Why doesn't my mysql query in php return more than one row?

Question

I am trying to return a list of users' first and last names in JSON format in PHP.

My PHP looks something like:

    $_Query = '
                    SELECT
                    Fname,Lname
                    FROM
                    users
                    WHERE
                    number = "'.$_REQUEST['number'].'"
                                                    ;';

    $SQLResult = mysql_query($_Query) or die(mysql_error());

    $_UserData = array();

    if(mysql_num_rows($SQLResult) <> 0){

            while($SQLRow = mysql_fetch_array($SQLResult)){
                    $_UserData['Fname'] = $SQLRow['Fname'];
                    $_UserData['Lname'] = $SQLRow['Lname'];
            }

    }
echo json_encode($_UserData);

I expect more than one row in the format

{"Fname":["First_name1","First_name2"],"Lname":["Last_name1","Last_name2"]}

of course. However the script returns the last row with the correct conditions

{"Fname":"Steve","Lname":"LastName"}

The MySQl server returns what it should with the same query.

+--------------+-------------+
| Fname        | Lname       |
+--------------+-------------+
| First_name1  | Last_name1  |
| First_name2  | Last_name2  |
+--------------+-------------+

Why is this happening and how should I go about fixing it? Thanks!

Answer 1

Try this:

while($SQLRow = mysql_fetch_array($SQLResult)){
    $_UserData[] = $SQLRow;
}

In your example you are essentially overwriting the $_UserData array for every new row, hence the missing rows.

UPDATE : This will make the JSON string look more the way you want it.

while($SQLRow = mysql_fetch_assoc($SQLResult)){
    $_UserData['Fname'][] = $SQLRow['Fname'];
    $_UserData['Lname'][] = $SQLRow['Lname'];
}

Answer 2

$_UserData['Fname'] = $SQLRow['Fname'];
$_UserData['Lname'] = $SQLRow['Lname'];

You're repeatedly setting the same two elements of _UserData, rather than appending new values to the array.

Answer 3

Your loop overrides the values in $_UserData array. You should append each row to the array.

In addition, please note that you should escape your input.

Your code should look like this:

$_Query = '
                SELECT
                Fname,Lname
                FROM
                users
                WHERE
                number = "'.(int)$_REQUEST['number'].'"
                                                ;';

$SQLResult = mysql_query($_Query) or die(mysql_error());

$_UserData = array();

if(mysql_num_rows($SQLResult) <> 0){

        while($SQLRow = mysql_fetch_array($SQLResult)){
                $_UserData[] = $SQLRow;
        }

}
echo json_encode($_UserData);

Answer 4

Try with this:

$_UserData = array();

if(mysql_num_rows($SQLResult) <> 0){
        $i = 0;
        while($SQLRow = mysql_fetch_array($SQLResult)){
                $_UserData[$i]['Fname'] = $SQLRow['Fname'];
                $_UserData[$i]['Lname'] = $SQLRow['Lname'];
                $i++;
        }

Answer 5

I'm sure you'll kick yourself when you realise that you've only assigned to 1 element of your $_UserData array.

Should be something like:

$_UserData = array();

if(mysql_num_rows($SQLResult) <> 0){

        while($SQLRow = mysql_fetch_array($SQLResult)){
                // Either...
                // Create a new entry in $_UserData with same structure and contents as $SQLRow
                $_UserData[] = $SQLRow;
                // OR more specific - say if the structures aren't identical
                // $DataRow=array();
                // $DataRow['Fname'] = $SQLRow['Fname'];
                // $DataRow['Lname'] = $SQLRow['Lname'];
                // $_UserData[] = $DataRow;
        }

}

Cheers