mysql_fetch_row() function does not return entire row

Question

I am seeing the echo of an entire row of a query:

echo "<ul>";
while($row = mysql_fetch_row($result))
{
   echo "<li>{$row[0]}</li>";
}
echo "</ul>";

When the query is processed there is only one column (client_id) displayed from the database instead of the entire row. However, when I use

print_r(mysql_fetch_assoc($result));

just before while loop I can see displayed:

Array (
    [client_id] => 1
    [contact_first_name] => carla
    [contact_last_name] => clausen
    [business_name] => Banana Belt Liquors
    [business_address] => 300 U.S. 24 Woodlan
    [business_address2] =>
    [business_city] => Woodland Park
    [business_state] => Colorado
    [business_zip] => 80863
    [business_areaCode_phone] => 719
    [business_phone] => 687
    [business_phone2] => 9757
    [business_url] => http://bananabeltliquors.com/
    [business_email] => [email protected]
)

Your line echo "<li>{$row[0]}</li>" is only echoing the first element of each row. That's why you don't see any more. — user1864610
– user1864610, Commented Jul 8, 2013 at 4:00
Look at your vardump... do you see any numeric keys in that array? Neither do I... Try echo "<li>{$row['contact_first_name']}</li>"; instead. And don't mix fetch_row and fetch_assoc. pick one style and go with it. Or use fetch_array, and get both. — Marc B
– Marc B, Commented Jul 8, 2013 at 4:01
This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. bg2.php.net/manual/en/function.mysql-fetch-row.php — d.danailov
– d.danailov, Commented Jul 8, 2013 at 4:12

Goutam Pal · Accepted Answer · 2013-07-08 04:04:07Z

1

mysql_fetch_row is fetching the entire row but as per your code you are only fetchin the first column from the entire rows.

echo "<ul>";
while($row = mysql_fetch_row($result))
{
   print_r($row);////// this will print the entire row
   echo "<li>{$row[0]}</li>";


    echo "<li>{$row[1]}</li>";///for other columns
    echo "<li>{$row[2]}</li>";
    ---------------------
    ----------------------
    echo "<li>{$row[8]}</li>";
}
echo "</ul>";

answered Jul 8, 2013 at 4:04

Goutam Pal

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Orangepill · Accepted Answer · 2013-07-08 04:05:32Z

0

you are only echoing out the first field for each of the records. If you wanted to see them all you would change your code to:

echo "<ul>";
while($row = mysql_fetch_row($result))
{
   echo "<li>".implode("<br/>", $row)."</li>";
}
echo "</ul>";

This will show all of fields separated by a break for each row in a list item.

If you wanted each field to be in it's own list element you could do

echo "<ul>";
while($row = mysql_fetch_row($result))
{
   echo "<li>".implode("</li><li>", $row)."</li>";
}
echo "</ul>";

answered Jul 8, 2013 at 4:05

Orangepill

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Comments

Nishu Tayal · Accepted Answer · 2013-07-08 04:14:32Z

0

$row will return entire row but $row[0] will have only first column. So You can count the number of columns and then iterate $row.

 echo "<ul>";
    while($row = mysql_fetch_row($result))
    {
       $len = count($row);
       for($i =0;$i<$len;$i++){
             echo "<li>{$row[$i]}</li>";
       }
    }
    echo "</ul>";

answered Jul 8, 2013 at 4:14

Nishu Tayal

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