如何让jQuery ajax执行错误功能

I have looked around and followed the instructions on this post but I still can not get the error function to execute. What I want to do is have my PHP script return either error or success along with a message. Eventually it will be data returned from a database that will be put inside of a div but for now I just need to get the error handling working. I am hoping someone here can help me out with this. I have included my code below.

我环顾四周并按照这篇文章的说明进行操作，但仍然无法执行错误功能。我想要做的是让我的PHP脚本返回错误或成功以及消息。最终它将是从数据库返回的数据，将被放入div中，但是现在我只需要让错误处理工作。我希望有人可以帮助我解决这个问题。我在下面提供了我的代码。

This is obviously my AJAX request.

这显然是我的AJAX请求。

function getProductInfo() {
    if($("#serialNumber").val() != '') {
        serialNumber = $("#serialNumber").serialize();
        $.ajax({
            type: "POST",
            url: 'post.php',
            data: serialNumber + "&getSerialNumber=" + 1,
            dataType: 'json',
            success: function(data, textStatus, jqXHR) {
                console.log("SUCCESS");
            },
            error: function(jqXHR, textStatus, errorThrown) {
                console.log("ERROR");
            }
        });
    }
}

This is my php function that will return the error and message as JSON

这是我的php函数，它将以JSON的形式返回错误和消息

function getSerialNumber() {
    $serial = $_POST['serial'];
    $product = new Inventory;

    if($product->GetSerial($serial)) {
        $productInfo = $product->GetSerial($serial)->first();
        echo '{"error": false, "message": "Successfully got serial number"}';
    } else {
        echo '{"error": true, "message": "failed to get serial number"}';
    }
}

As the current code stand it will only keep outputting SUCCESS regardless if it actually has an error or not.

就目前的代码而言，无论是否实际出现错误，它都只会继续输出SUCCESS。

2 个解决方案

#1

You need to send an http status code other than 200:

您需要发送200以外的http状态代码：

if($product->GetSerial($serial)) {
    $productInfo = $product->GetSerial($serial)->first();
    header('Content-Type: application/json', true, 200);
    die(json_encode(["error"=> false, "message"=> "Successfully got serial number"]));
} else {
    header('Content-Type: application/json', true, 400);
    die(json_encode(["error"=> true, "message"=> "failed to get serial number"]));
}

Also, when sending json, set the content type accordingly, and never try and manually build a json string, use the built in json_encode function instead, and its a good idea to use die() or exit() rather than echo, to avoid any accidental additional output

另外，在发送json时，相应地设置内容类型，并且永远不要尝试手动构建json字符串，而是使用内置的json_encode函数，并且最好使用die（）或exit（）而不是echo，以避免任何意外的额外输出

That said, although sending appropriate status codes seems like a good idea, you may well find it a lot easier to always return 200 and parse the response, and keep the error handler for unexpected errors

也就是说，尽管发送适当的状态代码似乎是一个好主意，但您可能会发现总是返回200并解析响应更容易，并保留错误处理程序以应对意外错误

2 个解决方案

#1

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