Question

当我使用AJAX调用PHP运行我的小javascript时，它总是返回一个JSON parsererror。

在PHP中我可以看到我的json填充了这样的数组： json编码：{"Year":"2012","Make":"Ford","Model":"Taurus","Error":"no error"}

This是网站：

当访问者离开左侧的字段VIN时，将调用AJAX，并且应自动填充右侧的3个字段。

function decode_my_vin(myvin) {

alert("javascript began");

dataoneID  = '00000';
dataoneKEY = 'x0x0x';

jQuery.ajax(
    {
    cache: false,
    type: 'POST',
    url: '/wp-content/themes/Impreza-child/vin-decoder.php',
	dataType: "json",
    data: { 
			'value1_VIN':	myvin,
			'value2_ID':	dataoneID,
			'value3_KEY':	dataoneKEY,
		},
	success: function(response) {
		
			var resultYear = response.Year;
			var resultMake = response.Make;
			var resultModel = response.Model;
			var resultMessage = response.Message;
			alert("success returned: Year " + resultYear + " Make " + resultMake + " Model " + resultModel + " Message " + resultMessage); 

		    document.getElementById("fld_7290902_1").value = resultYear;
	    	document.getElementById("fld_1595243_1").value = resultMake;
    		document.getElementById("fld_7532728_1").value = resultModel;
    		document.getElementById("fld_7532728_1").value = resultMessage;

			return;
		},
    error: function (jqXHR, exception) {
        var msg = '';
        if (jqXHR.status === 0) {
            msg = 'Not connect.\n Verify Network.';
        } else if (jqXHR.status == 404) {
            msg = 'Requested page not found. [404]';
        } else if (jqXHR.status == 500) {
            msg = 'Internal Server Error [500].';
        } else if (exception === 'parsererror') {
            msg = 'Requested JSON parse failed.';
        } else if (exception === 'timeout') {
            msg = 'Time out error.';
        } else if (exception === 'abort') {
            msg = 'Ajax request aborted.';
        } else {
            msg = 'Uncaught Error.\n' + jqXHR.responseText;
        }
		alert("Error in jquery: " + msg); 
        $('#post').html(msg);
	},
	complete: function(value) {
			alert("returned after complete: " + value);
		}
	});
}

<?php

    header('Content-Type: application/json');

    $vinResult = array(
      'Year'  => '2012',
      'Make'  => 'Ford',
      'Model' => 'Taurus',
      'Error' => 'No Error'
	);

    echo json_encode($vinResult);
?>

Answer 1

在ajax调用的错误函数中，它为$('#post').html(msg);说错误函数提供了错误。尝试使用jQuery('#post')代替$('#post')。

要解析您的回复，请使用var data = jQuery.parseJSON( response );，然后按data.Year等方式访问字段。

AJAX：从PHP返回错误：请求的JSON解析失败

1 个答案: