Q.all promise永远不会运行然后回调

Question

I am bit lost as to how Q.all works. I have the following code snippet ( jsfiddle ):

function callUrl(remoteEndPoint){
  return $.ajax({
    url: remoteEndPoint,
    method: "GET",
    dataType: "json",
    contentType: "application/json; charset=utf-8",
    success: function (data) {
      console.dir(data);
    },
    error: function (xhr, textStatus, errorThrown) {
            console.error("error");
    }
  });
}
Q.all([callUrl("https://httpbin.org/get"), callUrl("https://httpbin.org/undefined")]).then(function(){
    console.log("Done");
});

The first call to callUrl will succeed, the second will fail (will return an error as the url does not exist). I am expecting the Q.all to execute the then callback like what ajax.always does but this isn't the case.

Am I doing something wrong?

Answer 1

You need to chain a catch method and callback for that, because Q.all will return a rejected promise when it encounters a rejected promise.

You get called back concerning a rejected promise via catch (or the optional second callback you can pass to then ).

Answer 2

The 404 is an error that has propogated from the server, into your JS application.

use .catch

Q.all([callUrl("https://httpbin.org/get"), callUrl("https://httpbin.org/undefined")]).then(function(){
    console.log("Done");
}).catch(function(e){
    console.log("err",e);
});