MYSQLI_NUM error

Discussion in 'PHP' started by goppss, Feb 20, 2011.

  1. #1
    Hi All,

    I am experiencing the following error while running the code mentioned. Can anyone please help?

    Warning: mysqli_fetch_array() [function.mysqli-fetch-array]: The result type should be either MYSQLI_NUM, MYSQLI_ASSOC or MYSQLI_BOTH in /home7/example/public_html/test/archives/test9.php on line 83
    Code (markup):
    <?php
    require_once "../../mysql/connect/connect_to_mysql.php";
            
    $sql =  'SELECT linkurl, h1tag, writ, lastmodified, h2tag FROM newarchive ORDER BY id DESC';
            $result = mysqli_query($myConnection, $sql) or die (mysqli_error($myConnection));
            $row = mysqli_fetch_array($result) or die(mysql_error());
            mysqli_close($myConnection);
    ?>
    
    
    <html>
      <head>
        <title></title>
      </head>
      <body>
    	<?php
    	$counter = 2;					
    	while ($row = mysqli_fetch_array( $result, $counter <= 1048576 )) {    // line 83
    		echo "<div class='newtruc'><div class='bytitle'>";
    		echo '<a href="' . $row['linkurl'] . '">' . $row['h1tag'] . '</a><br />';
    		echo "</div>";
    		echo "<div class='byline'><span class='byline1'>";
    		echo "written by: ";
    		echo $row ['writ'];
    		echo "</span>";
    		echo " | ";
    		echo "<span class='byline2'>";
    		echo $row['lastmodified'];	
    		echo "</span></div>";
    		echo "<div class='byphrase'>";
    		echo $row ['h2tag'];
    		echo "</div></div>";
    		
    		$counter = $counter * 2;
    		}		 
    		?>
      </body>
    </html>
    Code (markup):
     
    goppss, Feb 20, 2011 IP
  2. tvoodoo

    tvoodoo Active Member

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    #2
    The 2nd parameter for mysqli_fetch_array can be one of the following : MYSQLI_ASSOC, MYSQLI_NUM, or MYSQLI_BOTH. So replace $counter <= ... with the appropriate of the 3 mentioned. Or better yet tell us what you are trying to do and maybe we can suggest a better solution for you.
     
    tvoodoo, Feb 21, 2011 IP