I have three HTTP calls that need I need to make in a synchronous manner and how do I pass data from one call to the other?
function first()
{
ajax()
}
function second()
{
ajax()
}
function third()
{
ajax()
}
function main()
{
first().then(second).then(third)
}
I tried to use the deferred for the two functions and I came up with a partial solution. Can I extend it to be for three functions?
function first() {
var deferred = $.Deferred();
$.ajax({
"success": function (resp)
{
deferred.resolve(resp);
},
});
return deferred.promise();
}
function second(foo) {
$.ajax({
"success": function (resp)
{
},
"error": function (resp)
{
}
});
}
first().then(function(foo){second(foo)})
In each case, return the jqXHR object returned by $.ajax()
.
These objects are Promise-compatible so can be chained with .then()
/.done()
/.fail()
/.always()
.
.then()
is the one you want in this case, exactly as in the question.
function first() {
return $.ajax(...);
}
function second(data, textStatus, jqXHR) {
return $.ajax(...);
}
function third(data, textStatus, jqXHR) {
return $.ajax(...);
}
function main() {
first().then(second).then(third);
}
Arguments data
, textStatus
and jqXHR
arise from the $.ajax()
call in the previous function, ie. first()
feeds second()
and second()
feeds third()
.
DEMO (with $.when('foo')
to deliver a fulfilled promise, in place of $.ajax(...)
).