在查找结果集上使用jquery find

Question

When I do a find look-up on my table for hidden fields, I am seeing my two hidden fields. However, I want to further refine these 2 fields by their IDs. I notice that when I use find on the entire table using the "contains" that I get my 2 fields. However, if I do a find on the find results from the hidden fields, it returns an empty set. Can anyone explain why this is the case?

    var table = sender.parentNode.parentNode.parentNode.parentNode;
     // this finds my 2 hidden fields
     var hidden_fields = $(table).find("input[type='hidden']");
     // this finds each of the 2 fields individually by ID
     var my_id_fieldA = $(table).find("[id*='hfMyIdFieldA']");
     var my_id_fieldB = $(table).find("[id*='hfMyIdFieldB']");

     // but this returns an empty set
     var my_id_fieldA = $(hidden_fields).find("[id*='hfMyIdFieldA']");

Answer 1

You're looking for the filter function, not find . find selects child elements while filter filters the current selection.

Also there's no reason to find the table like that... try something like this.

var $table = $(sender).closest("table")
  , $hidden_fields = $table.find("input[type='hidden']")
  , $my_id_fieldA = $hidden_fields.filter("[id*='hfMyIdFieldA']")
  , $my_id_fieldB = $hidden_fields.filter("[id*='hfMyIdFieldB']")
  , $my_id_fieldA = $hidden_fields.filter("[id*='hfMyIdFieldA']")
  ;