javascript - jQuery: avoid code duplication because of fadeOut() -


here's problem i've faced many times:

  • i have element may or may not visible
  • i need stuff element
  • if element visible, fadeout() element
  • once stuff done: fadein() element

the problem here: code looks this:

function showonlyelement(myel) {     $('body')         .children('div:not(.'+myel+')')         .fadeout()         .promise().done(function() {             var el=b.children('div.'+myel);             if (el.is(':visible')) {                 /* have hide before modifying */                 el.fadeout(function() {                     /* long code (a) modifying innerhtml of el */                     el.fadein();                 });             } else {                 /* again same long code (a) modifying innerhtml of el */                 el.fadein();             }         }); }

i want make clean there's not repetition of same long code (a)

how do (generic way of doing this)?

simple generic solution :

$('body')     .children('div:not(.'+myel+')')     .fadeout()     .promise().done(function() {         var el=b.children('div.'+myel);         function longcode(){             /* long code (a) modifying innerhtml of el */             el.fadein();         }         if (el.is(':visible')) {             el.fadeout(longcode);         } else {             longcode();         }     });

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