JQUERY：获取已选中和未选中复选框的 ID
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JQUERY: get the id's of the checked and not checked checkBox
提问by jayAnn
I have this not so many checkboxes:
我没有这么多复选框：
<input type="checkbox" id="mango" value="mango" /> <label>MANGO</label><br>
<input type="checkbox" id="santol" value="santol" /> <label>SANTOL</label><br>
<input type="checkbox" id="guava" value="guava" /> <label>GUAVA</label><br>
<input type="checkbox" id="lomboy" value="lomboy" /> <label>LOMBOY</label><br>
<input type="checkbox" id="apple" value="apple" /> <label>APPLE</label><br>
<input type="checkbox" id="orange" value="orange" /> <label>ORANGE</label><br>
...................<>
Now, what I'm trying to do is get all the ID's of check boxes with check, and no check and put in an array. Something like this:
现在，我正在尝试做的是获取复选框的所有 ID，并且不选中并放入数组中。像这样的东西：
"fruitsGranted":["apple","lomboy","orange"]; //check is true
"fruitsDenied":["mango","santol","guava"]; //check false
can please someone show me how to do it.? thanks.
请有人告诉我该怎么做。？谢谢。
回答by James Montagne
var someObj={};
someObj.fruitsGranted=[];
someObj.fruitsDenied=[];
$("input:checkbox").each(function(){
var $this = $(this);
if($this.is(":checked")){
someObj.fruitsGranted.push($this.attr("id"));
}else{
someObj.fruitsDenied.push($this.attr("id"));
}
});
回答by Mo Valipour
I think I got a shorter version here:
我想我在这里得到了一个较短的版本：
var idSelector = function() { return this.id; };
var fruitsGranted = $(":checkbox:checked").map(idSelector).get();
var fruitsDenied = $(":checkbox:not(:checked)").map(idSelector).get();
You can see it in action here: http://jsfiddle.net/XPMjK/3/
你可以在这里看到它的实际效果：http: //jsfiddle.net/XPMjK/3/
回答by John Kalberer
I would do this something like this:
我会这样做：
var all, checked, notChecked;
all = $("input:checkbox");
checked = all.filter(":checked");
notChecked = all.not(":checked)");
After that you can use jQuery.map to get the ids of each collection.
之后，您可以使用 jQuery.map 来获取每个集合的 id。
var checkedIds = checked.map(function() {
return this.id;
});
var notCheckedIds = notChecked.map(function() {
return this.id;
});
回答by karim79
This can be done using .map, though in this case it is not really different from using .each:
这可以通过 using 来完成.map，但在这种情况下它与 using 并没有真正的不同.each：
$(":checkbox").change(function() {
var notChecked = [], checked = [];
$(":checkbox").map(function() {
this.checked ? checked.push(this.id) : notChecked.push(this.id);
});
alert("checked: " + checked);
alert("not checked: " + notChecked);
});
