javascript - Return function multiple times instead of one time using .fadeOut(); -

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i don't know if suppoused this, guess is...

i thought @ first may wrong whole script , managed make new file on localhost , test fadeout(); function.

apparently returned function twice again went jsfiddle check happen there. , same thing happened. in case console.log(); returned twice.

  • what did , trying ?

well want return specified function, or in fiddle sample, specified console.log(); once. fadingout multiple elements (two, exact).

is there way that, instead of duplicating each element fadeout @ same time ?

sample return console.log(); twice.

settimeout(function () {      $( ".one, .two" ).fadeout(300, function () {         console.log("return function!");     });  }, 2000);

sample return console.log(); once.

settimeout(function () {      $( ".one" ).fadeout(300);      $( ".two" ).fadeout(300, function () {         console.log("return function!");     });  }, 2000);

fiddle preview: fiddle redirect

there 2 elements in collection, fadeout called twice, yes, it's supposed that.

you can have callback fire once regardless of number of elements in collection using returned promise animation , $.when , $.then

settimeout(function () {      $.when( $( ".one, .two" ).fadeout(300) ).then(function () {         console.log("return function!");     });  }, 2000);

fiddle


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