Javascript Gulp：如何将文件内容读入变量？
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时间：2020-08-23 03:57:02 来源:igfitidea点击:
Gulp: How do I read file content into a variable?
提问by OpherV
I have a gulp task that needs to read a file into a variable, and then use its content as input for a different function that runs on the files in the pipe. How do I do that?
我有一个 gulp 任务，需要将文件读入变量，然后将其内容用作在管道中的文件上运行的不同函数的输入。我怎么做？
Example psuedo-psuedo-code
示例伪伪代码
gulp.task('doSometing', function() {
var fileContent=getFileContent("path/to/file.something"); //How?
return gulp.src(dirs.src + '/templates/*.html')
.pipe(myFunction(fileContent))
.pipe(gulp.dest('destination/path));
});
回答by OpherV
Thargor pointed me out in the right direction:
Thargor 向我指出了正确的方向：
gulp.task('doSomething', function() {
var fileContent = fs.readFileSync("path/to/file.something", "utf8");
return gulp.src(dirs.src + '/templates/*.html')
.pipe(myFunction(fileContent))
.pipe(gulp.dest('destination/path'));
});
回答by Thargor
Is this what you're looking for?
这是你要找的吗？
fs = require("fs"),
gulp.task('doSometing', function() {
return gulp.src(dirs.src + '/templates/*.html')
.pipe(fs.readFile("path/to/file.something", "utf-8", function(err, _data) {
//do something with your data
}))
.pipe(gulp.dest('destination/path'));
});
