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首页 > 解决方案 > 为什么我的 AJAX JQUERY 试图将多个变量值传递给 php 服务器端总是返回错误？

javascript - 为什么我的 AJAX JQUERY 试图将多个变量值传递给 php 服务器端总是返回错误？

问题描述

我正在尝试通过 JavaScript 将多个值从客户端发送到服务器端。下面是我的 JavaScript JQUERY 函数：

// This function addToRentalCart(car) adds the requested
// car to a user's rental cart. It does this through the use
// of an AJAX call to the server with the required data.
// Since the required car data is already on the associated
// html webpage (index.html) all this function does is read it
// from there and send it to the server side process.
// Server side processing will then take care of the rest
// of the operation.

function addToRentalCart (car)
{
    var carPosition = 'carRow' + car;
    var carAvailabilityCol = 'Col10';
    var carAvailableElement = carPosition + carAvailabilityCol;
    var carAvailable = document.getElementById(carAvailableElement);
    
    if (carAvailable.innerHTML === 'N')
    {                                                         //If the car is not available
        alert("Sorry, the car is not available now. Please try other cars.");
    }
    else
    {                                                       //If the car is available
        //var carPictureFileCol = 'Col0';               //Set the column numbers of each desired variable
        var carMakeCol = 'Col1';
        var carModelCol = 'Col2';
        var carYearCol = 'Col3';
        var carPricePerDayCol = 'Col8';
        
        var carMakeElement = carPosition + carMakeCol;
        var carMake = document.getElementById(carMakeElement).innerHTML;   //Get the car make or brand
        
        var carModelElement = carPosition + carModelCol;
        var carModel = document.getElementById(carModelElement).innerHTML;      //Get the car model
        var carPictureFile = carModel + '.jpg';                           //Get the car picture file
        
        var carYearElement = carPosition + carYearCol;
        var carYear = document.getElementById(carYearElement).innerHTML;        //Get the car year
        
        var carPricePerDayElement = carPosition + carPricePerDayCol;
        var carPricePerDay = document.getElementById(carPricePerDayElement).innerHTML;
        carPricePerDay = carPricePerDay.substring(1);    //Get the price per day without the dollar sign
        $.ajax({
                type: "GET",
                url: "rentalCarsCart.php",
                data: {"carPicFile": carPictureFile, "carBrand": carMake, "carMod": carModel, 
                       "carYearMan": carYear, "carPPD": carPricePerDay},
                dataType: "json",
                success: function()
                {
                    alert("You have successfully added this car to your rental cart");
                },
                error: function()
                {
                    alert("error in Ajax call to cart url");
                },
            });
    }
}

即使根据 Apache Netbeans IDE 输出，所有值似乎都很好，但它会爆炸并始终显示错误警报。我什至使用断点和逐步方法在 Chrome 开发人员工具中对其进行了调试，所有值都很好。它在 JQUERY.JS 本身内部轰炸，而不是将 GET 请求发送到 PHP 后端。有谁知道为什么？我似乎找不到我的代码有什么问题。如果你能帮助我，我将不胜感激。

我的服务器端代码如下：

<!DOCTYPE html>
<html>
    <head>
        <meta charset="UTF-8">
        <title>Rental Car Shopping Cart</title>
        <style>
            h1 {text-align: center;}
            button:hover{cursor: pointer;}
        </style>
    <h1>Car Reservation</h1>
    </head>
    <body>
        <?php
            header("Access-Control-Allow-Origin: *");
            
            $carPicture = $_GET['carPicFile']; // get car picture filename
            $carMake = $_GET['carBrand'];     // get the make of the car
            $carModel = $_GET['carMod'];      // get the car model
            $carYear = $_GET['carYearMan']; // Get the car year of manufacture
            $carPricePerDay = $_GET['carPPD'];    // Get the car price per day
            echo ('<div>');
            echo ($carPricePerDay);
            echo ('</div>');
        ?>
    </body>
</html>

标签： javascriptphpjqueryjsonajax

解决方案

Phil 在上面的评论中是正确的，您不需要来自 PHP 服务器端编码的 dataType 响应，因此我将其删除。我还发现了 PHP 后端的另一个问题，虽然上面的 PHP 代码中没有显示，这最初是非常基本的，但后来变得更加复杂：

Instead of if (($carPicture!=0) && ($carMake!=0) && ($carModel!=0) && ($carYear!=0) && 
  ($carPricePerDay!=0))
            {
                session_start();         //etc
Try:  
if (isset($carPicture) && isset($carMake) && isset($carModel) && isset($carYear) && isset($carPricePerDay))
            {
                session_start();           //etc

原因是当变量未定义时，它们也等于零，或者我不确定为什么它决定使用底层版本。正逻辑似乎比 PHP 的负逻辑更好。多谢你们。

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