Return string of a line in a file with multiple line after a matching pattern

Question

I have a file with multiple lines in it. I am trying to get a string that appears after a specific pattern...

Here is the command i am trying to do:

sed -n -e 's/^.*DB_PASS=//p' <<< $(cat /root/dadosDB)

But instead of giving me only the line i want, its giving me everything after the pattern I specified...

What I expect:

Return only the output of the line that contains DB_PASS and ignore other lines.

What the command is doing:

Printing everything of the entire file after DB_PASS and ignoring everything that appears before that pattern.

Any help is apprreciate it!

Here is what I have inside dadosDB:

DB_USER=myusername
DB_PASS=mypassword
DB_NAME=mydatabase

I want to get only what is after DB_PASS= and stop at that line, ignoring the following ones.

Actual output of the sed:

$ sed -n 's/^.*DB_PASS=//p' <<< $(cat /root/dadosDB)
mypassword DB_NAME=mydatabase

What I need:

mypassword

Answer 1

The reason why you are getting this result is because you don't apply the sed command to the file directly, but to the result of a command substitution.

This means that the text processing is not actually applied to

DB_USER=myusername
DB_PASS=mypassword
DB_NAME=mydatabase

but to the result of the cat command applied to the file content, after an unquoted command substitution that squeezes the lines.

If you try

echo $(cat /root/dadosDB)

you will notice that the output is

DB_USER=myusername DB_PASS=mypassword DB_NAME=mydatabase

all on a single line, rather than on separate lines. If you place the command substitution in double-quotes, as in

echo "$(cat /root/datosDB)"

the newlines will be preserved.

The key points are that

1. you need neither the cat nor the command substitution, and
2. if you use the command substitution without placing it in double-quotes, you will lose the newlines (see this Q&A e.g.).⁽¹⁾

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TL;DR:

Use

sed -n -e 's/^.*DB_PASS=//p' /root/dadosDB

or (rather)

sed -n -e 's/^DB_PASS=//p' /root/dadosDB

to exclude lines with keywords that start differently, but end in DB_PASS.

Please note: You write "and stop at that line, ignoring the following ones." I took this as meaning that you want to print the word after DB_PASS, but not the following lines (that technically also come after DB_PASS. If you mean that there could be more lines starting with DB_PASS and you want to ensure matching only the first, you will need a slight adjustment:

sed -n -e '0,/DB_PASS/s/^DB_PASS=//p' /root/dadosDB

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⁽¹⁾ Note that the issue in your particular case is how your shell (I assume Bash) handles "Here-strings" with unquoted substitutions, so (as noted by @steeldriver) a better demonstration is

cat <<< "$(cat /root/dadosDB)"

vs.

cat <<< $(cat /root/dadosDB)

The behavior has changed with Bash v4.4: Before, the output of the two commands was different in the same way as the echo examples mentioned above. In later versions of Bash, the unquoted command substitution would no longer lead to word-splitting and consequent squeezing of multi-line output.

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If you are interested to learn more, you may want to look at

• When is double-quoting necessary?

Also, I would recommend checking your scripts with shellcheck, also available as standalone tool on many Linux distributions, to avoid syntax (and other) errors.

Answer 2

You can also use awk, which seems (to me) a more suited option than sed for the task:

$ awk -F'=' '$1 == "DB_PASS" { print $2 }' /root/dadosDB
mypassword

Another option is a combination of grep and cut:

$ grep "^DB_PASS=" /root/dadosDB | cut -d= -f2
mypassword

Note both commands treat the input as columns separated by =.

Answer 3

Try this:

sed -n "s/.*DB_PASS=\(.*\)/\1/p" <<< $(cat /root/dadosDB)

If it's just one line ->

echo "DB_USER=myusername DB_PASS=mypassword DB_NAME=mydatabase" | sed -n "s/.*DB_PASS=\(.*\)DB.*/\1/p"