如何使用PHP将现有值对推送到Array？

Question

我想将$ pair推送到现有阵列但不行...... 我已经使用了合并数组而且没有用。

  $pair = array(
        'status'    => 1
  );

    $numberof = mysqli_num_rows($result);

    if($numberof > 0)
    {
        // output data of each row
 while($row[] = $result->fetch_assoc()) {

 $tem = $row;

 $tem = $row['status']  = '1';


 $json = json_encode($tem);

这是最终结果：

{"0":{"id":"7","nome":"dfsdfsdfsdff","email":"and@gmail.com"},"status":1}

我只想要最终结果：

{"id":"7","nome":"dfsdfsdfsdff","email":"and@gmail.com","status":1}

Answer 1

$ TEM =阵列（）; $温度= array_merge（$ TEM，$对）;

然后编码$ temp

Answer 2

你在while循环中使用了数组。拜托，希望你能得到答案

    $pair = array(
        'status'    => 1
  );

    $numberof = mysqli_num_rows($result);

    if($numberof > 0)
    {
        // output data of each row
 while($row = $result->fetch_assoc()) { // instead $row[] use only $row



  $row['status']  = '1';


 $json = json_encode($row);

Answer 3

你试过while($row = $result->fetch_assoc()) { ... } 代替 while($row[] = $result->fetch_assoc()) { ... }？

Answer 4

$pair = array( 'status'    => 1);

$numberof = mysqli_num_rows($result);

if($numberof > 0) { 
    while($row[] = $result->fetch_assoc()) {
        $row['status']  = '1';
    } 
    $json = json_encode($row);
}