简化这个jQuery代码？ [英] Simplify this jQuery Code?

问题描述

我很确定有一种方法可以缩短它，但我不知道如何。

I'm pretty sure there is a way to shorten this, but I don't know how.

img[1] = $('#img1').attr('src');
img[2] = $('#img2').attr('src');
img[3] = $('#img3').attr('src');
img[4] = $('#img4').attr('src');

title[1] = $('#title1').text();
title[2] = $('#title2').text();
title[3] = $('#title3').text();
title[4] = $('#title4').text();

desc[1] = $('#description1').attr('value');
desc[2] = $('#description2').attr('value');
desc[3] = $('#description3').attr('value');
desc[4] = $('#description4').attr('value');

url[1] = $('#url1').attr('value');
url[2] = $('#url2').attr('value');
url[3] = $('#url3').attr('value');
url[4] = $('#url4').attr('value');

已经创建了数组，我只是决定将它们排除在外，因为这里不需要它。我只是基本上把一些数据拉到了DOM。

The arrays have already be created, i just decided to exclude them from this becuase it's not really needed here. I'm just basically pulling some data the DOM.

推荐答案

for(i=1;i<=4;i++) {
   img[i] = $('#img'+i).attr('src');
   title[i] = $('#title'+i).text();
   desc[i] = $('#description'+i).attr('value');
   url[i] = $('#url'+i).attr('value');
}

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