And*_*y E 12
jQuery只是一个JavaScript库,所以你可以使用这个console.log()函数:
console.log(myVar);
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(function($) {
$.log = function() {
if( 'console' in window )
console.log.apply(null, arguments);
};
}(jQuery));
$.log("hello world, I'm just kidding with this answer, please upvote it anyway!");
$.log("You should really just call console.log() directly");
$.log("Over and out!");
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