jquery return ajax success value

Asked May 05 '20 at 07:51

Active May 05 '20 at 08:04

Viewed 26 times

if ajax is success, then i have to return that value and use it outside function like this :

let result;

function getGameNameFromWikiApi(searchParams) {
    $.ajax({
        type: "GET",
        url: "https://ko.wikipedia.org/w/api.php?action=query&format=json&list=search&srprop=sectiontitle&srlimit=1",
        data: {
            srsearch: searchParams
        },
        dataType: "jsonp",
        jsonpCallback: "callbackData",
    }).done(function (result) {
        result = result.query.search[0].title
    }).fail(function (error) {
        console.log(JSON.stringify(error));
    });
}

console.log(result); // undefined

getGameNameFromWikiApi('something')

unfortunately, result return always undefined.

how can i get that success value?

edited May 05 '20 at 07:59

asked May 05 '20 at 07:51

did you call the getGameNameFromWikiApi funtion? – Kurohige May 05 '20 at 07:55
yes i call it ! – May 05 '20 at 07:59
if you want to print it then you need to add async:false in ajax variable – Mohanlal Prajapati May 05 '20 at 08:01
1

Does this answer your question? [Why is my variable unaltered after I modify it inside of a function? - Asynchronous code reference](https://stackoverflow.com/questions/23667086/why-is-my-variable-unaltered-after-i-modify-it-inside-of-a-function-asynchron) – Robert Mennell May 05 '20 at 08:03
2

@MohanPrajapati `async: false` is very rarely an acceptable solution, and it absolutely is not one in this situation. – Rory McCrossan May 05 '20 at 08:04

jquery return ajax success value

0 Answers0