[英]perl or sed command to find 2 strings and replace the first string with string 2
我有以下文本的多個文件。
文件1的內容:
Begin("tmp1","derived from file \"/disks/setup/ORIG_FILE1.txt\"");
文件2的內容:
Begin("tmp1","derived from file \"/disks/setup/ORIG_FILE2.txt\"");
文件3的內容:
Begin("tmp9","derived from file \"/disks/setup/ORIG_FILE4.txt\"");
等等.....
在perl或sed中是否有任何命令來查找帶有“ Begin”的行,並在該行中搜索“ ...”(即本例中的“ tmp1”)之間的第一個字符串並將其存儲在變量中,然后搜索/ .. \\“”之間的最后一個字符串,即(此處為ORIG_FILE1.txt),並將其存儲在變量2中,然后從變量2中刪除“ .txt”,然后將變量1替換為變量2並更新文件。
因此輸出文件如下所示:
文件1的內容:
Begin("ORIG_FILE1","derived from file \"/disks/setup/ORIG_FILE1.txt\"");
文件2的內容:
Begin("ORIG_FILE2","derived from file \"/disks/setup/ORIG_FILE2.txt\"");
文件3的內容:
Begin("ORIG_FILE4","derived from file \"/disks/setup/ORIG_FILE4.txt\"");
等等.....
有人可以幫我嗎?
嘗試這個:
perl -i.bak -pe '
if (/.*\/(.+)\.txt/) {
$file = $1;
s/(?<=").*?(?=")/$file/
}
' file1 file2 file3 ...
這將首先捕獲“ orig”文件名(假設它們都具有“ .txt”擴展名), 然后替換“ tmp”占位符。
在任何UNIX機器上的任何外殼中使用任何sed:
$ sed 's/\([^"]*"\)[^"]*\(.*\/\([^.]*\)\)/\1\3\2/' file
Begin("ORIG_FILE1","derived from file \"/disks/setup/ORIG_FILE1.txt\"");
Begin("ORIG_FILE2","derived from file \"/disks/setup/ORIG_FILE2.txt\"");
Begin("ORIG_FILE4","derived from file \"/disks/setup/ORIG_FILE4.txt\"");
通過gnu sed作為您在“ d”中的數據,
sed -E 's~^(Begin\(")\w+(".+/)(\w+)(\.\w+\\""\);)~\1\3\2\3\4~' d
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