More efficient way to write this Jquery code

Posted by adamwstl on Stack Overflow See other posts from Stack Overflow or by adamwstl
Published on 2010-04-20T21:10:29Z Indexed on 2010/04/20 21:13 UTC
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Is there a better, more efficient way to write this code? It's a make shift drop down menu that allows user to RSVP for multiple people. Sorry, it's kind of a mess, but I think what I'm doing is clear. If not, I'm at my computer and will respond quickly with more info need be.

 //There's got to be a better way to do this
 $('#guest_num_1').click( function() {
 $('#num_guests a#quant_guests').html("1")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6 ').hide()
 });

 $('#guest_num_2').click( function() {
 $('#num_guests a#quant_guests').html("2")
 $('.guest_name_2').fadeIn()
 $('.guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_3').click( function() {
 $('#num_guests a#quant_guests').html("3")
 $('.guest_name_2, .guest_name_3').fadeIn()
 $('.guest_name_4, .guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_4').click( function() {
 $('#num_guests a#quant_guests').html("4")
 $('.guest_name_2, .guest_name_3, .guest_name_4').fadeIn()
 $('.guest_name_5, .guest_name_6').hide()
 });

 $('#guest_num_5').click( function() {
 $('#num_guests a#quant_guests').html("5")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5').fadeIn()
 $('.guest_name_6').hide()
 });

 $('#guest_num_6').click( function() {
 $('#num_guests a#quant_guests').html("6")
 $('.guest_name_2, .guest_name_3, .guest_name_4, .guest_name_5, .guest_name_6').fadeIn()
 });

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