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Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
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Join Date: May 2006
Posts: 3
Reputation:
Solved Threads: 0
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
0
#1 May 31st, 2006
[code]
[php]
<?php
$SearchType = $_POST["SearchPlace"];
$host= 'localhost';
$user= 'user';
$passwd= 'password';
$database= 'dbase';
$table1= 'BUS';
$table2= 'STUDENT';
$connect= mysql_connect($host, $user, $passwd);
mysql_select_db($database);
$pquery = "select busNo from $table1 where place like '".$SearchType."'";
$squery = "select * from $table2 where busNo like '".$pquery."'";
$run = mysql_query($squery);
$num_results = mysql_num_rows($run);
if (!$SearchType )
{
echo "<b>You have not selected search details. Please go back and try again.</b>";
}
else
{
if (!$connect)
{
echo"</b>Error: Could not connect to database. Please try again later.</b>";
}
else
{
for ($i=0; $i<$num_results; $i++)
{
$row = mysql_fetch_array($run);
$IDNo= stripslashes($row["IDNo"]);
$sName= stripslashes($row["sName"]);
$startDate= stripslashes($row["startDate"]);
$endDate = stripslashes($row["endDate"]);
$sPhone= stripslashes($row["sPhone"]);
$OBox= stripslashes($row["OBox"]);
$busNo= stripslashes($row["busNo"]);
echo "<tr>\n";
echo "<td align=left>\n";
echo "<p> Student ID: ".$IDNo."";
echo "<br>Student Name: ".$sName."";
echo "<br>Starting Date: ".$startDate."";
echo "<br>Ending Date: ".$endDate."";
echo "<br>Student Phone: ".$sPhone."";
echo "<br>P.O.Box: ".$OBox."";
echo "<br>Bus number: ".$busNo."";
echo "</p>";
echo "</td></tr>\n";
echo "\n";
echo "\n";
echo "</table>\n";
}
}
}
?>
[inlinecode]
[/php]
Hi I am getting this warning and I don't know why?!!
:!:
please can you help me to find the error
[php]
<?php
$SearchType = $_POST["SearchPlace"];
$host= 'localhost';
$user= 'user';
$passwd= 'password';
$database= 'dbase';
$table1= 'BUS';
$table2= 'STUDENT';
$connect= mysql_connect($host, $user, $passwd);
mysql_select_db($database);
$pquery = "select busNo from $table1 where place like '".$SearchType."'";
$squery = "select * from $table2 where busNo like '".$pquery."'";
$run = mysql_query($squery);
$num_results = mysql_num_rows($run);
if (!$SearchType )
{
echo "<b>You have not selected search details. Please go back and try again.</b>";
}
else
{
if (!$connect)
{
echo"</b>Error: Could not connect to database. Please try again later.</b>";
}
else
{
for ($i=0; $i<$num_results; $i++)
{
$row = mysql_fetch_array($run);
$IDNo= stripslashes($row["IDNo"]);
$sName= stripslashes($row["sName"]);
$startDate= stripslashes($row["startDate"]);
$endDate = stripslashes($row["endDate"]);
$sPhone= stripslashes($row["sPhone"]);
$OBox= stripslashes($row["OBox"]);
$busNo= stripslashes($row["busNo"]);
echo "<tr>\n";
echo "<td align=left>\n";
echo "<p> Student ID: ".$IDNo."";
echo "<br>Student Name: ".$sName."";
echo "<br>Starting Date: ".$startDate."";
echo "<br>Ending Date: ".$endDate."";
echo "<br>Student Phone: ".$sPhone."";
echo "<br>P.O.Box: ".$OBox."";
echo "<br>Bus number: ".$busNo."";
echo "</p>";
echo "</td></tr>\n";
echo "\n";
echo "\n";
echo "</table>\n";
}
}
}
?>
[inlinecode]
[/php]
Hi I am getting this warning and I don't know why?!!
:!: please can you help me to find the error
•
•
Join Date: Feb 2006
Posts: 32
Reputation:
Solved Threads: 1
Re: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
0
#2 May 31st, 2006
Your query is probably failing. Try
[php]$run = mysql_query($squery) or die(mysql_error());[/php]
And see if you get an error.
Or echo out your SQL statment and trying running it against the database.
[php]$run = mysql_query($squery) or die(mysql_error());[/php]
And see if you get an error.
Or echo out your SQL statment and trying running it against the database.
•
•
Join Date: May 2006
Posts: 3
Reputation:
Solved Threads: 0
Re: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
0
#3 May 31st, 2006
Re: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource
0
#4 May 31st, 2006
[PHP]
$pquery = "select busNo from $table1 where place like '".$SearchType."'";
$squery = "select * from $table2 where busNo like '".$pquery."'";
[/PHP]
If $_POST['searchplace'] was given as "Foobar" (for want of a better example), $squery would be set to:
Try running that by itself :-|
$pquery = "select busNo from $table1 where place like '".$SearchType."'";
$squery = "select * from $table2 where busNo like '".$pquery."'";
[/PHP]
If $_POST['searchplace'] was given as "Foobar" (for want of a better example), $squery would be set to:
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select * from STUDENT where busNo like 'select busNo from BUS where place like 'Foobar''
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