Re: [PHP] Check to see if mysql_fetch_array result is empty

[Mike Tuller]

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:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 12:43 PM
To: [EMAIL PROTECTED]
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:[EMAIL PROTECTED]
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";
                }
        }
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