CODEWITHSUNDEEP
regexstringbashshellawk
Dharmendra
Dharmendra

Reputation: 394

Extract substring from a string using awk

I have string which can be one of the following two formats : 

dts12931212112 : some random message1 : abc, xyz
nodts : some random message2

I need to extract substring from these two string which doesn't have 'dts' part i.e. it should return :

some random message1 : abc, xyz
some random message2

I need to do this inside a bash script.

Can you help me with the awk command, which does this operation for both kind of strings?

Upvotes: 2

Views: 13055

Answers (2)

Jotne
Jotne

Reputation: 41460

Here is another awk

awk -F" : " '{$1="";sub(FS,"")}1' OFS=" : " file
some random message1 : abc, xyz
some random message2

Just remove first field when separated by :

Another version:

awk -F" : " '{print substr($0,index($0,$2))}' file
some random message1 : abc, xyz
some random message2

Get all data from second field and out separated by :

Upvotes: 1

Avinash Raj
Avinash Raj

Reputation: 174844

Through awk's gsub function.

$ awk '{gsub(/^[^:]*dts[^:]*:|:[^:]*dts[^:]*/, "")}1' file
 some random message1 : abc, xyz
 some random message2
$ awk '{gsub(/^[^:]*dts[^:]*:[[:blank:]]*|:[^:]*dts[^:]*/, "")}1' file
some random message1 : abc, xyz
some random message2

You could apply the same regex in sed also, but you need to enable -r --regexp-extended parameter.

^ asserts that we are at the start. [^:]* negated character class which matches any character but not of :, zero or more times. So this ^[^:]*dts[^:]*: would match the substring at the start which contain dts. It it won't touch if the substring is present at the middle. This :[^:]*dts[^:]* pattern matches the middle or last substring which has dts. Finally replacing the matched chars with an empty string will give you the desired output.

Update:

$ awk '{gsub(/^[^[:space:]]*dts[^[:space:]]*[[:space:]:]*|[[:space:]:]*[^[:space:]]*dts[^[:space:]]*/, "")}1' file
some random message1 : abc, xyz
some random message2

Upvotes: 2

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