如何修改Select-String输出格式：将行号置于路径括号内

原输出格式：

\\path-to-logfile\copy2xl.log:390:NOTE: At least one W.D format was too small for the number to be printed. The decimal may be shifted by the "BEST" format.

目标输出格式：

\\path-to-logfile\copy2xl.log(390):NOTE: At least one W.D format was too small for the number to be printed. The decimal may be shifted by the "BEST" format.

可以通过PowerShell管道结合ForEach-Object重构输出内容，具体命令如下：

# 替换为你的目标文件路径和匹配模式
Select-String -Path "\\path-to-logfile\copy2xl.log" -Pattern "W.D format was too small" | ForEach-Object {
    # 按目标格式拼接路径、行号和行内容
    "$($_.Path)($($_.LineNumber)):$($_.Line)"
}

核心逻辑：

• Select-String返回的匹配对象自带Path（文件路径）、LineNumber（行号）、Line（匹配行内容）属性
• 通过字符串插值将这三个属性按路径(行号):行内容的格式重新组合，即可得到目标输出

内容的提问来源于stack exchange，提问作者data null