Question

我无法找到我的代码有什么问题。从post_receiver.php打印json文件时，会相应地打印json。

从post_receiver.php打印的JSON

     <?php 
      session_start();
      ob_start();                    
      require_once('../../mysqlConnector/mysql_connect.php');
       $result_array = array();

   $query="SELECT COUNT(initID) AS count, urgency, crime, initID, TIMESTAMPDIFF( minute,dateanalyzed,NOW()) AS minuteDiff FROM initialanalysis WHERE commanderR='0' AND stationID='{$_SESSION['stationID']}';";

 $result=mysqli_query($dbc,$query);
   if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {

      array_push($result_array, $row);

      }

                            }
    echo json_encode($result_array);
                            ?>

上面的结果：

[{"count":"10","urgency":"Low","crime":"Firearm","initID":"5","minuteDiff":"329"}]

我的ajax代码：

$.ajax({
        method: 'POST',
        url: "post_receiver.php",
        data: {
            'count': count,
            'urgency': urgency
        },...

＆＃39;计数＆＃39;和紧急情况＆＃39;变量没有定义，我不熟悉JSON格式...

Answer 1

在success回调中，您会收到一个data字符串，其中包含回复。要将其解析为JSON，请使用json dataType设置：

$.ajax({
  method: 'POST',
  url: 'post_receiver.php',
  dataType: 'json',
  success: function (data) {
    // 'data' contains the parsed JSON
    console.log('Count:', data[0].count); // read the values from the JS object and log them to the console
    console.log('Urgency:', data[0].urgency);
  }
});

ajax，变量没有定义？

1 个答案: