看代码:

var file1 = "50.xsl";
var file2 = "30.doc";
getFileExtension(file1); //returns xsl
getFileExtension(file2); //returns doc

function getFileExtension(filename) {
    /*TODO*/
}

当前回答

以下解决方案足够快速和简短,可以用于批量操作并节省额外的字节:

 return fname.slice((fname.lastIndexOf(".") - 1 >>> 0) + 2);

下面是另一个单行非regexp通用解决方案:

 return fname.slice((Math.max(0, fname.lastIndexOf(".")) || Infinity) + 1);

对于没有扩展名(例如myfile)或以开头的名称,两者都可以正常工作。点(例如。htaccess):

 ""                            -->   ""
 "name"                        -->   ""
 "name.txt"                    -->   "txt"
 ".htpasswd"                   -->   ""
 "name.with.many.dots.myext"   -->   "myext"

如果你关心速度,你可以运行基准测试,检查提供的解决方案是否是最快的,而简短的解决方案是非常快的:

短的是如何工作的:

String.lastIndexOf method returns the last position of the substring (i.e. ".") in the given string (i.e. fname). If the substring is not found method returns -1. The "unacceptable" positions of dot in the filename are -1 and 0, which respectively refer to names with no extension (e.g. "name") and to names that start with dot (e.g. ".htaccess"). Zero-fill right shift operator (>>>) if used with zero affects negative numbers transforming -1 to 4294967295 and -2 to 4294967294, which is useful for remaining the filename unchanged in the edge cases (sort of a trick here). String.prototype.slice extracts the part of the filename from the position that was calculated as described. If the position number is more than the length of the string method returns "".


如果你想要更清晰的解决方案,将以同样的方式工作(加上额外的全路径支持),请检查以下扩展版本。这个解决方案比以前的一行程序要慢,但是更容易理解。

function getExtension(path) {
    var basename = path.split(/[\\/]/).pop(),  // extract file name from full path ...
                                               // (supports `\\` and `/` separators)
        pos = basename.lastIndexOf(".");       // get last position of `.`

    if (basename === "" || pos < 1)            // if file name is empty or ...
        return "";                             //  `.` not found (-1) or comes first (0)

    return basename.slice(pos + 1);            // extract extension ignoring `.`
}

console.log( getExtension("/path/to/file.ext") );
// >> "ext"

这三种变体都可以在客户端的任何web浏览器中工作,也可以在服务器端NodeJS代码中使用。

其他回答

fetchFileExtention(fileName) {
    return fileName.slice((fileName.lastIndexOf(".") - 1 >>> 0) + 2);
}
function extension(fname) {
  var pos = fname.lastIndexOf(".");
  var strlen = fname.length;
  if (pos != -1 && strlen != pos + 1) {
    var ext = fname.split(".");
    var len = ext.length;
    var extension = ext[len - 1].toLowerCase();
  } else {
    extension = "No extension found";
  }
  return extension;
}

/ /使用

扩展(“file.jpeg”)

总是返回扩展低cas,以便您可以检查它的字段更改 适用于:

file.JpEg

文件(无扩展名)

文件。(noextension)

Wallacer的回答很好,但还需要再检查一下。

如果文件没有扩展名,它将使用filename作为扩展名,这是不好的。

试试这个:

return ( filename.indexOf('.') > 0 ) ? filename.split('.').pop().toLowerCase() : 'undefined';
var extension = fileName.substring(fileName.lastIndexOf('.')+1);
function func() {
  var val = document.frm.filename.value;
  var arr = val.split(".");
  alert(arr[arr.length - 1]);
  var arr1 = val.split("\\");
  alert(arr1[arr1.length - 2]);
  if (arr[1] == "gif" || arr[1] == "bmp" || arr[1] == "jpeg") {
    alert("this is an image file ");
  } else {
    alert("this is not an image file");
  }
}