MySQL integer field is returned as string in PHP

Question

I have a table field in a MySQL database:

userid INT(11)

So I am calling it to my page with this query:

\"SELECT userid FR         


        

Answer 1

If prepared statements are used, the type will be int where appropriate. This code returns an array of rows, where each row is an associative array. Like if fetch_assoc() was called for all rows, but with preserved type info.

function dbQuery($sql) {
    global $mysqli;

    $stmt = $mysqli->prepare($sql);
    $stmt->execute();
    $stmt->store_result();

    $meta = $stmt->result_metadata();
    $params = array();
    $row = array();

    while ($field = $meta->fetch_field()) {
      $params[] = &$row[$field->name];
    }

    call_user_func_array(array($stmt, 'bind_result'), $params);

    while ($stmt->fetch()) {
      $tmp = array();
      foreach ($row as $key => $val) {
        $tmp[$key] = $val;
      }
      $ret[] = $tmp;
    }

    $meta->free();
    $stmt->close();

    return $ret;
}

Answer 2

I like mastermind's technique, but the coding can be simpler:

function cast_query_results($result): array
{
    if ($result === false)
      return null;

    $data = array();
    $fields = $result->fetch_fields();
    while ($row = $result->fetch_assoc()) {
      foreach ($fields as $field) {
        $fieldName = $field->name;
        $fieldValue = $row[$fieldName];
        if (!is_null($fieldValue))
            switch ($field->type) {
              case 3:
                $row[$fieldName] = (int)$fieldValue;
                break;
              case 4:
                $row[$fieldName] = (float)$fieldValue;
                break;
              // Add other type conversions as desired.
              // Strings are already strings, so don't need to be touched.
            }
      }
      array_push($data, $row);
    }

    return $data;
}

I also added checking for query returning false rather than a result-set.
And checking for a row with a field that has a null value.
And if the desired type is a string, I don't waste any time on it - its already a string.

---

I don't bother using this in most php code; I just rely on php's automatic type conversion. But if querying a lot of data, to then perform arithmetic computations, it is sensible to cast to the optimal types up front.

Answer 3

My solution is to pass the query result $rs and get a assoc array of the casted data as the return:

function cast_query_results($rs) {
    $fields = mysqli_fetch_fields($rs);
    $data = array();
    $types = array();
    foreach($fields as $field) {
        switch($field->type) {
            case 3:
                $types[$field->name] = 'int';
                break;
            case 4:
                $types[$field->name] = 'float';
                break;
            default:
                $types[$field->name] = 'string';
                break;
        }
    }
    while($row=mysqli_fetch_assoc($rs)) array_push($data,$row);
    for($i=0;$i<count($data);$i++) {
        foreach($types as $name => $type) {
            settype($data[$i][$name], $type);
        }
    }
    return $data;
}

Example usage:

$dbconn = mysqli_connect('localhost','user','passwd','tablename');
$rs = mysqli_query($dbconn, "SELECT * FROM Matches");
$matches = cast_query_results($rs);
// $matches is now a assoc array of rows properly casted to ints/floats/strings

Answer 4

No. Regardless of the data type defined in your tables, PHP's MySQL driver always serves row values as strings.

You need to cast your ID to an int.

$row = $result->fetch_assoc();
$id = (int) $row['userid'];

Answer 5

Easiest Solution I found:

You can force json_encode to use actual numbers for values that look like numbers:

json_encode($data, JSON_NUMERIC_CHECK) 

(since PHP 5.3.3).

Or you could just cast your ID to an int.

$row = $result->fetch_assoc();
$id = (int) $row['userid'];

Answer 6

For mysqlnd only:

 mysqli_options($conn, MYSQLI_OPT_INT_AND_FLOAT_NATIVE, true);

Otherwise:

  $row = $result->fetch_assoc();

  while ($field = $result->fetch_field()) {
    switch (true) {
      case (preg_match('#^(float|double|decimal)#', $field->type)):
        $row[$field->name] = (float)$row[$field->name];
        break;
      case (preg_match('#^(bit|(tiny|small|medium|big)?int)#', $field->type)):
        $row[$field->name] = (int)$row[$field->name];
        break;
    }
  }