Chain multiple “then” in jQuery.when

老子叫甜甜 提交于 2019-11-28 16:22:44
Kevin B

Use an anonymous function inside of .then and pass the parameters that you want to pass. I'm replacing .then with .done because you don't need .then in this case.

function do_something() {
    // some code

    return $.when(foo, bar, baz).done(function(_foo_2, _bar_2, _baz_2){
        do_something_else.apply(this,_foo_2);
    });
}

.then actually creates a new deferred object and sends that to the chain. Since you didn't return anything from .then, the new deferred object has no arguments. See this example:

$.when($.Deferred().resolve(2), $.Deferred().resolve(4))
.then(function(a,b) { 
    console.log(a,b); // 2,4
    return $.Deferred().resolve(a,b,6);
}).then(function(a,b,c) { 
    console.log(a,b,c); // 2,4,6
});

If you instead just used .done, it would work as expected.

$.when($.Deferred().resolve(2), $.Deferred().resolve(4))
.done(function(a,b) { 
    console.log(a,b);
}).done(function(a,b) { 
    console.log(a,b);
});

The most common use for .then is chaining ajax requests:

$.ajax({...}).then(function(){
    return $.ajax({...});
}).then(function(){
    return $.ajax({...});
}).then(function(){
    return $.ajax({...});
}).then(function(){
    return $.ajax({...});
});

which can also be easily done in a loop. Each .then will have access to the returned data from the previous request.

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