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Through bash, I am trying to extract part of a line from a file.

Currently I am using two sed commands back to back like so:

sed -n -e "s/^abc=//p" file | sed -n -e "s/\.//gp"

which can take in abc=1.2.3 and spit out 123. This got me thinking... Can I achieve this with just one command call? As in I want to find all strings in a file that match abc=<digit1>\.<digit2>\.<digit3> and spit out <digit1><digit2><digit3>.

EDIT:

Just to clarify, I want this to only print out lines that match. For instance, if I have the following file:

1.2.3.4
abc=quack
qtip=1.2.3
abc=1.2.3
abc = 4.5.6

running the command should only print 123

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5 Answers 5

Reset to default
2

You can use awk instead for removing . from part after abc=:

awk -F= '$1=="abc"{gsub(/\./, "", $2); print $2}' file
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Comments

1

You could try the below GNU sed command is the string abc=<digit1>\.<digit2>\.<digit3> present anywhere in a file,

sed -nr 's/.*abc=([0-9])\.([0-9])\.([0-9]).*/\1\2\3/p' file

OR

You could try the below sed command if the string abc= is at the start of a line.

sed -nr 's/^abc=([0-9])\.([0-9])\.([0-9]).*/\1\2\3/p' file

Example:

$ cat file
abc=1.2.3
foo abc=4.5.6
bar
$ sed -nr 's/.*abc=([0-9])\.([0-9])\.([0-9]).*/\1\2\3/p' file
123
456
$ sed -nr 's/^abc=([0-9])\.([0-9])\.([0-9]).*/\1\2\3/p' file
123
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Comments

0

This should work:

sed -E 's/abc=|([0-9])\./\1/g' file
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4 Comments

I am getting an error saying "/bin/sed: invalid option -- E". If I use "-e" I get a complaint about '\3' in conjunction with the 's' command. The man page does not include a '-E' option
Which version of bash are you using? I've updated my answer, so maybe that will fare you better.
I am using version 4.1.5 on a RHEL 5.9 box
@user972276, -E should work with that. Try replacing it with -r perhaps. echo 'abc=1.2.3.4.5' | sed -r 's/abc=|([0-9])\./\1/g'
0

You can also simply use tr:

$ tr -d [a-z.=] <<< abc=1.2.3
123

EDIT: I missed the part of the question where 'I want to find all strings in a file that match...' So this might or might not work depending on the content of the other, unwanted, lines.

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2 Comments

Nice one — a person that thinks outside the box!
For my specific use at the moment, it will only find one in the file, but for future reference its nice to know more generally how to do things :)
0

Here is an approach using awk that will work with any number of digits, not just three, separated by periods:

$ echo 'abc=1.2.3.4' | awk -F. -v OFS= '{sub(/.*=/, "", $1); print}'
1234
$ echo 'abc=1.2' | awk -F. -v OFS= '{sub(/.*=/, "", $1); print}'
12

Taking the awk command in parts:

  • -F.

    Use a period as the field separator. If the input is abc=1.2, for example, then awk sees two fields: abc=1 and 2.

  • -v OFS=

    This tells awk not to put any spaces between the fields when we print them out.

  • sub(/.*=/, "", $1)

    This removes the abc= part from the beginning of the line.

  • print

    This prints out the final line.

Selecting which lines to process

Suppose that we only want to process lines starting with abc= and followed only by numbers and periods. In that case:

$ awk -F. -v OFS= '/^abc=[0-9.]+$/ {sub(/.*=/, "", $1); print}' sample
123

where sample is the name of the file containing the sample lines in the updated question.

The sole change above is the addition of the pattern /^abc=[0-9.]+$/. This limits the commands which follow to only apply to lines matching this regular expression. Since /^abc=[0-9.]+$/ only matches lines that start with abc= followed by any combination of numbers of periods, only those lines are processed. Non-matching lines are ignored.

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2 Comments

This works to remove the abc= and also the dots, but it does it for every line in the file. I need it to print out only the lines that start with 'abc='
@user972276 That can be done. The answer is now updated according the current edit of the question.

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