problem in mysql_fetch_array()

Question

manzoor1 0 Newbie Poster

15 Years Ago

hi
i have problem , rhe code is working properly on local machine but when i upload it on net then it give the warning
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home7/wyzerpmc/public_html/wyzer/survey/survey.php on line 22

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<center>
<body leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
  <table width="800" border="0" cellspacing="0" cellpadding="0">
    <tr> 
      <td height="50" colspan="3" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
    <tr> 
      <td width="300" bgcolor="#99CCFF">&nbsp;</td>
      <td width="500"> 
        <?php 

ob_start();
include_once("connection.php");
	$sql="SELECT *
	FROM `question`";
	$quer=mysql_query($sql);
		while($row=mysql_fetch_array($quer))		{
		echo $row[uniqueid].") ".$row[stem]."<br>";
		$qid= $row[uniqueid];
				$sql1="select * from scale where uniqueid=$qid ;";
				$quer1=mysql_query($sql1);
				while($row1=mysql_fetch_array($quer1))
				{
				
				?>
        <input type="<?php echo $row[type];  ?>" name="<?php echo $row[questionID];  ?>" value="<?php echo $row1[label];  ?>"> 
        <?php echo $row1[label]; ?> <br> 
        <?php
				
				}
			echo "<hr>";		
		}


ob_end_flush();
?>
      </td>
      <td width="200" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
    <tr> 
      <td colspan="3" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
  </table>
</body>
</center>
</html>

php

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5 Replies
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Latest Post 15 Years Ago Latest Post by msikwal

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cwarn23 387 Occupation: Genius

15 Years Ago

while($row=mysql_fetch_array($quer)) {
echo $row[uniqueid].") ".$row[stem]."<br>";
$qid= $row[uniqueid];
$sql1="select * from scale where uniqueid=$qid ;";
$quer1=mysql_query($sql1);
while($row1=mysql_fetch_array($quer1))
{

?>
<input type="<?php echo $row[type]; ?>" name="<?php echo $row[questionID]; ?>" value="<?php echo $row1[label]; ?>"> 
<?php echo $row1[label]; ?> <br> 
<?php

}
echo "<hr>"; 
}

I am unsure if this is solved but to add to the info, if no rows are found the same error will be reported. To solve this, replace the code in the quote with the following:

if (mysql_num_rows($quer)>0) {
while($row=mysql_fetch_array($quer)) {
echo $row[uniqueid].") ".$row[stem]."<br>";
$qid= $row[uniqueid];
$sql1="select * from scale where uniqueid=$qid ;";
$quer1=mysql_query($sql1);
while($row1=mysql_fetch_array($quer1))
{

?>
<input type="<?php echo $row[type]; ?>" name="<?php echo $row[questionID]; ?>" value="<?php echo $row1[label]; ?>"> 
<?php echo $row1[label]; ?> <br> 
<?php

}
echo "<hr>"; 
}
}

nav33n commented: Absolutely.. +10

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shadiadiph 0 Junior Poster · Answer 1 · 2009-02-23T14:25:03+00:00

seems to be bit all over the place try this

$sql="SELECT * FROM `question`";
$quer=mysql_query($sql);
while($row=mysql_fetch_array($quer)) {

echo ".$row[uniqueid]." ".$row[stem]."<br>";
$qid= $row[uniqueid];
}

manzoor1 0 Newbie Poster · Answer 2 · 2009-02-24T19:00:34+00:00

manzoor1 0 Newbie Poster

15 Years Ago

thats good

shadiadiph 0 Junior Poster · Answer 3 · 2009-02-26T04:53:26+00:00

yes not sure if it was solved but he said that's good so i guess it solved the problem?? if this is solved pleae mark as solved thanks

msikwal 0 Newbie Poster · Answer 4 · 2009-02-28T17:27:13+00:00

hi
i have problem , rhe code is working properly on local machine but when i upload it on net then it give the warning
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home7/wyzerpmc/public_html/wyzer/survey/survey.php on line 22

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<center>
<body leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
  <table width="800" border="0" cellspacing="0" cellpadding="0">
    <tr> 
      <td height="50" colspan="3" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
    <tr> 
      <td width="300" bgcolor="#99CCFF">&nbsp;</td>
      <td width="500"> 
        <?php 

ob_start();
include_once("connection.php");
	$sql="SELECT *
	FROM `question`";
	$quer=mysql_query($sql);
		while($row=mysql_fetch_array($quer))		{
		echo $row[uniqueid].") ".$row[stem]."<br>";
		$qid= $row[uniqueid];
				$sql1="select * from scale where uniqueid=$qid ;";
				$quer1=mysql_query($sql1);
				while($row1=mysql_fetch_array($quer1))
				{
				
				?>
        <input type="<?php echo $row[type];  ?>" name="<?php echo $row[questionID];  ?>" value="<?php echo $row1[label];  ?>"> 
        <?php echo $row1[label]; ?> <br> 
        <?php
				
				}
			echo "<hr>";		
		}


ob_end_flush();
?>
      </td>
      <td width="200" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
    <tr> 
      <td colspan="3" bgcolor="#99CCFF">&nbsp;</td>
    </tr>
  </table>
</body>
</center>
</html>

use
while ($row = mysql_fetch_array($result, MYSQL_ASSOC))

second arg as MYSQL_ASSOC always ..........

problem in mysql_fetch_array()

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