使用awk或sed删除两个模式（包括模式）之间的线

Question

My output file is as below:

judi#cat file
---ABC---
word1
word2
word3
word4
word5
word6
---end_ABC---

---DEF---
line1
line2
line3
line4
---end_DEF---
judi#

I need to remove the lines in between the pattern ABC and end_ABC (inclusive the pattern, then replace with new content; the new content is in a file).

The content of the file varies, so I need to use only the pattern.

judi#file1
---ABC---
wordA1
wordA2
wordA3
---end_ABC---
judi#

Desired result has to be

judi#
---ABC---
wordA1
wordA2
wordA3
---end_ABC---

---DEF---
line1
line2
line3
line4
---end_DEF---
judi#

I tried this command:

sed '/ABC/,/end_ABC/{/ABC/!{/end_ABC/!d}}' file > file 2

But I get this error:

sed: command garbled: /ABC/,/end_ABC/{/ABC/!{/end_ABC/!d}}

Answer 1

sed '/end_ABC/a ##here' file | sed '/ABC/,/end_ABC/d' | sed '/##here/r file1' | sed '/##here/d' >file2  

output

judi#cat file
judi#file1
---ABC---
wordA1
wordA2
wordA3
---end_ABC---
judi#

---DEF---
line1
line2
line3
line4
---end_DEF---
judi#

a ##here is appending ##here after matching end_ABC .

r file1 is inserting text from file1 after finding pattern ##here .

Answer 2

Never use range expressions as they make trivial tasks very slightly briefer but even slightly more complicated tasks need a complete rewrite or duplicate conditions. Just use a flag:

awk '
NR==FNR { rep = rep $0 OFS; next }
/---ABC---/ { printf "%s", rep; inBlock=1 }
!inBlock
/---end_ABC---/ { inBlock=0 }
' file1 file