Javascript 如何将参数传递给 jQuery $.getJSON 回调方法？

How can I pass parameters to a jQuery $.getJSON callback method?

提问by Pure.Krome

I'm trying to use jQuery to call some custom API via Ajax/$.getJSON.

我正在尝试使用 jQuery 通过Ajax/$.getJSON.

I'm trying to pass a custom value into the Ajax callback method, but that value is not getting passed through and is actually getting overwritten. This is my code:

我正在尝试将自定义值传递到 Ajax 回调方法中，但该值并未传递，实际上已被覆盖。这是我的代码：

var locationType = 3;
var url = 'blah blah blah' + '&locationType=' + locationType;

$("#loading_status").show();

$.getJSON(url, null, function(results, locationType) {
    searchResults(results, locationType)
});

The value of locationTypeBEFORE I call the URL using AJAX is 3. But after the call returns the data successfully, the value for locationTypeis now success. This is because the method signature of the callbackis:

的价值locationType在我请使用AJAX是URL 3。但是调用成功返回数据后，值为locationTypenow success。这是因为回调的方法签名是：

callback(data, textStatus)A callback function that is executed if the request succeeds.

callback(data, textStatus) 请求成功时执行的回调函数。

How can I pass 1 or more parameters to a callback method?

如何将 1 个或多个参数传递给回调方法？

采纳答案by Nick Craver

You don't need to pass it in, just reference the variable you already have, like this:

您不需要传入它，只需引用您已有的变量，如下所示：

var locationType = 3;
var url = 'blah blah blah' + '&locationType=' + locationType;
$("#loading_status").show();
$.getJSON(url, null, function(results) {
    searchResults(results, locationType)
});

Also there's no need to pass nullif you don't have a data object, it's an optional parameter and jQuery checks if the second param is a function or not, so you can just do this:

此外，null如果您没有数据对象，则无需传递，它是一个可选参数，jQuery 检查第二个参数是否为函数，因此您可以这样做：

$.getJSON(url, function(results) {
    searchResults(results, locationType)
});

回答by Felix Kling

Warp in a function, e.g.

扭曲函数，例如

function getResults(locationType) {
    $.getJSON(url, null, function(results) {
        searchResults(results, locationType)
    });
}

But in you specific situation you don't even have to pass it, you can access the value directly in the callback.

但是在您的特定情况下，您甚至不必传递它，您可以直接在回调中访问该值。

回答by Darin Dimitrov

You could use the .ajaxmethod:

您可以使用.ajax方法：

var locationType = 3;
var url = 'blah blah blah' + '&locationType=' + locationType;
$.ajax({
    url: url,
    context: { lt: locationType },
    success: function(results) {
        searchResults(results, this.lt);    
    }
});

回答by Pekka

If you want to use locationType(whose value is 3) in the callback, simply use

如果您想在回调中使用locationType（其值为3），只需使用

function(results) { .....

thanks to closures, locationTypewill be automatically available in the callback.

多亏了closures，locationType将在回调中自动可用。

回答by user6450793

Could try:

可以试试：

function getResults(locationType) {
    $.getJSON(url, {p1:'xxx', p2: 'yyy'}, function(results) {
        searchResults(results, locationType)
    });
}