Linux sed command replace regex string

Question

I'm trying to edit a text file programmatically, which contains a line like this:

db.username="tata" db.password="toto" 

which I want to look like this:

db.username="fofo" db.password="bar" 

Here is the regex :

 ^\s*db.username="([^"]*) 

when i try to use this command

sed -i -E '^\s*db.username="([^"]*)' 'fofo' file.txt 

I'm getting the error:

sed: -e expression #1, char 1: unknown command: `^' 

Answer 1

you are mixing grep's regex and sed's substitution.

use either

sed -e '/db\.username/s/"[^"]*"/"foo"/' txt db.username="foo" db.password="toto" 

where

• /db\.username/ tell sed to operate on line whith db.username (. is special caracter in sed)
• s replace
• "[^"]*" a quote, any number of non quote, a quote by
• "foo"
• /db.username/ would match db_username, dbXusername ....

or

sed -e 's/db\.username="[^"]*"/db.username="foo"/' txt db.username="foo" db.password="toto" 

---

on a further note, sed can capture pattern

sed -e 's/db\.username="\([^"]*\)"/usename is \1/' txt usename is tata db.password="toto" 

• \( \) capture text
• \1 use first captured text

grep can too

grep -Eo '"[^"]*"' txt "tata" "toto"