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Home/ Questions/Q 8969199
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T17:33:34+00:00 2026-06-15T17:33:34+00:00

I want to put a sql query result into an array. I tried the

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I want to put a sql query result into an array.
I tried the code below but it shows the 1st record for $count times.
Obviously it’s something wrong at the “$dept[$i]= $row[‘name’];”.
But i have no idea how to fix it.
Somebody help please?

$sql="SELECT name FROM system_dept ORDER BY id";
$result=mysql_query($sql);
$row = mysql_fetch_array($result);
$count=mysql_num_rows($result);
if (!mysql_query($sql,$con))
{
    die('Error: ' . mysql_error());
}
else
{
    $dept = array();
    $i=0;

    for($i=0;$i<$count;$i++)
    {
        $dept[$i]= $row['name'];
        echo $dept[$i];
    }
}

Ok, i tried to use mysqli but it doesnt work.
the web server shows that:
MySQL client version: 4.1.22
PHP extension: mysql
Can mysqli works in mysql php extension?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T17:33:35+00:00Added an answer on June 15, 2026 at 5:33 pm

    you are only fetching one row to fetch more rows you’ll need to call fetch result again
    so just add this to your code and things should be fine:

    for($i=0;$i<$count;$i++)
        {
            $dept[$i]= $row['name'];
            echo $dept[$i];
            $row=mysql_fetch_array($result);
        }
    

    I would recommend using a while loop instead like the following:

    $sql="SELECT name FROM system_dept ORDER BY id";
    $result=mysql_query($sql);
    $count=mysql_num_rows($result);
    if (!mysql_query($sql,$con))
    {
        die('Error: ' . mysql_error());
    }
    else
    {
        $dept = array();
        while($row=mysql_fetch_array($result))
        {
          $dept[]=$row['name'];
          echo $row['name'];
         }
    }
    

    If at all possible look into mysqli and PDO, as they are both more efficient

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