Re: Check to see if mysql_fetch_array result is empty

From: Date: Wed, 12 Mar 2003 20:56:41 +0000
Subject: Re: Check to see if mysql_fetch_array result is empty
References: 1  Groups: php.general 
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That was it. Thanks. On Thursday, March 13, 2003, at 07:13 AM, Michael Roger C. Bianan wrote:
Hi, It's a logical error in your IF stmt. ----------- if ($num = 0) { echo "There are no matches for your search. Please try again."; } else { ----------- that should have been if ($num == 0) . Notice the two = (==). Thanks, Miches:) -----Original Message----- From: Mike Tuller [mailto:php@ce.anoka.k12.mn.us] Sent: Wednesday, March 12, 2003 12:43 PM To: miches@learningtogo.com Cc: php mailing list list Subject: Re: [PHP] Check to see if mysql_fetch_array result is empty Ok. I have it changed, but I have something wrong here, because when I there result is 0, it doesn't print out the message that I want. If I have it print out the result, it says 0. It has to be something simple, but I can't see what is wrong. if ($num = 0) { echo "There are no matches for your search. Please try again."; } else { while ($row = mysql_fetch_array( $db_query )) { echo "<tr> <td align=\"left\"><a href=\"editsoftwareasset.php?id=$row[asset_id]\">$row[asset_id]</a></td> </td> <td>$row[developer] </td> <td align=\"center\">$row[title] </td> <td align=\"center\">$row[version] </td> <td align=\"center\">$row[platform] </td> </tr>\n"; } } On Thursday, March 13, 2003, at 05:45 AM, Michael Roger C. Bianan wrote:
Mike, Use mysql_num_rows($db_query) ; - returns no of rows in the result set. - if none, returns 0. Thanks, Miches -----Original Message----- From: Mike Tuller [mailto:php@ce.anoka.k12.mn.us] Sent: Wednesday, March 12, 2003 11:17 AM To: php mailing list list Subject: [PHP] Check to see if mysql_fetch_array result is empty How can I check to see if a mysql_fetch_array is empty. I have a search page, and want to have it so that when there are no matches, it returns a message saying that there were no matches, and if there are, then display them. Here is what I have so far to give you an idea as to what I am trying to do. Thanks ----------------- $query = "SELECT * FROM software_assets WHERE $searchType LIKE '$search' "; $db_query = mysql_query($query, $db_connect) or die (mysql_error()); while ($row = mysql_fetch_array( $db_query )) { if $row = NULL { echo "There were no results that match your query. Please try again"; } else { echo "<tr> <td align=\"left\"><a href=\"editsoftwareasset.php?id=$row[asset_id]\">$row[asset_id]</a></ td> </td> <td>$row[developer] </td> <td align=\"center\">$row[title] </td> <td align=\"center\">$row[version] </td> <td align=\"center\">$row[platform] </td> </tr>\n"; } } --PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php --PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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