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supplied argument is not a valid MySQL-Link resource error

I'm getting an error on line 11, saying Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource. What does this mean?
<?php

$con = mysql_connect("localhost","","");
	if (!$con)
 	{
 	 die('Could not connect: ' . mysql_error());
 	 }

	mysql_select_db("", $con);

$query = mysql_query("SELECT description, cross_streets, title, town, feeornofee, lease, phone, contact, office, email, rent FROM apartments WHERE description LIKE '%$search%' ¦¦ cross_streets LIKE '%$search%' ¦¦ title LIKE '%$search%' ¦¦ town LIKE '%$search%' ¦¦ feeornofee LIKE '%$search%' ¦¦ lease LIKE '%$search%' ¦¦ phone LIKE '%$search%' ¦¦ contact LIKE '%$search%' ¦¦ contact LIKE '%$search%' ¦¦ office LIKE '%$search%' ¦¦ email LIKE '%$search%' ¦¦ rent LIKE '%$search%'", $db);

if ($query) 
{ 
echo "<table width=100% border=0 cellpadding=0 cellspacing=5 class=border_playlist_table><tr>";

while ($myrow = mysql_fetch_array($query)) { // Begin while 
$title = $myrow["title"]; 
$town = $myrow["town"];
$contact = $myrow["contact"]; 
$office = $myrow["office"]; 
$phone = $myrow["phone"]; 
$rent = $myrow["rent"]; 
echo "<tr class=sub_1><td colspan=4>$title : $town</td></tr> 
<tr bgcolor=#55556F><td><p><a href=$url target=_blank>$url</a></p></td> 
<td><p>$title</p></td> 
<td><p>$town</p></td> 
<td><p>$contact</p></td></tr> 
<tr bgcolor=#55556F><td colspan=4>$office</td></tr> 
<tr bgcolor=#55556F><td colspan=4>$phone</td></tr> 
<tr><td colspan=4>&nbsp;</td></tr>"; 
} // end while 
echo "</tr></table>"; 
} 
else{ 
echo "Search string is empty. <br> Go back and type a string to search"; 
} 
?>

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0
genesisvh
Asked:
genesisvh
1 Solution
 
phutsaCommented:
Hey. Did you ever try to insert you database name on "mysql_select_db("  your Database name  ", $con);" . if it's not work I think some value of your select seem don't have any value. try to check value again in mysql command.

0
 
genesisvhAuthor Commented:
I have put in the name I just excluded it here.
0
 
Dave BaldwinFixer of ProblemsCommented:
You have '$db' on line 11 but you have '$con' on line 3.  They should match.  '$con' is the 'Resource ID' from the connection.
0
 
phutsaCommented:
DaveBaldwin: Oh. It's true. Why I can't notice it. lol Oh It's a small mistake but make more trouble. LOL
0
 
Ray PaseurCommented:
You don't need the $db variable at all in the query.
0

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