javascript 删除完整路径，只保留文件名
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remove full path, keep filename only
提问by user1219691
Trying to remove the full url that is being returned to imgurl: Usually returns something like http://localhost/wordpress/wp-content/uploads/filename.jpgor http://localhost/wordpress/wp-content/uploads/images/filename.jpg
试图删除返回到 imgurl 的完整 url：通常返回类似http://localhost/wordpress/wp-content/uploads/filename.jpg或http://localhost/wordpress/wp-content/uploads/images 的内容/文件名.jpg
I'd like to strip off everything except filename.jpg and return it to ahng_photos_upload_image. Strip off everything to the last forward-slash. How can I do that with Jquery?
我想剥离除 filename.jpg 之外的所有内容并将其返回给 ahng_photos_upload_image。将所有内容剥离到最后一个正斜杠。我怎样才能用 Jquery 做到这一点？
window.send_to_editor = function(html) {
imgurl = jQuery('img',html).attr('src');
jQuery('#ahng_photos_upload_image').val(imgurl);
tb_remove();
}
回答by Andy
You don't need jQuery for that, just plain old JavaScript will do :)
你不需要 jQuery，只需要普通的旧 JavaScript 就可以了 :)
alert('http://localhost/wordpress/wp-content/uploads/filename.jpg'.split('/').pop());??
In your case:
在你的情况下：
var filename = imgurl.split('/').pop();
回答by Vivek Chandra
you can use a regular expression in order to achieve this..
您可以使用正则表达式来实现这一点..
var file = imgUrl.replace(/^.*[\\/]/, '');
Now the file would consist of only the file name ..
现在文件将只包含文件名 ..
回答by Pointy
If you're pretty confident that the URLs don't have funny stuff like hashes or parameters, a regex like this would do it:
如果您非常确信 URL 没有诸如哈希或参数之类的有趣内容，则可以使用这样的正则表达式：
var filename = imgurl.replace(/^.*\/([^/]*)$/, "");
Also: don't forget to declare "imgurl" with var, and you should probablyuse .prop()instead of .attr()if your version of jQuery is 1.6 or newer:
另外：不要忘记用 声明“imgurl” var，如果您的 jQuery 版本是 1.6 或更高版本，您可能应该使用.prop()而不是.attr()：
var imgurl = jQuery('img', html).prop('src');
Also jQuery internally turns the two-argument form of the function into this:
此外，jQuery 在内部将函数的两个参数形式转换为：
var imgurl = jQuery(html).find('img').prop('src');
so you might as well code it that way.
所以你也可以这样编码。
回答by David says reinstate Monica
One further option:
另一种选择：
var filename = imgurl.substring(imgurl.lastIndexOf('/') + 1);
回答by sdw
Note that if you don't know if you have forward or backward slashes, you are better off using the RE version of split:
请注意，如果您不知道是否有正斜杠或反斜杠，最好使用 RE 版本的 split：
"path".split(/[\/\]/).slice(-1)
回答by Ari
Here is an answer that will work when your file name is like ./file.jpg
这是一个当您的文件名类似于 ./file.jpg 时有效的答案
var extension = fileName.slice((fileName.lastIndexOf(".") - 1 >>> 0) + 2);
var baseName = fileName.replace(/^.*\/([^/]*)$/, "");
var path = fileName.replace(/(^.*\/)([^/]*)$/, "");
回答by piouPiouM
Try this one:
试试这个：
imgurl.split('/').slice(-1);
Edit:Look at the version of @Andywho uses the pop()method, the latter being faster than slice(-1).
编辑：查看使用该方法的@Andy 的版本pop()，后者比slice(-1).
回答by Luis
Here you have
在这里你有
var filename = imgurl.split('/').slice(-1);
Good luck!
祝你好运！
