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Hangwire

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  1. Hangwire
    Thank you all for your replies. Your feedback has been very helpful, and I'm going to re-do the entire project.
  2. Hangwire
    Thank you for your reply! Nah, I'm absolutely sure I didn't add them. This is the code of every php script in the whole project: Add-To-Database script: <?php $cocode = $_POST['cocode']; $coname = $_POST['coname']; mysql_connect ("pdb1.myhostingcompany.com", "countries", "*********") or die ('Error: '. mysql_error()); mysql_select_db ("countries") $query="INSERT INTO countries (cocode, coname) VALUES ('".$cocode."', '".$coname."')"; mysql_query($query) or die ('Error updating database'); echo " <html> <body> <center> Success! Country added. </center> </body> </html> " ; ?> Search-In-Database script: <?php mysql_connect ("pdb1.myhostingcompany.com", "countries","**********") or die (mysql_error()); mysql_select_db ("countries"); $term = $_POST['term']; $sql = mysql_query("select * FROM countries WHERE cocode = '$term'"); $num_rows = mysql_num_rows($sql); if ($num_rows == 0) { echo "No results found."; exit; } while ($row = mysql_fetch_array($sql)){ echo '<br/> Code: '.$row['cocode']; echo '<br/> Country: '.$row['coname']; echo '<br/><br/>'; } ?> I've searched the entire database for those names and couldn't find a thing. I've examined the entries for numbers 380, 401 and 403, everything looks normal. I can extract the database into an .sql, but I think that's a bit redundant. If anyone wants it, I can send it to them. Thank you for your time.
  3. Hangwire
    Hello everybody, An year ago while learning php I made a basic function to insert rows in a database and then a function to read from that database. Tl;dr I made a barcode reader. You enter the first three or two digits from the barcode of a product and you get what country it's produced in. A month ago I tried it out just by accident because some of my friends were wondering about where a product was made and my tool came to mind. When I enter 380 (Code for Bulgaria), Bulgaria pops up along with a line under it that says Jesus. When I enter 401 (One of the codes for Germany), Germany comes up along with three names - Leah, Nathaniel and Robert. 403 is another code for Germany. When I enter that, along with Germany comes up Destiny. I honestly have no explanation for this except that it's maybe some kind of SQL insert hack and someone tried to be funny. You can all test this out here: http://training.nbrain.net/searchform.html Thanks.
  4. Hangwire
    Thank you, I presume I can insert html code in the area between the comas?
  5. Hangwire
    Hello all. I'm using this code to go through the database and output certain user-defined numbers <?php mysql_connect ("pdb1.awardspace.com", "anastasov_db","moscow1945") or die (mysql_error()); mysql_select_db ("anastasov_db"); $term = $_POST['term']; $sql = mysql_query("select * FROM countries WHERE cocode = '$term'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Code: '.$row['cocode']; echo '<br/> Country: '.$row['coname']; echo '<br/><br/>'; } ?> Now how would I go about making a page that appears if the script can't find any results in the database? Thank you in advance
  6. Hangwire
    My bad, removed the wildcards at $term, it works fine now. Thanks everyone!
  7. Hangwire
    $sql = mysql_query("select * FROM countries WHERE cocode = '%$term%'"); This is the change I made, but now it doesn't give any output at all.
  8. Hangwire
    Hello all. First of all, if this isn't the right place for this topic, please excuse me and may the moderators/admins move it where it should belong. Second, I'm trying to make a mysql search - Everything works, but if say, a user searches for 34 it displays all numbers that have the number 34 in them. I want the results to be exact - if a user searches for 34, to display only 34, not 334, 5034 or 3401 (like it happens now). Here's the code: <?php mysql_connect ("**************", "*******","*****") or die (mysql_error()); mysql_select_db ("********"); $term = $_POST['term']; $sql = mysql_query("select * FROM countries WHERE cocode LIKE '%$term%'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Code: '.$row['cocode']; echo '<br/> Country: '.$row['coname']; echo '<br/><br/>'; } ?> If you have any questions, you're welcome to ask them in the thread. Thank you in advance
  9. Hangwire
    Thank you for your time, I fixed it myself. It was a very stupid error on my behalf, in the html input form I put the of it as "password", but $_POST was looking for "pass". Fixed it and everything works like a charm.
  10. Hangwire
    1) Wont bump, but there's another rule - All users must be courteous to others. 2) As I said in the first post, it was actually very late so... excuse me, in every post after the first one I presumed I had already mentioned it was concerning apache - didn't do that, but I gave an actual link, so it should have been pretty obvious. Either way, my mistake. Thank you so much for your help, I edited the directory and it's working like a charm! All the best!
  11. Hangwire
    Be a little more specific, what error does it exactly output?
  12. Hangwire
    I don't want to create a thread about another simple problem, but here it is... <?php $name = $_POST['name']; $pass = $_POST['pass']; // $ip = $_SERVER['ip']; mysql_connect ("localhost", "root", "****") or die ('Error: '. mysql_error()); mysql_select_db ("data"); $query="INSERT INTO data (name, pass) VALUES ('".$name."', '".$pass."')"; mysql_query($query) or die ('Error updating database'); In this case, the script only writes the name in the database, but not the pass. The database fields are identical, both being varchar's. So, what gives?
  13. Hangwire
    Not at all what I had in mind. I expect to see a reply in a good tone, just as I always refer to anybody else on this site. But when somebody is condescending... People should know more than just scripting languages. Let's go back on topic, though. Any ideas?
  14. Hangwire
    It's an apache error! That's the whole point! How can a php script output an "Object not found!" error from apache?! And, to Dan: 1) You know, there is more than 1 timezone on this planet. 2) Have you ever worked with apache, or do I need to be dead specific about everything? 3) Thank you.
  15. Hangwire
    Thank you, worked like a charm
  16. Hangwire
    Hi everybody, my goal is to get the IP of someone accesing the site, writing the time and date along with his IP into the database. Of course, I would be adding the script to my frontpage when I make it work. But I get this error: I've checked line 8, I dont find anything in it that is out of place. Here is the code: <?php $ip = $_SERVER['REMOTE_ADDR']; $date = date("m.d.y"); $time = time(); mysql_connect ("localhost", "root", "********") or die ('Error: '. mysql_error()); mysql_select_db ("ip"); $query = "INSERT INTO ipdo (time, date, ip) VALUES ('"$time"', '"$date"', '"$ip"')"; mysql_query($query) or die ('Error updating database'); echo "Database updated with: " .$ip. "" ; ?> This is the first script I write entirely on my own, so be gentle Help?
  17. Hangwire
    bump?
  18. Hangwire
    Indeed it was late. This is the function for posting an article <?php // Connect to database include("../settings.php"); $conn = mysql_connect($server, $user, $pass); if (!$conn) die ("Couldn't connect to the database"); mysql_select_db($db, $conn) or die (mysql_error()); // Data from the form $article_id = ($_POST['article_id']); $name = mysql_real_escape_string($_POST['name']); $website = mysql_real_escape_string($_POST['website']); $email = mysql_real_escape_string($_POST['email']); $comment = mysql_real_escape_string($_POST['comment']); $date = date("Y-m-d h:i:s"); $ip = $_SERVER["REMOTE_ADDR"]; // Then we check if everything is filled out if ($name == ""){ echo "<div style='margin:0 auto;width:200px;text-align:center;'> <img src='../images/caution.gif' alt='caution' /><br /> <p>Please fill in your name!</p> <p><a href='javascript: history.go(-1)' style='color:#f7941d;'>Click here</a> to go back..</p> </div>"; } elseif ($comment == ""){ echo "<div style='margin:0 auto;width:200px;text-align:center;'> <img src='../images/caution.gif' alt='caution' /><br /> <p>You <b>have</b> to fill in the comment!</p> <p><a href='javascript: history.go(-1)' style='color:#f7941d;'>Click here</a> to go back..</p> </div>"; } else{ $query = "INSERT INTO comments (article_id,name,website,email,comment,date,ip) VALUES ('$article_id','$name','$website','$email','$comment','$date','$ip')"; $result = mysql_query($query) or die ("Error = ".mysql_error().""); header("location: ../includes/ok2.php"); } mysql_close($conn); And this is what I found for displaying the article. function DisplayLatest(){ // Includes the settings file include("settings.php"); // Connect to database Connect(); // Fetch from database $query = "SELECT * FROM articles ORDER BY id DESC LIMIT ".$limit .""; $result = mysql_query($query); $numrows = mysql_num_rows($result); // If there are results, show them if (!$numrows == ""){ while($row = mysql_fetch_array($result)){ $id = $row['id']; $name = Secure($row['name']); $title = Secure($row['title']); $content = stripslashes($row['content']); $date = $row['date']; $ip = $row['ip']; $on = date('F j, Y', strtotime($row['date'])); $at = date('g:i a', strtotime($row['date'])); echo " <div class='article'> <h2><a href=\"article.php?id=$id\" title=\"Read more about $title\">$title</a></h2> <span class='author'>By: $name on $on @ $at</span><span class='comments'><a href='article.php?id=$id'>"; echo CheckComments($id); echo "</a></span> <div class='post'><p>"; echo $content; echo "</p></div></div>"; } } Again, I'm receiving an "Object not found" error when it's clearly written in the database - thats the whole issue.
  19. Hangwire
    Hi everybody. I'm using a very simple article script that displays news and so on where added. It also has an admin panel of sorts, very simple, from where you actually post the articles. I have this problem - The news show up on the front page just fine, but when I click the link to the article itself the server says that it cannot be found. I checked the database, checked the connections and everything. I also don't see the posting comment button show up, but everything is okay with the script and the database - perhaps a formatting problem? Anyway, you can try THIS for yourself to see what I'm talking about. Here is the code: function DisplayComments(){ // For the database settings include("settings.php"); // Then make the connection Connect(); // To know what article we look at $article = ($_GET['id']); // To make every second row another color $count = 0; // Fetch the results from database $query = "SELECT * FROM comments WHERE article_id LIKE '$article' ORDER BY id DESC"; $result = mysql_query($query); $numrows = mysql_num_rows($result); // If there are any results, show them if (!$numrows == ""){ while($row = mysql_fetch_array($result)){ $name = Secure($row['name']); $website = Secure($row['website']); $email = Secure($row['email']); $comment = htmlentities($row['comment']); $date = $row['date']; $on = date('F j, Y', strtotime($row['date'])); $at = date('g:i a', strtotime($row['date'])); $color = ($count % 2) ? $color1 : $color2; $border = ($count % 2) ? $border1 : $border2; echo "<div class='comm' style='background:$color;border-top:1px solid $border;border-bottom:1px solid $border;'>"; if (!$website == ""){ echo "<span class='by'><a href=\"$website\">$name</a></span>"; } else{ echo "<span class='by'>$name</span>"; } echo"<br /><span class='time'>$on @ $at</span><p>"; echo BBCode($comment); echo "</p></div>"; $count++; } } // If there are no results else{ echo "<p>There are no comments for this article yet.</p>"; } mysql_close(); } function CheckComments ($id){ // First include the settings file include("settings.php"); // Then we make the connection Connect(); // Now we do the counting $result = mysql_query("SELECT COUNT(*) as count FROM comments WHERE article_id LIKE $id") or die("Error fetching number in DB<br>".mysql_error()); $row = mysql_fetch_array($result); $numofrows = $row['count']; $total = ceil($numofrows / $archive); // To display 'comment' if there are more then one // Otherwhise display 'comments' if ($numofrows == 1){ $comm = "Comment"; } else{ $comm = "Comments"; } // If there are any comments, output the number of comments if (!$total == ""){ echo "$numofrows $comm"; } // Otherwhise we output a 'no' else{ echo "<br> No $comm"; } } That gives me an only an "Object not found". Also, as I said, I don't know how to add the comment button to each article as it currently doesn't show. Here is the comment form function: <?php // Runs the function the display only one article DisplayOne(); ?> <h4>Comments:</h4> <?php // Runs the funtion to display the comments that belongs to the article displayed DisplayComments(); ?> <?php echo "<h5>Add Your Comment</h5>"; // Variable to know what article we are commenting on $article_id = intval($_GET['id']); // Includes the comment form include("comments/form.php"); ?> I hope you can understand what I tried to say, as it's 4.40AM here and my brain isn't working very well. Thank you in advance!
