Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

[skidmark10]

I've never encountered this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

Does anyone knows where it stems from, and ways to fix it?

 

Here's my code just in case.

 

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID=" .$_POST['clientid']. ""or die ('Error: '.mysql_error ());

$query = mysql_query($sql);

echo "<table border='0'>
<tr>
<th>Client ID</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
</tr>";

while ($row = mysql_fetch_array($sql));
   {
   echo "<tr>";
   echo "<td>" . $row['Client_ID'] . "</td>";
   echo "<td>" . $row['Slim_Slender'] . "</td>";
   echo "<td>" . $row['Average'] . "</td>";
   echo "<td>" . $row['Athletic'] . "</td>";
   echo "<td>" . $row['Heavy'] . "</td>";
   echo "</tr>";
   }
echo "</table>";

mysql_close($con);
?> 

[cunoodle2]

I'm not sure but I do see 2 double quotes here..

 

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID=" .$_POST['clientid']. ""or die ('Error: '.mysql_error ());

[DavidAM]

This

while ($row = mysql_fetch_array($sql));

 

Should be this

while ($row = mysql_fetch_array($query));

[voip03]

yours

while ($row = mysql_fetch_array($sql));

 

remove ' ;'

while ($row = mysql_fetch_array($sql))
{

}

[voip03]

please mark as solved. the topic solved button can be found at the bottom left of the page

[skidmark10]

please mark as solved. the topic solved button can be found at the bottom left of the page

 

I am still receiving the error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource after making the corrections. Is there anything else I can do?

 

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID= " .$_POST['clientid']. "";

$query = mysql_query($sql);

echo "<table border='0'>
<tr>
<th>Client ID</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
<th>Partner's Body Type</th>
</tr>";

while ($row = mysql_fetch_array($query))
   {
   echo "<tr>";
   echo "<td>" . $row['Client_ID'] . "</td>";
   echo "<td>" . $row['Slim_Slender'] . "</td>";
   echo "<td>" . $row['Average'] . "</td>";
   echo "<td>" . $row['Athletic'] . "</td>";
   echo "<td>" . $row['Heavy'] . "</td>";
   echo "</tr>";
   }
echo "</table>";

mysql_close($con);
?> 

[voip03]

case sensitive issue

check your  table name and check the query

[Pikachu2000]

please mark as solved. the topic solved button can be found at the bottom left of the page

 

What if it isn't solved?

[Pikachu2000]

The query is failing for some reason, but you have no error-handling logic in place to catch that and report any errors. Is that all of the code, or is there more to it?

[skidmark10]

The query is failing for some reason, but you have no error-handling logic in place to catch that and report any errors. Is that all of the code, or is there more to it?

 

How can I put in the error-handling logic?

This is all of my php code (minus my database connection code), I'll post my html code so you can see the form.

 

HTML

<form name="input" action="fetchclientprofile.php" method="GET">
<center> Client ID <input type="text" name="clientid" /><br /> </center>
<center> <input type="submit" value="Retrieve" /> </center>

 

PHP

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID= " .$_POST['clientid']. "";

 

$query = mysql_query($sql);

 

echo "<table border='0'>

<tr>

<th>Client ID</th>

<th>Partner's Body Type</th>

<th>Partner's Body Type</th>

<th>Partner's Body Type</th>

<th>Partner's Body Type</th>

</tr>";

 

while ($row = mysql_fetch_array($query))

  {

  echo "<tr>";

  echo "<td>" . $row['Client_ID'] . "</td>";

  echo "<td>" . $row['Slim_Slender'] . "</td>";

  echo "<td>" . $row['Average'] . "</td>";

  echo "<td>" . $row['Athletic'] . "</td>";

  echo "<td>" . $row['Heavy'] . "</td>";

  echo "</tr>";

  }

echo "</table>";

 

mysql_close($con);

?>

[skidmark10]

Fixed it

 

Instead of

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID=" .$_POST['clientid']. ""or die ('Error: '.mysql_error ());

 

It should be

$sql = "SELECT * FROM `Partners_Body_Types` WHERE Client_ID='$_POST[clientid]'"