在声明它们之后,可以在这里重新评估字符串吗?
问题描述:
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。在声明它们之后,可以在这里重新评估字符串吗?
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?
答
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
。
答
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"
& $controllerContent
输出:
@Controller
@EnableWebMvc
public class HelloNorway {
}
答
感谢您的帮助的人。任何有兴趣,这是我想出了...
Param(
[Parameter(Mandatory=$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
+0
一些导入必须被添加,我也可以用它来创建测试类... powershell真的很有趣。 – Patrick
非常不错的主意脚本。 –