I have a text file as shown below:
where $ tells about EOL (end of line) location I have used to illustrate that.
53t83t5 5 gejgi3 gg 4gij503 $ /* rtdrfsetsrhs $ ryhrdhrh $ rhyrdhyyyyyyyyyyyrhyrshrh$ ryhrhrh */$ $ 345dfeb terfgb$ /*srdtfgyhgfs*/ $ $ $
Now I have to replace every line of
/* */ this block with EOL placed at start of the line as shown below, Notice the position of $ for Line 3 that means if there are some characters including whitespace we should not skip it instead we should place EOL at the position of
/* forward slash.
53t83t5 5 gejgi3 gg 4gij503 $ $ $ $ $ $ 345dfeb terfgb$ /*srdtfgyhgfs*/ $ $ $
I was able to detect the block using
sed '/\/\*/,/\*\//d inputFile even I was able to delete the whole block also but I wonder can we do the above one using sed command in .sed script.
/*...*/ this is used for illustration of a pattern block we can have different pattern enclosing blocks also, like
!!...!! however I want to know a snippet for
/*...*/, I can handle all others on my own.