Using ES2017 you should have this as the function declaration.
async function foo() { var response = await $.ajax({url: '...'}) return response;}And executing it like this.
(async function() { try { var result = await foo() console.log(result) } catch (e) {}})()Or the Promise syntax.
foo().then(response => { console.log(response)}).catch(error => { console.log(error)})Stack Snippet that demonstrates the code above.
// The function declaration:async function foo() { var response = await $.ajax({ url: 'https://jsonplaceholder.typicode.com/todos/1' }) return response;}// Execute it like this:(async function() { try { var result = await foo() console.log(result) } catch (e) {}})()// Or use Promise syntax:foo().then(response => { console.log(response)}).catch(error => { console.log(error)}).as-console-wrapper { max-height: 100% !important; top: 0; }<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>