Question

我在变量中保存了几个div。我想将相同的活动处理程序绑定到一堆div。我该怎么做？

var div1 = $('#selecto');
var div2 = $('#matic');
var div3 = $('#hello');
$(???).live('click', function() {
   console.log('one of those divs was clicked');
});
// I tried (unsuccessfully) replacing ??? with [div1, div2, div3]

Answer 1

由于.live()的工作方式，您无法组合多个jQuery对象，然后调用.live()。您需要一个选择器。所以要么：

$('#selecto, #matic, #hello').live('click', function() {
   console.log('one of those divs was clicked');
});

或预先定义函数并单独附加：

function log() {
   console.log('one of those divs was clicked');
}

div1.live('click', log);
div2.live('click', log);
div3.live('click', log);

Answer 2

为什么不：

$('#selecto, #matic, #hello').live('click', function() {
   console.log('one of those divs was clicked');
});

Answer 3

由于div1已经是一个jQuery对象，你可以做到

div1.live('click', function() {
   console.log('one of those divs was clicked');
});

没有换行是在jQuery选择器$（div1）

中

Answer 4

您不需要live，只需使用click

即可

div1.add(div2, div3).click(function(e){ ... });

如果您不想重写ID，可以这样做：

var div1 = $('#selecto');
var div2 = $('#matic');
var div3 = $('#hello');

var selector = [];

div1.add(div2).add(div3).each(function(i, val) {
    selector.push("#" + val.id);
});

$(selector.join(", ")).live("click", function(e) {
    console.log(this.id + ' was clicked');
});

演示：http://jsfiddle.net/roberkules/bvpNX/

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4 个答案: