我可以在打开文件夹时自动启动任务吗?

Can I automatically start a task when a folder is opened?

VS 代码是否支持在启动时启动 gulp-watch 任务?我想在打开编辑器时启动观察器。

不,对不起。这是一个很好的建议(能够 运行 启动时的任务),你能在 GitHub 的 microsoft/vscode 回购中提出问题吗? (抱歉,我还不能 post 超过 2 个链接)?

同时,这里有几个选项:

我创建了一个可以做到这一点的扩展。 是"Blade Runner"。试试这个! Blade Runner on Visual Studio Code Market Place

这是一个旧线程,但问题仍然是最新的。 Blade Runner 不适合我的需求,其他任何扩展也不适合。

我创建了一个扩展 AutoLaunch,它可以 运行 来自 tasks.json 的特定任务或来自 launch.json 的特定调试配置。

此功能是在 v1.30 中添加的:Release notes: run task on folder open

Run on folder open

If you have a task that you always run when you open a folder, you can configure it to run automatically when that folder is opened by configuring its "runOn" property. We found this useful for making sure no one breaks our new strict null checks in the VS Code repository. By adding "runOn": "folderOpen", everyone who allows tasks to be run automatically will get markers when they violate the strict null check:

示例:

{
    "type": "npm",
    "script": "strict-null-check-watch",
    "label": "TS - Strict Null Checks",
    "isBackground": true,
    "problemMatcher": {
        "base": "$tsc-watch",
        "owner": "typescript-strict-null",
        "applyTo": "allDocuments"
    },
    "runOptions": {
        "runOn": "folderOpen"
    }
}

示例 2:

{
  "label": "Tasks: copy3",
  "type": "shell",
  "command": "gulp",
  "args": [
    "copy3",
    "--file",
    "${fileBasename}"
  ],
  "problemMatcher": [],
  "runOptions": {
    "runOn": "folderOpen"
  }
},

正如 Mark 所建议的,您可以使用 "runOn": "folderOpen",但我认为 运行 一项 Gulp 任务并不那么简单。我建议尝试按照以下方式进行操作。

{
    "label": "Watch",
    "type": "shell",
    "command": "gulp watch",
    "runOptions": {
        "runOn": "folderOpen"
    }
)