javascript - Ajax 调用不断失败，但没有错误

Question

所以我的 ajax 调用工作正常，我在 Web 检查器中看到它旁边的绿灯，我看到发送回的正确参数和返回的 JSON 对象。但是，它会自动转到“错误”。我添加了一些错误处理，但已经没有想法了..

有人看到什么了吗？

Javascript

    function getMoreOlder(id) { 
        $.ajax({
            url: "getmore.php",
            type : "POST",
            dataType : "json",
            data: {
                postid : id,
                type: "later",
            },
           success: function(response){
               alert("Success");
              for (var i=0; i<json.length; i++) {
               //     alert("Loop");
                    }
          },
          error: function(jqXHR,error, errorThrown){
                   if(jqXHR.status&&jqXHR.status==400){
                        alert(jqXHR.responseText); 
                   }else{
                       alert("Something went wrong");
                   }
          }   
         });
    }

PHP 脚本

        $id = $_POST['postid'];
        $type = $_POST['type'];

        if($type == "later") {
        $sql = "CALL `sp_BlogPosts_PageByPostID`('" . $id ."')";
        $exec = mysqli_query($conn, $sql);

        while($fetch = mysqli_fetch_assoc($exec)) { 
           $obj = array(
             "PostID" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['PostID']))),  
             "Title" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['title']))),     
             "Content" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['content']))),     
             "Email" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['email']))),    
             "Username" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['username']))),    
             "Created_At" => stripslashes(substr(trim(mysqli_real_escape_string($conn, $fetch['created_at'])), 0 , 10)));
              echo json_encode($obj);
        }

使用示例

  <button onClick="getMoreOlder(1113);"> Load More Posts </button>

返回示例

    {"PostID":"1177","Title":"Off track","Content":"Hi This is a Post", "Email":"me@yahoo.com", "Username":"Abc123", "Created_At":"12-2-14"}

要点

• 在 POST 中发送正确的变量

• POST返回正确信息

• 总是返回错误出了问题!

• Web Inspector 显示 POST 请求的 200 代码

Answer 1

问题是你不断地回显 json_encode，这将导致无效的 json 输出。

例如:

{"PostID":"1177","Title":"Off track"}{"PostID":"1177","Title":"Off track"}

这是无效的 json。因此，将所有数据放入数组/对象中，然后 echo json_encode

$objArr = array();
while($fetch = mysqli_fetch_assoc($exec)) { 
    $objArr[] = array(
        "PostID" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['PostID']))),  
        "Title" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['title']))),     
        "Content" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['content']))),     
        "Email" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['email']))),    
        "Username" => stripslashes(trim(mysqli_real_escape_string($conn, $fetch['username']))),    
        "Created_At" => stripslashes(substr(trim(mysqli_real_escape_string($conn, $fetch['created_at'])), 0 , 10))
    );          
}

echo json_encode($objArr);

或者按照评论中的建议一次性获取所有数据并回显

echo json_encode($exec->fetch_all(MYSQLI_ASSOC));

此外，您的 for 循环使用了错误的变量

for (var i=0; i<json.length; i++) {

应该是

for (var i=0; i<response.length; i++) {