Question

我拿着一系列不同孩子的形式。每个人都在那里有一个检查箱。我试图做的是,如果检查其中的检查箱,就删除其中的一种表格。

<form id="placement-form">
</form>
<form id="placement-form1">
</form>
<form id="placement-form2">
</form>

我写了在“Qu”中的短功能,在理论上,如果检查箱是“检查”的,就应当删除表格,但我可以让它工作。 j 详细情况如下:

$(function() {
    $( .deleteSelected ).click(function() { //fire when the button is clicked
        $( #placement-form:checkbox:checked ).each(function() {
            $( #placement-form ).remove();                
        });
    });
});

It s probably a little muddled but I m still getting to grips with jQuery with a combination of the documentation and a bit of Googling..!

EDIT:尽管我刚才已经实现了这一文字,但现在只有一种形式(安插版)的Im。

Answer 1

审判

$(function() {
    $( .deleteSelected ).click(function() { //fire when the button is clicked
        $( form input:checkbox ).each(function() {
          var checkbox = $(this);  
          if(checkbox.is( :checked )) checkbox.parents( form ).remove();        
        });
    });
});

http://www.un.org。方便

$( .deleteSelected ).click(function(){
    $( form input:checked ).parents( form ).remove();
});

http://jsfiddle.net

Answer 2

$( #placement-form ) will find only an element with that ID, not others. You should write something like:

$( .deleteSelected ).click(function() {
    $(this).parents( form ).remove();
});

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