在声明它们之后，可以在这里重新评估字符串吗？

Question

我正在尝试使用PowerShell进行小型代码生成任务。我想让脚本生成一些java类和接口。

在脚本中我想声明一个字符串变量。例如：

$controllerContent = @" 
package controller; 

import org.springframework.stereotype.Controller; 
import org.springframework.web.servlet.config.annotation.EnableWebMvc; 

@Controller 
@EnableWebMvc 
public class $completeName { 

} 
"@ 

声明后我想将它传递给函数，我计算变量$completeName。但我不知道用什么方法替换字符串中的变量。我必须使用-replace吗？或者有其他方法吗？

Answer 1

我通常使用格式这样的任务字符串。所有你需要做的是更换$completeName与{0}，你可以格式化字符串中的任何时间：

$controllerContent = @" 
package controller; 

import org.springframework.stereotype.Controller; 
import org.springframework.web.servlet.config.annotation.EnableWebMvc; 

@Controller 
@EnableWebMvc 
public class {0} {{ 

}} 
"@ 

现在你可以使用命名类：

$controllerContent -f 'MyController' 

注意：唯一downsite到这是，你需要像我的例子中所示的那样转义大括号。因此，您的选择是否使用格式字符串或-replace。

Answer 2

我通常使用类似上面的答案的格式字符串，但您也可以将其包装在scriptblock中并在需要时调用。这不需要对文本本身进行任何修改。例如：

$completeName = "HelloWorld" 

$controllerContent = { @" 
@Controller 
@EnableWebMvc 
public class $completeName { 

} 
"@ } 

& $controllerContent 
#Save the string: $str = & $controlledContent 

输出：

@Controller 
@EnableWebMvc 
public class HelloWorld { 

} 

改变变量：

$completeName = "HelloNorway" 

& $controllerContent 

输出：

@Controller 
@EnableWebMvc 
public class HelloNorway { 

} 

Answer 3

感谢您的帮助的人。任何有兴趣，这是我想出了...

Param(
    [Parameter(Mandatory=$True)] 
    [string]$name 
) 
$sourceDir = "..." 
$classNameGeneration = {param($name,$type)Return "$name"+ ($type.Substring(0,1)).ToUpper()+($type.Substring(1))} 
$interfaceNameGeneration = {param($name,$type)Return "$name"+($type.Substring(0,1)).ToUpper()+($type.Substring(1))+"IF"} 

$controllerName = & $classNameGeneration -name $name -type controller 
$serviceInterfaceName = & $interfaceNameGeneration -name $name -type service 
$serviceName = & $classNameGeneration -name $name -type service 
$daoInterfaceName = & $interfaceNameGeneration -name $name -type dao 
$daoName = & $classNameGeneration -name $name -type dao 

function create($name, $type, $content) { 
    $dir = $sourceDir+"\$type" 
    $fileName = $name+".java" 
    $filePath = "$dir\$fileName" 
    cd $dir 
    New-Item $filePath -ItemType file 
    Set-Content $filePath $content 
} 

"[INFO]create controller class..." 
$controllerContent = @" 
package controller; 

import org.springframework.stereotype.Controller; 
import org.springframework.web.servlet.config.annotation.EnableWebMvc; 
import org.springframework.beans.factory.annotation.Autowired; 

@Controller 
@EnableWebMvc 
public class {0} {{ 

@Autowired 
private {1} service; 

}} 
"@ 
$controllerContent = $controllerContent -f $controllerName,$serviceInterfaceName 
create $controllerName controller $controllerContent 

"[INFO]Create service interface..." 
$serviceInterfaceContent = @" 
package service; 

public interface {0} {{ 

}} 
"@ 
$serviceInterfaceContent = $serviceInterfaceContent -f $serviceInterfaceName 
create $serviceInterfaceName service $serviceInterfaceContent 

"[INFO]Create service class..." 
$serviceContent = @" 
package service; 

import org.springframework.beans.factory.annotation.Autowired; 

public class {0} implements {1} {{ 

@Autowired 
private {2} dao; 
}} 
"@ 
$serviceContent = $serviceContent -f $serviceName,$serviceInterfaceName,$daoInterfaceName 
create $serviceName service $serviceContent 
"[INFO]Create dao interface..." 
$daoInterfaceContent = @" 
package dao; 

public interface {0} {{ 

}} 
"@ 
$daoInterfaceContent = $daoInterfaceContent -f $daoInterfaceName 
create $daoInterfaceName dao $daoInterfaceContent 
"[INFO]Create dao class..." 
$daoContent = @" 
package dao; 

public class {0} implements {1} {{ 

}} 
"@ 
$daoContent = $daoContent -f $daoName,$daoInterfaceName 
create $daoName dao $daoContent