选中所有必需的单选按钮时启用按钮

我是这样做的:

$(document).ready(function () {
    var $form = $("form"),
        $button = $form.find("button"),
        $radios = $form.find('input[type="radio"]'),
        $requiredRadios = $radios.filter('[data-req="1"]');
    $requiredRadios.on("click", function () {
        var numValid = 0;
        $requiredRadios.each(function () {
            var $this = $(this),
                myName = $this.attr("name"),
                isYes = ($this.val() === "1"),
                $target = (isYes ? $this : $radios.filter('input[name="' + myName + '"][value="1"]'));
            if ($target.prop("checked")) {
                numValid++;
            }
        });
        if (numValid === $requiredRadios.length) {
            $button.prop("disabled", false);
        } else {
            $button.prop("disabled", true);
        }
    });
});

http://jsfiddle.net/aCheQ/1/

它找到所有必需的单选按钮，并为每个按钮绑定一个click处理程序。当单击其中一个按钮时，将遍历所有必需的单选按钮，并计算是否单击了足够的"Yes"单选按钮。如果是，则启用该按钮，否则禁用该按钮。我相信它可以稍微清理一下，或者更有效率，但它似乎是人们所期望的。

$(function() {
    $("input[type='radio']").click(function(){
        $requires = $('input:radio[data-req=1][value=1]');
        var checked = 0;
        $requires.each(function () {
            if($(this)[0].checked)
                $("button").attr("disabled", (++checked < $requires.length));
        });
    });
});

jsFiddle代码

只需添加以下javascript，代码注释中有解释

$('input[type="radio"]').change(function ()
{
   var req,reqYes;
   req=$('input[type="radio"][data-req="1"][value="1"]').length;
   reqYes=$('input:checked[type="radio"][data-req="1"][value="1"]').length;
   document.getElementsByName("btn")[0].disabled= (req!=reqYes);
   return true;
 }
);
/* - Is onchange because it happens after the click once the radiobutton state has been changed 
   - Is getElementsByName()[0] because you used the name instead the ID (The id is unique, but not the name so it returns an array of objects)
   - No need to do a foreach since you can check the Radiosbutton required and the Required with yes answer, the foreach is done by Jquery
**/

此处fiddle here http://jsfiddle.net/egorbatik/pVh4U/

如果每个组只包含yes/no，那么您只需检查以确保选中的yes无线电的数量是所需无线电总数的一半

$(function () {
    var req = $('input[data-req=1]');
    req.change(function () {
        // if the checked radios with value=1 amount == half of the total required (because only Yes or No can be checked)
        var allchecked = req.filter('[value=1]:checked').length == req.length/2
        // disable of not all checked - enable if all checked
        $('button').prop('disabled', !allchecked);
    });
});

小提琴

如果您想检查组的数量，您可以循环并获得不同的输入名称

$(function () {
    var req = $('input[data-req=1]');
    var groups = [];
    req.each(function(i,v){
        // if group name not in array add it
        if($.inArray(v.name,groups) == -1){
            groups.push(v.name);
        }
    });
    req.change(function () {
        // if the checked amount == number of groups
        var allchecked = req.filter('[value=1]:checked').length == groups.length;
        // disable of not all checked - enable if all checked
        $('button').prop('disabled', !allchecked);
    });
});

小提琴

想想这个——你也可以根据data-req=1和value=1的元素数量来检查选中YES的数量

$(function () {
    var req = $('input[data-req=1]');
    req.change(function () {        
        var allchecked = req.filter('[value=1]:checked').length == req.filter('[value=1]').length;
        $('button').prop('disabled', !allchecked);
    });
});

小提琴

完全测试代码和工作

    $("input[type='radio']").click(function(){
    var enableButton = false;
    var totalChecked = 0;
  var reqElements  =  $('input:radio[data-req =1]');
    var reqElementsCount= $("input:radio[data-req =1][value=1]").size();
    $(reqElements).each(function (index, ele) {
        if($(ele).is(":checked") && ele.value === "1"){
           totalChecked++;
        }
    });
    if(totalChecked == reqElementsCount){ 
        $("button").removeAttr("disabled");
    }
    else{
         $("button").attr("disabled",true);
    }
});

demo at JSFiddle