AEDELGOD Posted March 25, 2010 Share Posted March 25, 2010 Fatal error: Function name must be a string in [XXX] on line 36 <? include("include/session.php"); ?> <html> <title>Dykes Publishing Group :: Data Edit</title> <head><meta http-equiv="content-type" content="text/html; charset=UTF-8"><link href="style.css" rel="stylesheet" type="text/css"></head> <body> <center> <? if(isset($_SESSION['useredit'])){ unset($_SESSION['useredit']); echo "<h1>User Account Edit Success!</h1>"; echo "<p><b>$session->username</b>, your account has been successfully updated. " ."<a href=\"index.php\">Main Page</a>.</p>"; } else{ ?> <? if($session->logged_in){ ?> <h1>User Account Edit : <? echo $session->username; ?></h1> <? if($form->num_errors > 0){ echo "<td><font size=\"2\" color=\"#ff0000\">".$form->num_errors." error(s) found</font></td>"; } ?> <?php $sql = "SELECT * FROM SR_ WHERE username = '".mysql_real_escape_string($POST['username'])."'"; $result = $msql_query($sql, $link); $row = mysql_fetch_array($result); ?> <tr><td> <?php $row['field_1']; ?> </td><td> <?php $row['field_2']; ?></td></tr> <? } } ?> </center> </body> </html> Quote Link to comment Share on other sites More sharing options...
irkevin Posted March 25, 2010 Share Posted March 25, 2010 $result = $msql_query($sql, $link); Should be $result = $mysql_query($sql, $link); What is on line 36? Quote Link to comment Share on other sites More sharing options...
AEDELGOD Posted March 25, 2010 Author Share Posted March 25, 2010 Thanks for noticing that but that did not resolve my problem, I pasted the entire file, and it gives me the same problem, that line you pointed out was line 36 yet it still throws at me Fatal error: Function name must be a string in /xxx.php on line 36. :\ Quote Link to comment Share on other sites More sharing options...
irkevin Posted March 25, 2010 Share Posted March 25, 2010 oops . instead of $result = $mysql_query($sql, $link); Just put the one below $result = mysql_query($sql, $link); There shouldn't be a dollar sign in front of mysql_query(), because it is a function Quote Link to comment Share on other sites More sharing options...
AEDELGOD Posted March 25, 2010 Author Share Posted March 25, 2010 That did something, Now I get the errors: Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /***.php on line 36 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /***.php on line 38 :\ Quote Link to comment Share on other sites More sharing options...
irkevin Posted March 25, 2010 Share Posted March 25, 2010 Just append or die(mysql_error()) with it to get a clear explanation of what's happening.. try code below $result = mysql_query($sql, $link) or die('Error : '.mysql_error()); Quote Link to comment Share on other sites More sharing options...
AEDELGOD Posted March 25, 2010 Author Share Posted March 25, 2010 Now it shows: Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /***.php on line 36 Error : and it doesnt display any of my data i think the line: $sql = "SELECT * FROM SR_ WHERE username = '".mysql_real_escape_string($POST['username'])."'"; may be wrong is there another way i could approach that? I want to be able to display and edit the mysql information based on the username of the session. Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted March 25, 2010 Share Posted March 25, 2010 Read the error - mysql_query(): supplied argument is not a valid MySQL-Link resource There is no mysql link resource (connection) in the $link variable at the time the query was executed. Either your code is not attempting to make a connection to the database server at all or it did not set the $link variable or the code that is attempting to make the connection failed. What is in the include("include/session.php"); file? Quote Link to comment Share on other sites More sharing options...
irkevin Posted March 25, 2010 Share Posted March 25, 2010 Well maybe he didnt make a connection to the database You might need to add those code $link = mysql_connect('localhost','username','password'); mysql_select_db(yourdbname',$link); Quote Link to comment Share on other sites More sharing options...
AEDELGOD Posted March 25, 2010 Author Share Posted March 25, 2010 Thanks for your help, I thought I had that declared in session.php but I guess i forgot. and wouldn't you mean: $link = mysql_connect('localhost','username','password'); mysql_select_db('yourdbname',$link); You left out a ' The only problem I have now is my data wont display. Quote Link to comment Share on other sites More sharing options...
irkevin Posted March 25, 2010 Share Posted March 25, 2010 Oops sorry, was in a hurry when typing lol.. it's 23:35 here.. guess i'm half asleep.. HAHA Quote Link to comment Share on other sites More sharing options...
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