How to change jquery ajax to synchronization

In jquery, ajax can set async to false to make it synchronized; by default, ajax in jquery is an asynchronous request, that is, "async:true", by setting the parameter "asycn:false", The syntax is "$.ajax({async: false})".

The operating environment of this article: Windows 10 system, jquery version 3.6.1, Dell G3 computer.

How to change jquery ajax to synchronization

Ajax in jquery is an asynchronous request by default, that is, async:true, you can make it synchronous by setting the parameter asycn:false

ajax defaults to an asynchronous request; in ajax, you can determine whether it is an asynchronous request based on the async value. If the value of async is false, it means that the ajax request is synchronous. If the value of async is true, it means that the ajax request is asynchronous. By default, the value of async is "true", so ajax is an asynchronous request by default.

If you want to synchronize async, set it to false (the default is true)

var html = $.ajax({
url: “some.php”,
async: false
}).responseText;

Or set the Ajax attribute globally

$.ajaxSetup({
async: false
});

Then use post, get will be synchronized

The example is as follows:

Randomly generate a 10-digit integer and compare it with the back-end database. If there is this random number in the back-end database number, regenerate one, if not, return this number.

This requirement involves front-end and back-end interaction, so it is inevitable to use ajax, so I wrote such a piece of code at the beginning.

//randID是封装的生成随机数的函数
function userID() {
    let ranid = parseInt(randID(1000000000, 10000000001));
    let data = null;
    $.ajax({
        type: 'post',
        url: './php/findID.php',
        data: 'id=' + ranid,
        success: function(res) {
            res = JSON.parse(res);
            isok = res.length;
            if (isok != 0) {
                userID();
            } else {               
               return ranid;
            }
        }
    })
}
console.log(userID());

Output result:

jquery.ajax solution

• at Declare a local variable under the function (outside ajax)

• Treat ajax as synchronous processing (modification method of jquery.ajax: add this code to async: false)

• Return the declared local variable

function userID() {
    let ranid = parseInt(randID(1000000000, 10000000001));
    //声明的局部变量
    let data = null;
    $.ajax({
        type: 'post',
        url: './php/findID.php',
        data: 'id=' + ranid,
        //将ajax改为同步操作
        async: false,
        success: function(res) {
            res = JSON.parse(res);
            isok = res.length;
            if (isok != 0) {
                console.log(ranid);
                userID();
            } else {
                data = ranid;
            }
        }
    })
    //返回这个局部变量
    return data;
}
console.log(userID());

Output result:

Related tutorial recommendations: jQuery video tutorial

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