我正在尝试注释掉大量文件中的一些代码
文件全部包含以下内容:
stage('inrichting'){
steps{
build job: 'SOMENAME', parameters: param
build job: 'SOMEOTHERNAME', parameters: param
echo 'TEXT'
}
}
steps{ }
中的内容是可变的,但始终存在于0..N'echo'和0..N'build job'
我需要这样的输出:
//stage('inrichting'){
// steps{
// build job: 'SOMENAME', parameters: param
// build job: 'SOMEOTHERNAME', parameters: param
// echo 'TEXT'
// }
//}
PowerShell有什么好的方法吗?我用pattern.replace尝试了一些东西,但是距离还很远。
$list = Get-ChildItem -Path 'C:\Program Files (x86)\Jenkins\jobs' -Filter config.xml -Recurse -ErrorAction SilentlyContinue -Force | % { $_.fullname };
foreach ($item in $list) {
...
}
答案 0 :(得分:1)
这有点棘手,因为您要查找整个部分,然后在其中的所有行中添加注释标记。如果您的结构允许,我可能会用switch -regex
编写一个临时解析器(计算大括号可能会使事情变得更健壮,但在所有情况下都很难正确使用)。如果代码足够常规,则可以将其简化为以下内容:
stage('inrichting'){
steps{
... some amount of lines that don't contain braces
}
}
然后我们可以检查在开始处是否出现了两条固定线,并最终检查了两条带有大括号的线:
foreach ($file in $list) {
# lines of the file
$lines = Get-Content $file
# line numbers to comment out
$linesToComment = @()
# line number of the current block to comment
$currentStart = -1
# the number of closing braces on single lines we've encountered for the current block
$closingBraces = 0
for ($l = 0; $l -le $lines.Count; $l++) {
switch -regex ($lines[$l]) {
'^\s*stage\('inrichting'\)\{' {
# found the first line we're looking for
$currentStart = $l
}
'^\s*steps\{' {
# found the second line, it may not belong to the same block, so reset if needed
if ($l -ne $currentStart + 1) { $currentStart = -1 }
}
'^\s*}' {
# only count braces if we're at the correct point
if ($currentStart -ne -1) { $closingBraces++ }
if ($closingBraces -eq 2) {
# we've reached the end, add the range to the lines to comment out
$linesToComment += $currentStart..$l
$currentStart = -1
$closingBraces = 0
}
}
}
}
$commentedLines = 0..($lines.Count-1) | % {
if ($linesToComment -contains $_) {
'//' + $lines[$_]
} else {
$lines[$_]
}
} | Set-Content $file
}
未经测试,但总体思路可能可行。
更新:已修复并经过测试