully updated. if($result2){ echo "$dogname has been given a clean bill of health by the vet, who hands you a certificate to that effect.<br> You will now be able to enter this dog in training!"; echo " $100"; echo " has been deducted from your account in veterinary fees.";} else{ echo "Oops, didn't work";} } include('footer.php'); Just did view source as you suggested and the form looks like its being reproduced fine... at least I think so. Ought it to be showing the value of the options as ""$dogid\"? I'm pretty sure it ought to be showing those as the IDs it has got from the database there? Welcome to the Vets (n.b. not yet working)<br> Here the trained veterinarians will check your dogs over and give them certificates of clean health,<br> which will need to be renewed regularly in order to compete in training and races.<bR><br>Select a dog for a basic checkup<br><form action="basiccheckup.php" method=post> <select name=\"$dogid\"><option value=\"$dogid\">My Favrit</option><option value=\"$dogid\">hg</option><option value=\"$dogid\">jhju</option><option value=\"$dogid\">Wanda</option><option value=\"$dogid\">Hippe</option><option value=\"$dogid\">o</option><option value=\"$dogid\">Carlos</option><option value=\"$dogid\">Violette</option></select> <input type=submit name=submit value="Basic Health Check? $100"> </form><bR>Select a dog for a special pre-race checkup<br><form action=preracecheckup.php method=post> <select name=name><option value="dog2">My Favrit</option><option value="dog2">hg</option><option value="dog2">jhju</option><option value="dog2">Wanda</option><option value="dog2">Hippe</option><option value="dog2">o</option><option value="dog2">Carlos</option><option value="dog2">Violette</option></select> <input type=submit name=submit value="Pre-race Health Check? $150"> </form><hr>

Thanks, but now it just doesn't send anything through again :[ I can't work out where the missing link is. When, on page one, I echo '$dogid' it works (well, it echoes the ID of the last one on the dropdown list), but then it doesn't work when I try to echo '$dogname' on the next page.

Thing is I've actually tried this (sorry I didn't say, I've been sitting trying all sorts of combinations) and all it sends is literally $dogid ... as in the variable written out!

Hi all, I'm having difficulty getting the information to send properly from this form. Basically I want the "value" of the option to be equal to the ID number fetched from the database ($row['id']), however I can't seem to get it to send across to the next page. This is the form to send the number: // Get all the data from the dogs table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); echo '<bR><br>Select a dog for a basic checkup<br><form action="basiccheckup.php" method=post> <select name="DogID">'; while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table $dogid = $row['id']; echo '<option value=$dogid name=$dogid>'; echo $row['name']; echo '</option>'; } echo '</select>'; ... and this is how I'm trying to collect the data on the next page // fetch info from form $dogname = $_POST['$dogid']; echo $dogname; I've never made a form where what's being sent is a PHP variable before, and can't seem to get it to work! I'd really appreciate it if anybody is able to help. Please let me know if you need more details! Thanks

Hey guys, I'm trying to populate a multi dropdown list from a database where it finds all of the dogs (bit of a silly topic, I know, but whatever!) belonging to a certain owner and lists them all in a dropdown list. All I can seem to manage is this, which displays every single dog listed under the owner name, but it gives every single one their own dropdown list, of which they are the sole occupant. I really want them all to just sit in the same dropdown list. Can anybody help me work out how to do it? I've included the PHP below that I'm using currently... although, as I said, it doesn't work! // Get all the data from table $result = mysql_query("SELECT * FROM dogs WHERE owner=".$_SESSION['id']) or die(mysql_error()); while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo '<form action=rader.php method=post> Name: <select name=name> <option value="$row[\'id\']">'; echo $row['id']; echo $row['name']; echo '</option></select><br><br></form>'; ?> Thanks in advance!

Ah, I had no idea about that! Thanks a bunch... changed the word 'key' as you suggested x)

Hey guys, I'm trying to make a lost password tool for my site which basically works like so: you enter in your username, you get sent an email with a random key in it and at the same time a temporary record is created in the database of your account ID number and the random key. You can then use these bits of information to set a new password. I have a similar system used to validate people's accounts when they join, and so I assumed it'd be a relatively straightforward job of just editing it to work slightly differently. However, for some reason it's not working. Whenever I try to make the following work, I get this error: Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'key) VALUES ('4', '32a63b0d6f450d320821a51965e1b455')' at line 1 Those values are the ID number and random key. When I change it so that the key doesn't get sent, the whole thing works perfectly (apart from the fact that the key is not in the database, of course....) so that's definitely where the problem seems to be! Can anybody help me work out what's wrong? I've been staring at it for ages and failed to spot the error! //Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect to server"); mysql_select_db("$db_name")or die("cannot select DB"); // value for username sent from form $username=$_POST['username']; // create a random key to insert into lostpasswords table $key=md5(uniqid(rand())); // retrieve the information from the players table for that username $sql="SELECT * FROM $tbl_name WHERE username='$username'"; $result=mysql_query($sql); // if the username has been found, the row must be 1 row // keep value in variable name "$count" $count=mysql_num_rows($result); // compare if $count =1 row if($count==1){ $rows=mysql_fetch_array($result); // retrieve the player ID and email address $id=$rows['id']; $email_to=$rows['email']; // Insert data into the lostpasswords table for pickup $sql2 = @mysql_query("INSERT INTO lostpasswords (id, key) VALUES ('$id', '$key')") or die("Error: ".mysql_error()); $result2=mysql_query($sql2); Thanks!