Question

我需要替换一组参数。当我这样做时，第二个参数（dir2）+++的Link1没有被替换（因为在Link1中，两个dir2都有相同的值）+++。我该如何解决这个问题？

＆＃13;

var myArray = ["data/", "downloads/", "images/", "leftover/", "new2me/", "news/"];

$.each(myArray, function(key, value) {

  $('.submenu').each(function() {
    $(this).attr('href', ($(this).attr('href').replace("http://www.nu.nl/", "http://www.telegraaf.nl/").replace(value, 'latestdata/')));
  });

});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<li><a class="submenu" href="test.php?url1=http://www.nu.nl/&dir1=images/&url2=http://www.nu.nl/&dir2=images/" target="showme">Link1</a></li>
<li><a class="submenu" href="test.php?url1=http://www.nu.nl/&dir1=downloads/&url2=http://www.nu.nl/&dir2=new2me/" target="showme">Link2</a></li>
<li><a class="submenu" href="test.php?url1=http://www.nu.nl/&dir1=data/&url2=http://www.nu.nl/&dir2=leftover/" target="showme">Link3</a></li>

＆＃13;

编辑：见上文...... +++

之间

Answer 1

$(document).ready(function(){
var myArray = ["data/", "downloads/", "images/", "leftover/", "new2me/", "news/"];
//alert(myArray[0]);
for(i=0;i <myArray.length;i++){
$.each(myArray, function(key, value) {
$('.submenu').each(function() {
$(this).attr('href', ($(this).attr('href').replace("http://www.nu.nl/", "http://www.telegraaf.nl/").replace(value, 'latestdata/')));
});
});
}
});

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1 个答案: