如何从gulp调用执行PowerShell脚本？ [英] How to invoke executing a PowerShell script from gulp?

问题描述

  var msbuild = require（'gulp-msbuild'）;我们正在使用gulp来构建和部署我们的应用程序。 
 gulp.task（'build'，['clean']，function（）{
 return gulp.src（'../../*。sln'）
 .pipe（ msbuild（{
 toolsVersion：14.0，
 targets：['Rebuild']，
 errorOnFail：true，
 properties：{
 DeployOnBuild：true，
 DeployTarget：'Package'，
 PublishProfile：'Development'
}，
 maxBuffer：2048 * 1024，$ b $ stderr：true，
 stdout：true，
 fileLoggerParameters：'LogFile = Build.log; Append; Verbosity = detailed'，
}））; 
}）; 
  

然而，在构建之后，我必须调用PowerShell脚本文件publish.ps1，我该如何调用它在吞噬？

解决方案

我还没有测试过这个，但是如果你将两者结合起来，它会看起来像这样。

  var gulp = require（'gulp'）只运行缺省任务，该任务使用run-sequence来管理依赖项顺序。 ，
 runSequence = require（'run-sequence'），
 msbuild = require（'gulp-msbuild'），
 spawn = require（child_process）。spawn，
儿童; 
 $ b gulp.task（'default'，function（）{
 runSequence（'clean'，'build'，'powershell'）; 
}）; 
 $ b gulp.task（'build'，['clean']，function（）{
 return gulp.src（'../../*。sln'）
 .pipe（msbuild（{
 toolsVersion：14.0，
 targets：['Rebuild']，
 errorOnFail：true，
属性：{
 DeployOnBuild：true， 
 DeployTarget：'Package'，
 PublishProfile：'Development'
}，
 maxBuffer：2048 * 1024，
 stderr：true，
 stdout：true ，
 fileLoggerParameters：'LogFile = Build.log; Append; Verbosity = detailed'，
}））; 
}）; 
 
 gulp.task（'powershell'，function（callback）{
 child = spawn（powershell.exe，[c：\\temp\\helloworld。 （Powershell Data：+ data）; 
}）; 
 child.stderr.on（data，function（data）{
 console.log（Powershell Errors：+ data）; 
}）; 
 child.on（退出，函数（）{
 console.log（Powershell脚本完成）; 
}）; 
 child.stdin.end（）; //结束输入
回调（）; 
}）; 
  

编辑

<使用参数调用powershell文件

  var exec = require（child_process）。exec; 
 $ b gulp.task（powershell，function（callback）{
 exec（
Powershell.exe -executionpolicy remotesigned -File file.ps1，
 function （err，stdout，stderr）{
 console.log（stdout）; 
 callback（err）; 
} 
）; 
}）; 
  

Powershell file.ps1位于解决方案的根目录中

Write-Host'hello'

编辑2

好的，再试一次。你可以把params /参数放在file.ps1中吗？

$ $ $ $ $ $ $ $ $ $ $ $ $写入输出$ arg1;
写入输出$ arg2;

Write-Stuff -arg1hello-arg2See Ya

编辑3

通过吞咽任务传递参数::

  gulp.task（'powershell'，function（callback）{
 exec（Powershell.exe -executionpolicy remotesigned。。\\file.ps1; Write-Stuff -arg1 '我的第一个参数'-arg2'第二个'here'，函数（err，stdout，stderr）{
 console.log（stdout）; 
 callback（err）
}）; 
}）; 
  

更新file.ps1以删除

函数Write-Stuff（[string] $ arg1，[string] $ arg2）{
 Write-Output $ arg1; 
写入输出$ arg2; 
} 
  

I am using gulp to build and deploy our application.

var msbuild = require('gulp-msbuild');
gulp.task('build', ['clean'], function () {
return gulp.src('../../*.sln')
    .pipe(msbuild({
        toolsVersion: 14.0,
        targets: ['Rebuild'],
        errorOnFail: true,
        properties: {
            DeployOnBuild: true,
            DeployTarget: 'Package',
            PublishProfile: 'Development'
        },
        maxBuffer: 2048 * 1024,
        stderr: true,
        stdout: true,
        fileLoggerParameters: 'LogFile=Build.log;Append;Verbosity=detailed',
    }));
});

However after build I have to call a PowerShell script file "publish.ps1", how can I call it in gulp?

解决方案

I haven't tested this but if you combine the two it would look something like this. just run the default task, which uses run-sequence to manage the dependency order.

    var gulp = require('gulp'),
        runSequence = require('run-sequence'),
        msbuild = require('gulp-msbuild'),
        spawn = require("child_process").spawn,
        child;

    gulp.task('default', function(){
        runSequence('clean', 'build', 'powershell');
    });

    gulp.task('build', ['clean'], function () {
        return gulp.src('../../*.sln')
            .pipe(msbuild({
                toolsVersion: 14.0,
                targets: ['Rebuild'],
                errorOnFail: true,
                properties: {
                    DeployOnBuild: true,
                    DeployTarget: 'Package',
                    PublishProfile: 'Development'
                },
                maxBuffer: 2048 * 1024,
                stderr: true,
                stdout: true,
                fileLoggerParameters: 'LogFile=Build.log;Append;Verbosity=detailed',
            }));
    });

    gulp.task('powershell', function(callback){
        child = spawn("powershell.exe",["c:\\temp\\helloworld.ps1"]);
        child.stdout.on("data",function(data){
            console.log("Powershell Data: " + data);
        });
        child.stderr.on("data",function(data){
            console.log("Powershell Errors: " + data);
        });
        child.on("exit",function(){
            console.log("Powershell Script finished");
        });
        child.stdin.end(); //end input
        callback();
    });

EDIT

Call a powershell file with parameters

var exec = require("child_process").exec;

gulp.task("powershell", function(callback) {
    exec(
        "Powershell.exe  -executionpolicy remotesigned -File  file.ps1",
        function(err, stdout, stderr) {
            console.log(stdout);
            callback(err);
        }
    );
});

Powershell file.ps1 in the root of your solution

Write-Host 'hello'

EDIT 2

OK, one more try. Can you put the params/arguments in file.ps1?

function Write-Stuff($arg1, $arg2){
    Write-Output $arg1;
    Write-Output $arg2;
}
Write-Stuff -arg1 "hello" -arg2 "See Ya"

EDIT 3

Pass the params from the gulp task::

gulp.task('powershell', function (callback) {
    exec("Powershell.exe  -executionpolicy remotesigned . .\\file.ps1; Write-Stuff -arg1 'My first param' -arg2 'second one here'" , function(err, stdout, stderr){
       console.log(stdout); 
       callback(err)
    });
});

Update file.ps1 to remove

function Write-Stuff([string]$arg1, [string]$arg2){
    Write-Output $arg1;
    Write-Output $arg2;
}

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