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Home/ Questions/Q 8034039
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T01:53:20+00:00 2026-06-05T01:53:20+00:00

I have the following code: <?php include ‘db.php’; $con = mysql_connect($host,$user,$pass); $dbs = mysql_select_db($databaseName,

  • 0

I have the following code:

<?php 
include 'db.php';
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);

$tableName = "tbl_title";
$result = mysql_query("SELECT id, fld_title FROM $tableName");

$data = array();
while ( $row = mysql_fetch_row($result) )
{
    //$data[] = $row;
    //$data['item'][] = $row;
    //$data['item'][] = $row['id'].value;
    $id = $row['id'];
    $fld_title = $row['fld_title'];
    $data = array( "id" => $id, "fld_title" => $fld_title);

    //echo $row['id'] . " " . $row['fld_title'];
            //echo '<option value="'.$row->id.'">'.$row->fld_title.'</option>'; 
}
echo json_encode( $data );  
    //  echo $row[0] . " " . $row[1];
?>

I am trying to return the results in a formatted way, however, even though I get results from the query, $id and $fld_title don’t get set to a value. What am I missing?

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  1. Editorial Team
    Editorial Team
    2026-06-05T01:53:22+00:00Added an answer on June 5, 2026 at 1:53 am

    There’s no need to reference each column. $row is an array already. I’ve changed to using mysql_fetch_assoc because that one gives named columns.

    $data = array();
    while ( $row = mysql_fetch_assoc($result) )
    {
        $data[] = $row;
    }
    echo json_encode( $data );
    
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