Ajax+php json回答错误

Question

我的网站注册表有问题。使用 json 格式的交换从服务器获取“null”。

代码：

JS部分代码：

$("#customer-form").submit(function() { 
        var form_data = {
          customer_name: $("#customer_name").val(),
          customer_dir: $("#customer_dir").val(),
          confirm_dir: $("#confirm_dir").val(),
          customer_fname: $("#customer_fname").val(),
          customer_address: $("#customer_address").val(),
          customer_inn: $("#customer_inn").val(),
          customer_kpp: $("#customer_kpp").val(),
          customer_ogrn: $("#customer_ogrn").val(),
          customer_okpo: $("#customer_okpo").val(),
        }; 
        $.ajax({
        type: "POST",
        url: "customer_form.php",
        data: form_data,
        dataType: "json",
        success: function(data) { 
          console.log(data.status);
                },      
        error: function(xhr, status) {
               console.log("Error"+" "+xhr.responseText +" "+ status)
               }
           }); 
      });

PHP

<?php   
    $customer_fname=trim($_POST["customer_fname"]);
    $customer_fname=strip_tags($customer_fname); 

    $confirm_dir=trim($_POST["confirm_dir"]);
    $confirm_dir=strip_tags($confirm_dir);
    $confirm_dir = md5($confirm_dir);

    $customer_name=trim($_POST["customer_name"]);
    $customer_name=strip_tags($customer_name);

    $customer_dir = trim($_POST["customer_dir"]);
    $customer_dir=strip_tags($customer_dir);
    $customer_dir=md5($customer_dir);

    $customer_address=trim($_POST["customer_address"]);
    $customer_address=strip_tags($customer_address);

    $customer_inn=trim($_POST["customer_inn"]);
    $customer_inn=strip_tags($customer_inn);

    $customer_kpp=trim($_POST["customer_kpp"]);
    $customer_kpp=strip_tags($customer_kpp);

    $customer_ogrn=trim($_POST["customer_ogrn"]);
    $customer_ogrn=strip_tags($customer_ogrn);

    $customer_okpo=trim($_POST["customer_okpo"]);
    $customer_okpo=strip_tags($customer_okpo);

    if($customer_dir==$confirm_dir){ 
        $con = new mysqli("localhost", "root", "", "berezka");
        $sql = "SELECT * FROM CUSTOMER where customer_name='".$customer_name."'";
        $result = $con->query($sql);

        if($row = $result->fetch_assoc()){
           $response = array("status"=>"Login has been already given");
           echo json_encode($response);
        } 
        else{ 
            $sql_2 = "INSERT INTO customer (`customer_id`, `customer_fname`, `customer_name`, `customer_dir`, `customer_address`, `customer_inn`, `customer_kpp`, `customer_ogrn`, `customer_okpo`) VALUES ('', '$customer_fname', '$customer_name', '$customer_dir', '$customer_address', '$customer_inn', '$customer_kpp', '$customer_ogrn', '$customer_okpo')";
            mysqli_query($con, $sql_2);
            $response = array("status"=>"Registration is ok");
            echo json_encode($response);
           // mysqli_free_result($result);
           // mysqli_close($con);
       }  
    }  
    else{   
        $response = array("status"=>"Password doesnt match");
        echo json_encode($response);   
    } 
?>

这是我在 firefox firegub 或 Chrome 控制台中的错误： 未捕获的类型错误：无法读取 null 的属性“状态”

我试图使用 JSON.Parse 等。试图通过对象、数组或其他格式制作 json 消息。

Answer 1

请添加标题信息：

header('Content-Type:application/json'); 

在发送响应之前。