  20. Hangwire
    Alright, well I've decided to let that old script go and try with this: <?php function check_port($port) { $conn = @fsockopen("127.0.0.1", $port, $errno, $errstr, 0.2); if ($conn) { fclose($conn); return true; } } function server_report() { $report = array(); $svcs = array('21'=>'FTP', '22'=>'SSH', '25'=>'SMTP', '80'=>'HTTP', '110'=>'POP3', '143'=>'IMAP', '3306'=>'MySQL'); foreach ($svcs as $port=>$service) { $report[$service] = check_port($port); } return $report; } $report = server_report(); ?> <table> <tr> <td>Service</td> <td>Status</td> </tr> <tr> <td>FTP</td> <td><?php echo $report['FTP'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>SSH</td> <td><?php echo $report['SSH'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>SMTP</td> <td><?php echo $report['SMTP'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>HTTP</td> <td><?php echo $report['HTTP'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>POP3</td> <td><?php echo $report['POP3'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>IMAP</td> <td><?php echo $report['IMAP'] ? "Online" : "Offline"; ?></td> </tr> <tr> <td>MySQL</td> <td><?php echo $report['MySQL'] ? "Online" : "Offline"; ?></td> </tr> </table> So far so good, it works like a charm on its own but I can't implement it into anything. For example, when I put <?php include("server2.php"); ?> In the right place on my website, it doesn't show anything. The script is in the same folder... Help?
  21. Hangwire
    bump?
  22. Hangwire
    Thank you for the quick reply! I took good look at what you gave me, I tried it, but again, to no avail. I added <?php include(server2.php); ?> on the place i needed it, but it doesn't show up, but when I open server2.php from the server I can see it works and displays everything properly. My server2.php (the script) is in the main directory of apache (htdocs). I also tried to put it in the folder of the current index im working on, but again - it doesn't work. Help?
  23. Hangwire
    Hi all! Yet again I have a problem. I really couldn't find what the heck's wrong with this thing. It's extremely simple and still... I'm trying to put a php script in my website to monitor the status of my FTP and HFS servers. Help? <TITLE>Server Status</TITLE> <?php $data .= " </div> <center> <div style=\"border-bottom:1px #999999 solid;width:480;\"><b> <font size='1' color='#3896CC'>Server Status</font></b> </div> </center> <br>"; //configure script $timeout = "1"; //set service checks $port[1] = "80"; $service[1] = "Apache"; $ip[1] = "localhost"; $icon[1] = "habbo.jpg"; $alt[1] = "Apache"; $port[2] = "8080"; $service[2] = "HFS Server"; $ip[2] = "localhost"; $icon[2] = "web.png"; $alt[2] = "HFS Server"; $port[3] = "21"; $service[3] = "FTP Server"; $ip[3] = "localhost"; $icon[3] = "web.png"; $alt[3] = "FTP Server"; // // NO NEED TO EDIT BEYOND HERE // UNLESS YOU WISH TO CHANGE STYLE OF RESULTS // //count arrays $ports = count($port); $ports = $ports + 1; $count = 1; //beggin table for status $data .= "<table width='480' border='1' cellspacing='0' cellpadding='3' style='border-collapse:none' bordercolor='#CCCCCC' align='center'>"; while($count < $ports){ if($ip[$count]==""){ $ip[$count] = "localhost"; } $fp = @fsockopen("$ip[$count]", $port[$count], $errno, $errstr, $timeout); if (!$fp) { $data .= "<tr><td width='18' align='center'><img src='http://127.0.0.1/stat/$icon[$count]' title='$alt[$count]' alt='$alt[$count]'></td><td>$service[$count]</td><td width='18' align='center'><img src='http://127.0.0.1/stat/off.png'></td></tr>"; } else { $data .= "<tr><td width='18' align='center'><img src='http://127.0.0.1/stat/$icon[$count]' title='$alt[$count]' alt='$alt[$count]'></td><td>$service[$count]</td><td width='18' align='center'><img src='http://127.0.0.1/stat/on.png'></td></tr>"; fclose($fp); } $count++; } //close table $data .= "</table><div>"; echo $data; ?> The output of this is... Also, does somebody have a link or something to a tutorial/pre-made script (although I'd like to follow instructions about it) about server statistics? I want to have data on how much a certain partition on the server has free or full. Thank you in advance.
  24. Hangwire
    Problem is i've got my mind at seperate places, you could say im not that concentrated. Stupid mistake on my behalf, oh well. Thank you.
  25. Hangwire
    Parse error: syntax error, unexpected T_VARIABLE in D:\Program Files\xampp\xampp\htdocs\search.php on line 34 I'm starting to lose hope about this thing...
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