如何在PHP中以正确的JSON格式输出？

Question

我要输出的MySQL表通过以下方式的一些记录与resCode和resText在JSON：

{ 
    "data":[ 
      { 
       "id":"44", 
       "month":"January", 
       "income":"2500", 
       "expanse":"0" 
      }, 
      { 
       "id":"45", 
       "month":"February", 
       "income":"5500", 
       "expanse":"400" 
      }, 
      { 
       "id":"47", 
       "month":"March", 
       "income":"25000", 
       "expanse":"11000" 
      } 
     ], 
    "resCode":"200", 
    "resText":"SUCCESS" 
} 

因为我很新的PHP数组，我无法在服务器侧部格式化为输出如上。另外我不知道如何输出resCode和resText。

同样的任何帮助将不胜感激。

PHP部分：

<?php 
include_once 'includes/db_connect.php'; 
?> 

<?php 
$stmt = $mysqli->prepare("SELECT * FROM records"); 
     $stmt->execute(); // Execute the prepared query. 
     $stmt->store_result(); 

     // get variables from result. 
     $stmt->bind_result($id, $month, $income, $expanse); 
     while ($stmt->fetch()) { 
      $data[]=array(id=>$id, month=>$month, income=>$income, expanse=>$expanse); 
     } 
      $response["data"] = $data; 
      $response["resCode"] = "200"; 

      echo json_encode($response); 
?> 

Answer 1

while ($stmt->fetch()) { 
    $output["data"][]=array('id'=>$id, 'month'=>$month, 'income'=>$income, 'expanse'=>$expanse);   
} 
$output["resCode"] = '200'; 
$output["resText"] = 'SUCCESS'; 
echo json_encode($output); 

Answer 2

试试这个

$stmt->bind_result($id, $month, $income, $expanse); 
    $response = array(); 
    $data = array(); 
    while ($stmt->fetch()) { 
     $data[]=array(id=>$id, month=>$month, income=>$income,expanse=>$expanse); 
    } 
     $response["data"] = $data; 
     $response["resCode"] = "XXX"; 
     echo json_encode($response);