Error in display script

[ianco]

Hi all

 

I'm getting the error

mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/zxq.net/s/c/i/scienceproof/htdocs/new/incs/XMashEdit2.php on line 12

 

I want a form where the previous info shows up in the text area so that it can be edited and saved but I cant't see where i'm going wrong :'(

 

mysql_select_db("scienceproof_zxq_sciencetest", $con);
$options = $_POST["thing"];
$result="SELECT XMashTitle, XMashContent FROM XMash WHERE XMashTitle = $options";




while ($row=mysql_fetch_array($result)) {
    $content = $row['XMashContent'];
}





mysql_close($con);

?>
<p>Editing <b><?=$options?></b></p>
<form action="XMashEdit.php" method="post">
<textarea rows="10" cols="30" name="content"><?=$content?></textarea>
<input type="submit" />
</form>

[PFMaBiSmAd]

Your code does not have a mysql_query() statement in it to execute the query.

[ianco]

Even with it, it doesnt work

 

Updated script:

mysql_select_db("scienceproof_zxq_sciencetest", $con);
$options = $_POST["thing"];
$sql="SELECT XMashContent FROM XMash WHERE XMashTitle = $options";
$result=mysql_query($sql);



while ($row=mysql_fetch_array($result)) {
    $content = $row['XMashContent'];
}





mysql_close($con);

?>
<p>Editing <b><?=$options?></b></p>
<form action="XMashEdit.php" method="post">
<textarea rows="10" cols="100" name="content"><?=$content?></textarea>
<input type="submit" />
</form>

[PFMaBiSmAd]

If you are still getting the same error, it means that you query is failing due to an error. For debugging purposes, echo mysql_error(); on the next line after the line with the msyql_query() statement to find out why the query is failing.

[ianco]

That outputs

 

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 '!' at line 1

 

doesn't scream anything obvious to me

[PFMaBiSmAd]

I'm going to guess that XMashTitle is a character/string data type? If so, $options needs to be enclosed in single-quotes inside th query to make it a string.

 

You also need to use mysql_real_escape_string() on $options before you put it into the query to prevent sql injection and prevent any other special characters in it (such as a quote) from breaking the sql syntax.

[ianco]

Bingo!

 

Thanks for all the help