简单的gradle构建文件构建错误

Question

我正在尝试几个gradle基础知识.这是我的gradle文件"build.gradle"的样子:

task hello
{
    doLast
    {
        println 'Hello World!'
    }
}

这会导致以下错误:

D:\DevAreas\learn-gradle>gradle -q hello

FAILURE: Build failed with an exception.

* Where:
Build file 'D:\DevAreas\learn-gradle\build.gradle' line: 2

* What went wrong:
Could not compile build file 'D:\DevAreas\learn-gradle\build.gradle'.
> startup failed:
  build file 'D:\DevAreas\learn-gradle\build.gradle': 2: Ambiguous expression could be a parameterle
ss closure expression, an isolated open code block, or it may continue a previous statement;
   solution: Add an explicit parameter list, e.g. {it -> ...}, or force it to be treated as an open
block by giving it a label, e.g. L:{...}, and also either remove the previous newline, or add an exp
licit semicolon ';' @ line 2, column 1.
     {
     ^

  1 error


* Try:
Run with --stacktrace option to get the stack trace. Run with --info or --debug option to get more l
og output.

如果我像这样对构建文件进行微小的修改

[请注意我已将括号从第二行移到第一行]

task hello{
    doLast
    {
        println 'Hello World!'
    }
}

我看到了输出

Hello World!

没有问题.

括号中的括号是一个大问题吗？通过将括号放在第二行,我做错了什么？

Answer 1

与使用分号推断的其他语言一样,换行符在Groovy中有所不同.第一个片段被解析为task hello; { ... },这是不明确的(无法确定第二个语句是块还是闭包),因此无效的Groovy语法.不管怎么说,这不是你想要的; 您希望闭包与hello任务相关联.为了避免这种意外,我建议遵循Java大括号样式.