我用express 3在node.js中创建了一个文件上传函数。
我想抓取图像的文件扩展名。所以我可以重命名文件，然后附加文件扩展名。
app.post('/upload', function(req, res, next) {
var is = fs.createReadStream(req.files.upload.path),
fileExt = '', // I want to get the extension of the image here
os = fs.createWriteStream('public/images/users/' + req.session.adress + '.' + fileExt);
});
如何在node.js中获得图像的扩展名?
当前回答
const path = require('path');
function getExt(str) {
const basename = path
.basename(str)
// Patch: for hidden files
// Removes all dots at the beginning of a line
.replace(/^(\.+)/i, '');
const firstDot = basename.indexOf('.');
const lastDot = basename.lastIndexOf('.');
const extname = path.extname(basename).replace(/(\.[a-z0-9]+).*/i, '$1');
if (firstDot === lastDot) {
return extname;
}
return basename.slice(firstDot, lastDot) + extname;
}
const files = [
'/home/charlike/bar/.hidden.tar.gz~', // ".tar.gz"
'/home/charlike/bar/file.tar.gz~', // ".tar.gz"
'/home/charlike/bar/file.tar.gz+cdf2', // ".tar.gz"
'/home/charlike/bar/file.tar.gz?quz=zaz', // ".tar.gz"
];
const fileAndExt = files.map((file) => [ file, getExt(file) ]);
console.log(JSON.stringify(fileAndExt, null, 2));
其他回答
使用substr()方法比使用split() & pop()方法更有效
在这里看看性能差异:http://jsperf.com/remove-first-character-from-string
// returns: 'html'
var path = require('path');
path.extname('index.html').substr(1);
Update August 2019 As pointed out by @xentek in the comments; substr() is now considered a legacy function (MDN documentation). You can use substring() instead. The difference between substr() and substring() is that the second argument of substr() is the maximum length to return while the second argument of substring() is the index to stop at (without including that character). Also, substr() accepts negative start positions to be used as an offset from the end of the string while substring() does not.
更新
由于原来的答案，extname()已添加到路径模块，请参阅Snowfish答案
最初的回答:
我使用这个函数来获得文件扩展名，因为我没有找到一种更简单的方法(但我认为有):
function getExtension(filename) {
var ext = path.extname(filename||'').split('.');
return ext[ext.length - 1];
}
你必须要求'path'才能使用它。
另一个不使用path模块的方法:
function getExtension(filename) {
var i = filename.lastIndexOf('.');
return (i < 0) ? '' : filename.substr(i);
}
例如，您可以使用path.parse(path)
const path = require('path');
const { ext } = path.parse('/home/user/dir/file.txt');
这个解决方案支持查询字符串!
var Url = require('url');
var Path = require('path');
var url = 'http://i.imgur.com/Mvv4bx8.jpg?querystring=true';
var result = Path.extname(Url.parse(url).pathname); // '.jpg'
我相信您可以执行以下操作来获取文件名的扩展名。
var path = require('path')
path.extname('index.html')
// returns
'.html'
如果你想获得一个文件名的所有扩展名(例如filename.css.gz => css.gz)，试试这个:
const ext = 'filename.css.gz'
.split('.')
.filter(Boolean) // removes empty extensions (e.g. `filename...txt`)
.slice(1)
.join('.')
console.log(ext) // prints 'css.gz'
