每当相应的文件发生变化时,我想捆绑我的测试并开始业力,显示任何失败的测试。
我目前有监视任务:
gulp.task('default', ['browserify', 'css','runTests'], function () {
gulp.watch('./src/js/**/*.js', ['browserify']);
gulp.watch('./src/js/**/*.js', ['runTests']);
});
启动runTests.js
var testFile = [
'./src/components/tests/suite.js'
];
// bundle tests
var cmd = child.spawn('browserify', ['-e', './src/components/tests/suite.js', '-t', 'reactify', '-t']);
cmd.on('close', function (code) {
//cmd finished start karma
gulp.src(testFiles)
.pipe(karma({
configFile: 'karma.conf.js',
action: 'run'
}))
.on('error', function(err) {
throw err;
});
});
我的控制台目前出现错误:
[17:04:27] Starting 'runTests'...
[17:04:27] Finished 'runTests' after 2.73 ms
events.js:85
throw er; // Unhandled 'error' event
^
Error: spawn browserify ENOENT
at exports._errnoException (util.js:746:11)
at Process.ChildProcess._handle.onexit (child_process.js:1053:32)
at child_process.js:1144:20
at process._tickCallback (node.js:355:11)
at Function.Module.runMain (module.js:503:11)
at startup (node.js:129:16)
at node.js:814:3
Process finished with exit code 1
我可以获得一个基本命令来工作child = spawn("ls");
但不是browserify命令,任何人都可以帮忙吗?
答案 0 :(得分:0)
您可以在gulp任务中使用browserify API。有关使用browserify进行通配的一些提示,请查看substack/node-browserify#1170。然后你可以在karam配置文件数组中包含输出文件。
// Disclaimer: This code is untested
var files = require("glob").sync('./src/js/**/*.js');
files.push('./src/components/tests/suite.js');
var browserify = require("browserify")(files, {transform: 'reactify'});
// your can do more config on the browserify instance
// Run browserify bundle and output to a file
var myFile = require('fs').createWriteStream('myOutput.txt');
browserify.bundle().pipe(myFile);