Question

I want to create a sed command to find in a file the 1st line that matchs a pattern and the delete all that line or replace it all with some other text. I dont want to match all the line because the rule is to match part of it.

how can i do it with sed?

for instance:

myline 1 is pretty
line 2 is ugly
myline 111 is nice

I want to delete the first line that contains “1 is”

UPDATE: My line may have characters like “/” and “<”

Regards
Fak

Answer 1

I tend to use awk for more complicated tasks, it’s a bit more powerful than sed, having proper looping and selection constructs (expanded for readability, you can compress it back to one line if you wish):

pax$ echo '
xyz
myline 1 is pretty
line 2 is ugly
myline 111 is nice'  | awk '
     /1 is/ {if (f) {print} else {f = 1}}
    !/1 is/ {print}'

xyz
line 2 is ugly
myline 111 is nice

For any line not matching the pattern (!/1 is/), it just prints it.

For lines that do match the pattern, it prints them all if the flag f is set (it’s initially not set). When the flags not set and it encounters a matching line, it sets the flag and doesn’t print it. This basically deletes the first matching line as desired.

If you want to modify the first matching line rather than deleting it, you just insert code alongside the f = 1 to do that, something like:

pax$ echo '
xyz
myline 1 is pretty
line 2 is ugly
myline 111 is nice'  | awk '
     /1 is/ {if (f) {print} else {f = 1;print "URK!!!"}}
    !/1 is/ {print}'

xyz
URK!!!
line 2 is ugly
myline 111 is nice

---

To use a shell variable in awk, you can use the -v option to pass it in as a real awk variable:

pax$ export xyzvar=URK ; echo '
xyz
myline 1 is pretty
line 2 is ugly
myline 111 is nice'  | awk -vmyvar=$xyzvar '
     /1 is/ {if (f) {print} else {f = 1;print myvar}}
    !/1 is/ {print}'

xyz
URK
line 2 is ugly
myline 111 is nice