cpharry

Hi, I have got everything working except one thing. When the query that i run returns with a H i want it to change the echoed H's colour to red and the other option which is an A if that is returned echo it green. <?php $ally = str_replace("H","<p color= #FF0000 >H</p>",$row['ally']); $ally = str_replace("A","<p color='green'>A</p>",$row['ally']); echo $ally; ?> This is what i have alread but it isnt working. Any help appriciated.

How do i split up the search results from the different tables because if i have two results both with the id 1 when the user clicks on the link which uses the id to get the data from the table it will get confused, so how do i split the results up into say an echoed table for the different tabled results if that makes sense. Character Result Table ---------------------------- Character result id 1 Location Result Table --------------------------- Location result id 1 And currently whenever i search for something the entire two tables just echo all the data in them from the name column not what i searched for. <? $search = $_POST["search"]; $connection = mysql_connect("localhost","wowbasec","PASSWORD REMOVED"); mysql_select_db("wowbasec_class", $connection); $query = "select * from characters, location WHERE characters.name='$search' or location.name='$search' "; $result = mysql_query($query, $connection); if (!$result) { echo 'Invalid query: ' . mysql_error(). "<br>"; echo 'SQL Query: ' . $query. "<br>"; die(); } while ($r = mysql_fetch_assoc($result)) { // Begin while $ts = $r["name"]; ?> <table width="1000" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><a href="view.php?id=<?php echo $r['id'];?>"> <? echo $r['name'] ?></td> </tr> </table> <? } ?>

Just realised that if i put the 'name' like that instead of just name it doesnt give any errors but it also doesnt give any results which there are two.

This is the code now... <? $search = $_POST["search"]; $db = "`char`"; $connection = mysql_connect("localhost","wowbasec","PASSWORD REMOVED"); mysql_select_db("wowbasec_class", $connection); $query = "select * from `char`, location WHERE name='$search'"; $result = mysql_query($query, $connection); if (!$result) { echo 'Invalid query: ' . mysql_error(). "<br>"; echo 'SQL Query: ' . $query. "<br>"; die(); } while ($r = mysql_fetch_assoc($result)) { // Begin while $ts = $r["name"]; $ts1 = $r["id"]; ?> <table width="1000" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><a href="viewchar.php?id=<?php echo $ts1['id']; $ts1 = $_POST['id']; ?>"> <? echo $ts ?></td> </tr> </table> <? } ?> But i get this... Invalid query: Column 'name' in where clause is ambiguous SQL Query: select * from `char`, location WHERE name='harry'

<? $search = $_POST["search"]; $db = "`char`"; $connection = mysql_connect("localhost","wowbasec","PASSWORD REMOVED"); mysql_select_db("wowbasec_class", $connection); $query = "select * from `char`, location WHERE name='$search'"; $result = mysql_db_query("wowbasec_class", $query); while ($r = mysql_fetch_assoc($result)) { // Begin while $ts = $r["name"]; $ts1 = $r["id"]; ?> <table width="1000" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td><a href="viewchar.php?id=<?php echo $ts1['id']; $ts1 = $_POST['id']; ?>"> <? echo $ts ?></td> </tr> </table> <? } ?>

Hi, I have a search form, i need it to search upto 4 different tables in the same database for whatever the person is searching for. At current whenever i put the tables like `char`, location its comes up with.... Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/wowbasec/public_html/search.php on line 13 I need some help if anyone can do that Thanks

Thanks working now will change the db name too thanks

What i want is i want to display... #Locations #Characters etc. But whenever i put in $result2 = mysql_query("SELECT * FROM char"); $rows2 = mysql_num_rows($result2); Extra i get this error... Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/wowbasec/public_html/index.php on line 10

Hi, As youve probably notice im not that good with php however i know a bit. mysql_select_db("wowbasec_class", $connection); $result = mysql_query("SELECT * FROM location"); $result1 = mysql_query("SELECT * FROM char"); $rows = mysql_num_rows($result); I have this at the moment but if i want to put in another count for another table in the same database how do i do that without any errors?

Does anyone know what that error message means? Its know inserting the information but gives that error.

rckehoe when i changed to what you put i get the 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 '1' at line 1

Nope still same error rckehoe.

Hey everyone, I have a form that goes to a processing page, there is a drop down menu on the form that gets locations from a database. The code for the form page is first and then process page. <? // Connect database mysql_connect("localhost","USERNAME TAKEN OUT","PASSWORD TAKEN OUT"); mysql_select_db("wowbasec_class"); // If submitted, check the value of "select". If its not blank value, get the value and put it into $select. if(isset($select)&&$select!=""){ $select=$_GET['select']; } ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <title>Untitled Document</title> </head> <body> <table width="70" border="0" align="center" cellpadding="0" cellspacing="0" bgcolor="#00FFFF"> <tr> <td height="34"><div align="center">Character Add Page </div></td> </tr> <tr> <td height="208"><form action="addcharfile.php" method="post" name="form1"> <table width="500" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td width="191"><div align="right">Name:</div></td> <td width="309"><input name="name" type="text" id="name"></td> </tr> <tr> <td><div align="right">Location:</div></td> <td><select name="location" id="location"> <option value="">--- Select ---</option> <? // Get records from database (table "name_list"). $list=mysql_query("select * from location order by id asc"); // Show records by while loop. while($row_list=mysql_fetch_assoc($list)){ ?> <option value="<? echo $row_list['id']; ?>" <? if($row_list['id']==$select){ echo "selected"; } ?>><? echo $row_list['name']; ?></option> </select> </td> </tr> <tr> <td><div align="right">Health:</div></td> <td><input name="health" type="text" id="health"></td> </tr> <tr> <td><div align="right">Mana:</div></td> <td><input name="mana" type="text" id="mana"></td> </tr> <tr> <td><div align="right">Notes:</div></td> <td><textarea name="notes" id="notes"></textarea></td> </tr> <tr> <td colspan="2"><div align="center"> <input type="submit" name="Submit" value="Add"> </div></td> </tr> </table> </form> </td> </tr> <tr> <td> </td> </tr> </table> </body> </html> <? // End while loop. } ?> <? $connection = mysql_connect("localhost","USERNAME TAKEN OUT","PASSWORD TAKEN OUT"); mysql_select_db("wowbasec_class", $connection); // Get values from form. $name=$_POST['name']; $location=$_POST['location']; $health=$_POST['health']; $mana=$_POST['mana']; $notes=$_POST['notes']; // Insert all parameters into database. // The id field is auto increment. You don't have to insert any value $sql = mysql_query("insert into char(name, location, health, mana, notes) values('$name', '$location', '$health', '$mana', '$notes')"); if (!mysql_query($sql,$connection)) { die('Error: ' . mysql_error()); } echo "1 record added"; // Close database connection mysql_close(); ?> Any help is apreciated, Thanks.

Thanks everyone i have solved the problem thanks to the posts here. After looking carfully a spark lit in my head Thanks once more.

No still get the from thing saying [email protected] There has to be a way that i can make that change from that into WoW Base.