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Nodejs脚本,用于取消注释某些代码行或用文件中的内容替换注释

regex node.js javascript

0

Omar · 技术社区 · 7 年前

我正在寻找一个nodejs脚本,它将取代 IN 和 OUT

/* api_start ====
app.use('/api',require('./api/api'));
app.use('/sms',require('./api/sms'));
==== api_end */

或

// app.use('/api',require('./api/api'));
// app.use('/sms',require('./api/sms'));

出来

app.use('/api',require('./api/api'));
app.use('/sms',require('./api/sms'));

基本上只是取消对这些代码行的注释。

只需替换如下评论, /* api_here */ 包含来自文件的内容,该文件将包含来自在上面( api.js

发件人: https://stackoverflow.com/a/25072070/3066142

// require:
var replace = require("replace");

// use:
replace({
    regex: "//\s+app\.use\('/api',require\('\./api/api'\)\);", // string to be replaced
    replacement: "test worked!!", // replacement string
    paths: ['dist/server.js'], // path/to/your/file'
    recursive: true,
    silent: true,
});

1 回复 | 直到 5 年前

1

Frederic Klein neocanable 7 年前

工作代码笔(包括更多的块注释等): https://codepen.io/Frederic-Klein/pen/KbXxeB?editors=0012

用于使用正则表达式匹配和替换给定示例的示例代码 (...) 和非捕获 (?:...)

const content = `
/* api_start ====
app.use('/api',require('./api/api'));
app.use('/sms',require('./api/sms'));
==== api_end */
// app.use('/api',require('./api/api'));
// app.use('/sms',require('./api/sms'));
`;

const regex_block_comments = /^\/\*.*\n((?:app.*\n)*)(?:.*\*\/\n)/gm;
// match lines
// starting with /*
// having the following lines start with app, capture all those lines as one group ($1 reference)
// having a line ending with */ afterwards, as non-capturing group
// repeat globally (g) and match multiline (m) string

const regex_comment = /^\/\/\s*(app.*;\n)/gm;
// match lines
// starting with //
// ignore following whitespaces
// having text "app" afterwards, capture from app to line-ending as group ($1 reference)

const replacement = `$1`;

newcontent = content.replace(regex_block_comments,replacement);
newcontent = newcontent.replace( regex_comment,replacement);

// content after replacements
console.log(newcontent);

要了解更多关于正则表达式和构建模式的信息,我建议 https://regexr.com/ https://regex101.com/ .