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
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);
}
一个简单的解决方案，不需要require，解决了多个周期的扩展问题:
var filename = 'file.with.long.extension';
var ext = filename.substring(filename.indexOf('.'));
//ext = '.with.long.extension'
或者如果你不想要前导点:
var filename = 'file.with.long.extension';
var ext = filename.substring(filename.indexOf('.')+1);
//ext = 'with.long.extension'
确保测试文件也有扩展名。
下面的函数分割字符串并返回名称和扩展名，无论扩展名中有多少个点。如果没有扩展名，则返回空字符串。以点和/或空格开头的名字也可以。
function basext(name) {
name = name.trim()
const match = name.match(/^(\.+)/)
let prefix = ''
if (match) {
prefix = match[0]
name = name.replace(prefix, '')
}
const index = name.indexOf('.')
const ext = name.substring(index + 1)
const base = name.substring(0, index) || ext
return [prefix + base, base === ext ? '' : ext]
}
const [base, ext] = basext('hello.txt')
// you can send full url here
function getExtension(filename) {
return filename.split('.').pop();
}
如果您正在使用express，请在配置中间件时添加以下行(bodyParser)
app.use(express.bodyParser({ keepExtensions: true}));
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.
