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Bug #5529 Supplied argument is not a valid MySQL result resource
Submitted: 2000-07-12 08:38 UTC Modified: 2000-07-12 08:59 UTC
From: jagas at computerconsultant dot net Assigned:
Status: Closed Package: MySQL related
PHP Version: 4.0 Release Candidate 2 OS: Windows NT4
Private report: No CVE-ID: None
View Developer Edit
 [2000-07-12 08:38 UTC] jagas at computerconsultant dot net 
        if ($dbh =  mysql_connect('localhost','root',''))
        {
                $strSQL = "select authorsname,bodytext from stories where id = $storyid";

					mysql_select_db('mydb');
					$stmt = mysql_query($strSQL, $dbh);
        
					$row = mysql_fetch_array($stmt);
    }

Compiled with MySQL

I get an error when I call "$row = mysql_fetch_array($stmt)" . The error is:
Supplied argument is not a valid MySQL result resource in .... on line 9.

This function worked perfectly in PHP 3.

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 [2000-07-12 08:48 UTC] jagas at computerconsultant dot net 
Sorry. My sql statement was wrong. The variable $storyid had no value hence ...
 [2000-07-12 08:59 UTC] stas at cvs dot php dot net 
buggy bug report, close it
 [2003-01-16 13:45 UTC] joachim dot desmedt at meestersites dot be 
I have the same problem.

$db = mysql_connect("xxx", "xxx", "xxx") or die ("Sorry.
mysql_select_db("name", $db);
$SQL_statement = "SELECT * FROM iks WHERE voornaam= $naam";
$resultset = mysql_query($SQL_statement);

while($data = mysql_fetch_array($resultset)){
echo $data['voornaam']

}

But I'm sure that $naam has a value.
Help me please!
 [2003-01-16 13:45 UTC] joachim dot desmedt at meestersites dot be 
I have the same problem.

$db = mysql_connect("xxx", "xxx", "xxx") or die ("Sorry.
mysql_select_db("name", $db);
$SQL_statement = "SELECT * FROM iks WHERE voornaam= $naam";
$resultset = mysql_query($SQL_statement);

while($data = mysql_fetch_array($resultset)){
echo $data['voornaam']

}

But I'm sure that $naam has a value.
Help me please!
 
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