jQuery失败：抛出错误的parsererror：SyntaxError：意外的JSON输入结束

Question

我有一个联系表单，它通过ajax传递给PHP脚本。处理表单后，ajax将根据从PHP脚本中的json_encode（）函数收到的响应执行一些操作。问题是我收到以下错误信息：

parsererror 语法错误：JSON输入意外结束

的readyState：4 的responseText： “” 状态：200 状态文本： “OK”

当dataType是ajax调用中的文本，而PHP脚本只是简单地回显一条文本消息，然后代码工作正常，但使用json，我得到了上述错误。

我尝试过头（“Content-Type：application/json”）和JSON.parse（）没有成功。我在头文件中添加了charset =“UTF-8”，并在传递给json_encode的数组上尝试了encode_utf8（）函数，但似乎没有任何效果。

我在下面发布相关文件的代码。任何帮助解决这个问题将不胜感激。

contact.php

<form action="" method="post" name="contactForm" id="contactForm"> 

    <div class="form-row">   
     <div id="contactFormResponse"></div> 

     <div class="form-col"> 
      <label for="orderNumer">Order Number</label> 
      <input type="text" name="orderNumber" id="orderNumber" value="<?php echo ($_POST['orderNumber']); ?>" /> 
     </div> 

     <div class="form-col"> 
      <label for="comment">Comment *</label> 
      <textarea name="message" id="comment" maxlength="2000" rows="5"><?php echo ($_POST['comment']); ?></textarea> 
     </div> 

     <div class="form-col"> 
      <label for="title">Title *</label> 
      <select name="title" id="title"> 
       <option value="" <?php if ($_POST['title'] == "") {echo "selected='selected'";} ?>>Select a title...</option> 
       <option value="ms" <?php if ($_POST['title'] =="ms") {echo "selected='selected'";} ?>>Ms</option> 
       <option value="miss" <?php if ($_POST['title'] == "miss") {echo "selected='selected'";} ?>>Miss</option> 
       <option value="mrs" <?php if ($_POST['title'] == "mrs") {echo "selected='selected'";} ?>>Mrs</option> 
       <option value="mr" <?php if ($_POST['title'] == "mr") {echo "selected='selected'";} ?>>Mr</option> 
       <option value="other" <?php if ($_POST['title'] == "other") {echo "selected='selected'";} ?>>Other</option> 
      </select> 
     </div> 

     <div class="form-col"> 
      <label for="firstName">First Name *</label> 
      <input type="text" name="firstName" id="firstName" value="<?php echo ($_POST['firstName']); ?>" /> 
     </div> 

     <div class="form-col"> 
      <label for="surname">Surname *</label> 
      <input type="text" name="surname" id="surname" value="<?php echo ($_POST['surname']); ?>" /> 
     </div> 

     <div class="form-col"> 
      <label for="email">Email Address *</label> 
      <input type="text" name="email" id="email" value="<?php echo ($_POST['email']); ?>" /> 
     </div> 

     <div class="form-col"> 
      <input type="submit" name="submitContactForm" id="submitContactForm" value="Submit" class="btn" /> 
     </div> 

    </div> 
</form> 

jsCode.js

// process contact form 
$("#contactForm").submit(function(e) { 

    e.preventDefault(); 

    // some jQuery validation goes here... 

    $.ajax({ 
     type:"POST", 
     url: "functions.php", 
     dataType: "json", 
     data: new FormData(this), 
     //data: $('form').serialize(), 
     processData: false, 
     contentType: false, 
     success:function(response) { 
      if(response.status === "OK") { 
       $("#contactFormResponse").html("<div class='alert alert-success' id='message'></div>"); 
       $("#message").html(response.message).fadeIn("100"); 
       $("#contactForm")[0].reset(); 
       $(window).scrollTop(0); 
      } else if (response.status === "error") { 
       $("#contactFormResponse").html("<div class='alert alert-danger' id='message'></div>"); 
       $("#message").html(response.message).fadeIn("100"); 
       $(window).scrollTop(0); 
      } 
     }, 
     error:function(jqXHR, textStatus, errorThrown) { 
      console.log("JQuery failed: " + textStatus + " with error thrown: " + errorThrown); 
      console.log(jqXHR); 

     } 
    }); 
}); 

的functions.php

// send email 
function sendMessage() { 

    if (isset($_POST["submitContactForm"])) { 
     if (!$_POST["comment"]) { 
      $error .= "<br />Comment is required."; 
     } 

     if (!$_POST["firstName"]) { 
      $error .= "<br />First name is required."; 
     } 

     // validation for other form fields goes here... 

     if ($error) { 
      echo json_encode(array("status" => "error", "message" => "There were error(s)in your form: " . $error)); 

     } else { 

      $to   = "email@domain.com"; 
      $subject  = "Message from the website"; 
      $order_number = $_POST["orderNumber"]; 
      $comment  = $_POST["comment"]; 
      $title   = $_POST["title"]; 
      $first_name = $_POST["firstName"]; 
      $surname  = $_POST["surname"]; 
      $email_address = filter_var($_POST["email"], FILTER_SANITIZE_EMAIL); 
      $headers  = "From: " . $title . " " . $first_name . " " . $surname . " <" . $email_address . " >"; 
      $message  = "Order Number: " . $order_number . "/r/n" . "Topic: " . $topic . "/r/n" . "Comment: " . $comment; 

      $result = mail($to, $subject, $message, $headers); 

      if (!$result) { 

       echo json_encode(array("status" => "error", "message" => "Message failed.")); 

      } else { 

       echo json_encode(array("status" => "OK", "message" => "Message sent.")); 

      } 
     } 
    } 
} 

Answer 1

你是不是在解析您的success功能JSON响应，你需要使用$.parseJSON(response) like below

success:function(res) { 
var response=$.parseJSON(res); 
       if(response.status === "OK") { 
        $("#contactFormResponse").html("<div class='alert alert-success' id='message'></div>"); 
        $("#message").html(response.message).fadeIn("100"); 
        $("#contactForm")[0].reset(); 
        $(window).scrollTop(0); 
       } else if (response.status === "error") { 
        $("#contactFormResponse").html("<div class='alert alert-danger' id='message'></div>"); 
        $("#message").html(response.message).fadeIn("100"); 
        $(window).scrollTop(0); 
       } 
      },