Jquery - How can I return correct value from function with multiple ajax requests?

AlperUluses
June 13, 2023
198 views
2 votes
3 Answers

I have a function and its have nested two ajax call. I want to return a value inside from second ajax call when ajax call is successed. i think because function ending but the ajaxs call still not end. But i didn’t solve it.

I prepared a example code below:

        
    var patato = function(){
            $.ajax({
                type: 'POST',
                url: 'https://jsonplaceholder.typicode.com/posts',
                dataType: 'json',
                data: {
                    title: 'foo',
                    body: 'bar',
                    userId: 1,
                },
                success:function(res){
                    console.log("Work 1")
                    $.ajax({
                        type: 'POST',
                        url: 'https://jsonplaceholder.typicode.com/posts',
                        dataType: 'json',
                        data: {
                            title: 'foo',
                            body: 'bar',
                            userId: 1,
                        },
                        success:function(res){
                            console.log("Work 2")
                            return true;
                        }
                    })
                }
            })
            console.log("Çalıştı 3")
        }
        var patatos = patato();
        if(patatos) {
            console.log("Patato true")
        }else{
            console.log("Patato false")
        }

<script src="https://code.jquery.com/jquery-3.7.0.min.js" integrity="sha256-2Pmvv0kuTBOenSvLm6bvfBSSHrUJ+3A7x6P5Ebd07/g=" crossorigin="anonymous"></script>

Answers

Your function should return a promise since you are doing some async work (fetching).

Also I strongly recommend to use async/await to simplify logic and improve readability/maintainability:

async function patato() {

    console.log("Çalıştı 3")

    const res = await $.ajax({
        type: 'POST',
        url: 'https://jsonplaceholder.typicode.com/posts',
        dataType: 'json',
        data: {
            title: 'foo',
            body: 'bar',
            userId: 1,
        }
    });

    console.log("Work 1")

    //use res if needed...

    const res2 = await $.ajax({
        type: 'POST',
        url: 'https://jsonplaceholder.typicode.com/posts',
        dataType: 'json',
        data: {
            title: 'foo',
            body: 'bar',
            userId: 1,
        },
    });

    console.log("Work 2")

    //use res2 if needed...

    return true;

}

patato().then(patato => console.log(`Patato ${patato}`));

<script src="https://code.jquery.com/jquery-3.7.0.min.js" integrity="sha256-2Pmvv0kuTBOenSvLm6bvfBSSHrUJ+3A7x6P5Ebd07/g=" crossorigin="anonymous"></script>

First thing that I wanted to mention is that it will not make much sense to make sequential requests if you second request don’t need the response from the first one. On that note, I will assume you have a use case that requires two API call executed one after the other.

Here you have two options: use callbacks or async/await syntax.
Here you can find a good example of the usage of both: How to return an Ajax result using async/await?

For your particular example here’s how you could do this with callbacks:

        
    const patato = function(){
            return $.ajax({
                type: 'POST',
                url: 'https://jsonplaceholder.typicode.com/posts',
                dataType: 'json',
                data: {
                    title: 'foo',
                    body: 'bar',
                    userId: 1,
                }
            }).then(function(res1){ 
                console.log("Work 1")
                    return $.ajax({
                        type: 'POST',
                        url: 'https://jsonplaceholder.typicode.com/posts',
                        dataType: 'json',
                        data: {
                            title: 'foo',
                            body: 'bar',
                            userId: 1,
                        }
                    }).then(function(res2){
                        console.log("Work 2")
                        return res1 && res2;
                    })
            })
        }
        patato().then(patatos => {
          if(patatos) {
              console.log(patatos)
          }else {
              console.log("Patato false")
          }
        })

<script src="https://code.jquery.com/jquery-3.7.0.min.js" integrity="sha256-2Pmvv0kuTBOenSvLm6bvfBSSHrUJ+3A7x6P5Ebd07/g=" crossorigin="anonymous"></script>

Or you could use async/await syntax as well (IMO this is easier to follow than the callback one):

   

    const patato = async function(){
            console.log("Work 1");
            const res1 = await $.ajax({
                type: 'POST',
                url: 'https://jsonplaceholder.typicode.com/posts',
                dataType: 'json',
                data: {
                    title: 'foo',
                    body: 'bar',
                    userId: 1,
                }
            });

            console.log("Work 2");
            const res2 = await $.ajax({
                type: 'POST',
                url: 'https://jsonplaceholder.typicode.com/posts',
                dataType: 'json',
                data: {
                    title: 'foo',
                    body: 'bar',
                    userId: 1,
                }
            });
            return res1 && res2;

        }
        async function test() {
            let result = await patato();
            console.log(result);
        }
        test();

<script src="https://code.jquery.com/jquery-3.7.0.min.js" integrity="sha256-2Pmvv0kuTBOenSvLm6bvfBSSHrUJ+3A7x6P5Ebd07/g=" crossorigin="anonymous"></script>

You can also run both AJAX jobs in parallel if the output of the first job is not required for the second one:

async function patato() {

    console.log("Çalıştı 3")
    await $.when( $.ajax({type:'POST',url: 'https://jsonplaceholder.typicode.com/posts',
                         dataType:'json',data: {title: 'foo1',body: 'bar1',userId: 1}}),
                  $.ajax({type: 'POST',url: 'https://jsonplaceholder.typicode.com/posts',
                         dataType:'json',data: {title: 'foo2',body: 'bar2',userId: 2}})
    ).done(function([r1],[r2]){console.log("Both AJAX jobs are finished.",r1,r2)})
    console.log("End of patato.");
    return true;
}

patato().then(pa => console.log(`Patato returns ${pa}`));

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

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Jquery – How can I return correct value from function with multiple ajax requests?

Answers