vmicchia Posted June 9, 2010 Share Posted June 9, 2010 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($row["cushionName"]) "</label>"; } ?> </form> </div> </div> </body> </html> I have run the query in mySQL and it returns the correct data so that doesn't seem to be the problem. Thanks for any help. Link to comment https://forums.phpfreaks.com/topic/204277-error-help/ Share on other sites More sharing options...
micah1701 Posted June 9, 2010 Share Posted June 9, 2010 on line 17 (if I counted backwards correctly)... if($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. Link to comment https://forums.phpfreaks.com/topic/204277-error-help/#findComment-1069913 Share on other sites More sharing options...
syed Posted June 9, 2010 Share Posted June 9, 2010 use this instead while ( $row = mysql_fetch_array($result) { print $row['field']; } Link to comment https://forums.phpfreaks.com/topic/204277-error-help/#findComment-1069915 Share on other sites More sharing options...
vmicchia Posted June 9, 2010 Author Share Posted June 9, 2010 on line 17 (if I counted backwards correctly)... if($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. Link to comment https://forums.phpfreaks.com/topic/204277-error-help/#findComment-1069917 Share on other sites More sharing options...
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