"SyntaxError:JSON.parse:unexpected character"将多个变量从AJAX传递给PHP时出错

Question

我正在使用AJAX将变量从表单传递到PHP页面以处理数据库中的数据.

用户点击按钮后,会触发以下JavaScript:

$(document).ready(function() {

    $("#myForm").submit(function(event) {

        /* validate the fields */
        var firstDate= "11/10/2014"
        var secondDate = "10/10/2014"
        var myString = "some Text";
        var myArray = ["name1", "name2", "name3", "123-123-33gf"];

        processIT(firstDate, secondDate, muString, myArray);

    });/* end of submit */

});

function processIT(firstDate, secondDate, muString, myArray) {
    var response = ""; 
    $(function () {
        $.ajax({
            url: 'api.php',           // the script to call to get data
            type: "POST", 
            data: {
                firstDate: firstDate, 
                secondDate : secondDate , 
                myString : myString , 
                myArray : myArray , 
            },                 // you can insert url argumnets here to pass to api.php
            dataType: 'json',         // return data format
            success: function(data) { //
                alert(data);
             },
             error: function (jqXHR, textStatus, errorThrown){
                 console.log(textStatus, errorThrown);
             },
        });
    });
    return response;
}

api.php页面包含以下内容

<?php 


    if ( isset($_POST["firstDate"]) && !empty($_POST["firstDate"])){
        $response .= "<p>firstDate= " . $_POST["firstDate"] . "</p>"; 
    }
    else $response .= " 1 ";
    if ( isset($_POST["secondDate"]) && !empty($_POST["secondDate"])){
        $response .= "<p>secondDate = " . $_POST["secondDate"] . "</p>";
    }
    else $response .= " 2 ";
    if ( isset($_POST["myString"]) && !empty($_POST["myString"])){
        $response .= "<p>myString = " . $_POST["myString"] . "</p>";
    }
    else $response .= " 3 ";
    if ( isset($_POST["myArray"]) && !empty($_POST["myArray"])){
        $response .= "<p>myArray = " . $_POST["myArray"] . "</p>";
    }
    else $response .= " 4 ";

echo json_encode($response);
?>

但是,当我单击该按钮时,我收到以下错误:

SyntaxError:JSON.parse:JSON数据的第1行第1列的意外字符

但是,如果我将POST更改为GET,我可以看到传递的变量,但仍然会得到相同的错误.

我有什么想法我做错了吗？

Answer 1

您的PHP文件未输出有效的JSON响应,这JSON.parse就是抛出错误的原因.PHP代码中存在许多错误,这些错误包含在输出中,从而产生无效的JSON响应.

console.log("firstDate" + $_POST["firstDate"]);

这不是有效的PHP代码.PHP没有console.log().它有echo.PS你.用来连接PHP中的字符串,而不是+.

$_POST["secondDate "]
$_POST["myString "]
$_POST["myArray "]

这些钥匙.最后没有空间.他们应该是:

$_POST["secondDate"]
$_POST["myString"]
$_POST["myArray"]

最后,$_POST["myArray"]是一个数组.您无法将其连接到字符串.试试这个:

$response .= "<p>myArray = ".implode(', ', $_POST["myArray"])."</p>";