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Start your free trialMichael Baker
7,731 Pointstask 2...
What am I doing wrong here?
var $required = $(".required"); var values = $required.map(function(){ return $(this).val(); return $.inArray("Andrew",values) != -1; });
2 Answers
Jacob Miller
12,466 PointsjQuery will not execute any code that follows a return
, so you need to move $.inArray("Andrew",values) != -1;
outside of the function and onto the next line.
Michael Baker
7,731 PointsThanks!