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  1. #1
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    Having a problem with mysql_fetch_array()...

    So, here it is:
    On my computer I have Apache2 with PHP5..
    I'm writing all my codes on this system.

    So I built a simple code with mysql_fetch_array().. On my Comp. it works great,
    but when i uploaded it to some hosting company (who uses php5 too) it dosent work.

    Here is my code:
    PHP Code:
    <?php
        $link 
    mysql_connect ($dbhost$dbusername$dbpass)
            or die (
    "Can not connect to MySQL");
        if (
    mysql_select_db ($dbname)) {
            
    $result mysql_query ("Select * from GuestBook order by gb_index desc");
            if (
    mysql_affected_rows() == 0) {
                echo 
    "<font face='Arial' size='2'>אין כרגע רעיונות.</font>\n";
            } else {
                while (
    $row mysql_fetch_array ($result)) {

                    echo 
    "<html dir='rtl'>\n";
                    echo 
    "<p>\n";
                    echo 
    "<table dir='rtl' align='center' cellspacing='5' bgcolor='#e9e9e9' border=0 width='70%'style='font: 8pt verdana, geneva, lucida, arial, helvetica, sans-serif;'>\n";
                    echo 
    "<tr><td align=right>\n";
                    echo 
    "<b>מאת:</b>\n";
                    echo 
    "<br>".$row["name"]." <br>\n";
                    echo 
    "<b> פורסם ב: </b>\n";
                    echo 
    "<br>".$row["date"]." <br><br>\n";
                    echo 
    "<b> הרעיון: </b>\n";
                    echo 
    "<br>".$row["message"]."\n";
                    echo 
    "</tr></td></table>\n";
                    echo 
    "</html>\n";

                }
            }
        } else {
            echo 
    "<font face='Arial' size='2'>An error occured will trying to select the database.</font>\n";
        }
        
    mysql_close ($link);
    ?>
    Returns the next error:
    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/yazamim/webapps/htdocs/ideas/admin1.php on line 43

    Line 43:
    PHP Code:
    while ($row mysql_fetch_array ($result)) { 
    Can anyone help me?

  2. #2
    Regular Coder
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    change this line
    PHP Code:
    $result mysql_query ("Select * from GuestBook order by gb_index desc"); 
    to

    PHP Code:
    $result mysql_query ("Select * from GuestBook order by gb_index desc") or die (mysqlL_error()); 
    That error tends to mean there is no data in $result so you could check that my putting this line

    PHP Code:
    print mysql_num_rows($result); 
    just before the while loop

    that won't fix the problem but might help find out what the problem is

  3. #3
    New to the CF scene
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    Thank You!
    I have managed to find the problem..
    The table name at the database wasn't correct (case-sensitive)

    Thanks Again

  4. #4
    Senior Coder CFMaBiSmAd's Avatar
    Join Date
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    Location
    Denver, Colorado USA
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    That error tends to mean there is no data in $result
    The error means the query failed (syntax error, typo's or incorrect table/column names, no database selected, no connection to database server...) and returned a FALSE value. Executing a mysql function using the FALSE value instead of a resource result is what causes the error. A query that succeeds returns a valid resource result even if the number of rows in the result is zero.

    Executing a mysql_fetch_array($result) on a resource result with zero rows returns a FALSE value. Executing a mysql_num_rows($result) on a resource result with zero rows returns a ZERO value.

    Executing a mysql_fetch_array($result) or a mysql_num_rows($result)following a failed query, results in the error.
    Finding out HOW to do something is called research, i.e. keep searching until you find the answer. After you attempt to do something and cannot solve a problem with it yourself, would be when you ask others for help.


 

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