mysql_fetch_array(): - Dev Articles

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January 20th, 2004, 03:40 AM

sickofcoding

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Join Date: Jan 2004

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mysql_fetch_array():

hello everybody,
i've been stuck with this for a while and cannot find a way to solve it; i'll really appreciate your help. please take a look at it since i'm new to php. thank you!
my error is :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/NcDB/mailForm.php on line 37

<?php require_once('Connections/connectdb.php'); ?>


<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=windows-874">
</head>

<body>
<?
mysql_select_db($database_connectdb, $connectdb);
$sql0= "select ServiceGr from ServiceType where ServiceID=$serid";
$searchSql0 = mysql_query($sql0,$connectdb) or die(mysql_error());
$resultSearch0=mysql_fetch_array($searchSql0);

mysql_select_db($database_connectdb, $connectdb);
if('$resultSearch0[ServiceGr]'=="1")
{
$sql= "SELECT LinkID from Link where ServiceID=1";
$searchSql = mysql_query($sql,$connectdb) or die(mysql_error());
$resultSearch=mysql_fetch_array($searchSql);

mysql_select_db($database_connectdb, $connectdb);
$sql1= "SELECT ContactID from ContactMap where GroupID=$resultSearch[LinkID]";
$searchSql1 = mysql_query($sql1,$connectdb) or die(mysql_error());
$resultSearch1=mysql_fetch_array($searchSql1);

mysql_select_db($database_connectdb, $connectdb);
$sql2= "select ContactEmail from Contact where ContactID=$resultSearch1[ContactID]";
$searchSql2 = mysql_query($sql2,$connectdb) or die(mysql_error());
$resultSearch2=mysql_fetch_array($searchSql2);
//echo"$resultSearch2[ContactEmail]";
}
?>
<?php
// line 37 is here
while($resultSearch2=mysql_fetch_array($searchSql2 ))
{
?>
<form name="form1" method="post" action="">
<table width="75%" border="0">
<tr>
<td width="26%">Mail From:</td>
<td width="74%"> </td>

</tr>
<tr>
<td>Mail To:</td>
<td> </td>
<?php echo" $resultSearch2[ContactEmail]";?>
</tr>
<tr>
<td>Bcc:</td>
<td> </td>
</tr>
<tr>
<td>Subject:</td>
<td> </td>
</tr>
<tr>
<td colspan="2"><textarea name="textarea"></textarea></td>
</tr>
<tr>
<td> </td>
<td> <input name="Sent" type="submit" value="Sent">
<input type="submit" name="Clear" value="Clear"></td>
</tr>
</table>
</form>
<?php
}
?>
</body>
</html>

January 20th, 2004, 07:30 AM

dhouston

Contributing User

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Join Date: May 2003

Location: Tennessee

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Typically, this means that you're not connected to the database or that you haven't passed the db handle to the calling function. Have you assigned the mysql_connect() return to a variable? If so, try passing that variable in your mysql_fetch_array() as the second parameter.

January 20th, 2004, 09:18 AM

MadCowDzz

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Dev Articles Frequenter (2500 - 2999 posts)

Join Date: Jan 2003

Location: Toronto, Canada

Posts: 2,890

Time spent in forums: 1 Week 16 h 4 m 48 sec
Reputation Power: 8

Which line is line 37?

In the past I've found the problem usually relates to a messed up SQL statement...

try adding:
or die("Error ".mysql_errno().": ".mysql_error()."\nQuery: $sql");

to the end of your mysql_query commands... also, change the $sql variable to your variables containing the SQL statement

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