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Error Help


vmicchia

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Hi I am working on a pretty simple script at the moment. I keep getting this error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/stay/public_html/index.php on line 26.

 

Here is my code

!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<div id="outerContainer" style="width:1024px; border:thin; border-color:#000000">
<div id="imageContainer" style="width:500px;">
</div>
<div id="formContainer">
<form name="selectCushion" id="selectCushion" enctype="multipart/form-data" action="fabricSelect.php">
<?php include("includes/openDbConn.php");
$sql="SELECT cushionName, cushionDescription, cushionPrice, cushionImagePath FROM cushions";
//echo $sql;
$result = mysql_query($sql);
//echo $result;
//echo "This works";
if($result = 0){
	$num_results = 0;
	//echo "Past If";
}else{
	//echo "Past else";
	//echo $num_results;
	$num_results = mysql_num_rows($result);
}
for($i = 0; $i < $num_results; $i++){
	$row = mysql_fetch_array($result);
	echo "<input type='image' src='".trim($row["cushionImagePath"])."' height='' width='' alt='".trim($row["cushionDescription"])."'/>";
	//echo "<label>" trim($row["cushionName"]) "</label>";
} ?>

</form>
</div>
</div>
</body>
</html>

 

I have run the query in mySQL and it returns the correct data so that doesn't seem to be the problem.

Thanks for any help.

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on line 17 (if I counted backwards correctly)...

 

if($result = 0){

 

needs to be:

 

if($result == 0){ //with 2 equal signs

 

Instead of checking if $result is equal to zero, you're re-assigning the value of $result to 0.

 

Thanks so much for catching that. I copied some of the code from an old project so I guess I just expected it to be working and right. Much appreciated.

 

use this instead

 

while ( $row = mysql_fetch_array($result) {

  print $row['field'];

}

 

I will try that I appreciate the advice very much.

 

 

It works great now. Much appreciated.

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https://forums.phpfreaks.com/topic/204277-error-help/#findComment-1069917
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