如何将参数传递给 jQuery $.getJSON 回调方法? [英] How can I pass parameters to a jQuery $.getJSON callback method?

问题描述

我正在尝试使用 jQuery 通过 Ajax/$.getJSON 调用一些自定义 API.

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

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

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:

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

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

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

locationType 在我使用 AJAX 调用 URL 之前的值是 3.但是在调用成功返回数据后，locationType 的值现在是 success.这是因为回调的方法签名是:

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

callback(data, textStatus)一个回调执行的函数，如果请求成功.

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

如何将 1 个或多个参数传递给回调方法?

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

推荐答案

不需要传入，直接引用你已有的变量，像这样:

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

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

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

Also there's no need to pass null if 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:

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

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