The simplest way to “subtrack” a string variable in terminal?
Let's assume that we've a variable $test which holds a value asd 123 - what is the most simplest way to cut, for example, asd part?
I've googled this for awhile and surprised that I couldn't find an answer.
command-line
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
add a comment |
Let's assume that we've a variable $test which holds a value asd 123 - what is the most simplest way to cut, for example, asd part?
I've googled this for awhile and surprised that I couldn't find an answer.
command-line
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
add a comment |
Let's assume that we've a variable $test which holds a value asd 123 - what is the most simplest way to cut, for example, asd part?
I've googled this for awhile and surprised that I couldn't find an answer.
command-line
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
Let's assume that we've a variable $test which holds a value asd 123 - what is the most simplest way to cut, for example, asd part?
I've googled this for awhile and surprised that I couldn't find an answer.
command-line
command-line
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
asked Jan 4 at 6:03
Cecily MillerCecily Miller
10115
10115
add a comment |
add a comment |
1 Answer
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In Bourne-like shells, such as dash ( /bin/sh on Ubuntu ), bash and ksh there is something known as parameter expansion, and is a feature specified by POSIX standard. Specifically, to quote dash manual:
${parameter%word} Remove Smallest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the suffix matched by the pattern deleted.
${parameter%%word} Remove Largest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the suffix matched by the pattern deleted.
${parameter#word} Remove Smallest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the prefix matched by the pattern deleted.
${parameter##word} Remove Largest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the prefix matched by the pattern deleted.
Thus, to replace asd part you can do:
$ var="asd 123"
$ echo ${var#asd*}
123
To remove 123 you can do:
$ echo ${var%*123}
asd
bash goes even further with this with another form: ${parameter/string/replacement} and ${parameter//string/replacement}. First one replaces first occurrence of string. Second form replaces all occurrences. For instance:
$ echo ${var//123/}
asd
$ echo ${var//asd/}
123
Note that according with the syntax, the replacement part is empty string
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In Bourne-like shells, such as dash ( /bin/sh on Ubuntu ), bash and ksh there is something known as parameter expansion, and is a feature specified by POSIX standard. Specifically, to quote dash manual:
${parameter%word} Remove Smallest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the suffix matched by the pattern deleted.
${parameter%%word} Remove Largest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the suffix matched by the pattern deleted.
${parameter#word} Remove Smallest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the prefix matched by the pattern deleted.
${parameter##word} Remove Largest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the prefix matched by the pattern deleted.
Thus, to replace asd part you can do:
$ var="asd 123"
$ echo ${var#asd*}
123
To remove 123 you can do:
$ echo ${var%*123}
asd
bash goes even further with this with another form: ${parameter/string/replacement} and ${parameter//string/replacement}. First one replaces first occurrence of string. Second form replaces all occurrences. For instance:
$ echo ${var//123/}
asd
$ echo ${var//asd/}
123
Note that according with the syntax, the replacement part is empty string
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
add a comment |
In Bourne-like shells, such as dash ( /bin/sh on Ubuntu ), bash and ksh there is something known as parameter expansion, and is a feature specified by POSIX standard. Specifically, to quote dash manual:
${parameter%word} Remove Smallest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the suffix matched by the pattern deleted.
${parameter%%word} Remove Largest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the suffix matched by the pattern deleted.
${parameter#word} Remove Smallest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the prefix matched by the pattern deleted.
${parameter##word} Remove Largest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the prefix matched by the pattern deleted.
Thus, to replace asd part you can do:
$ var="asd 123"
$ echo ${var#asd*}
123
To remove 123 you can do:
$ echo ${var%*123}
asd
bash goes even further with this with another form: ${parameter/string/replacement} and ${parameter//string/replacement}. First one replaces first occurrence of string. Second form replaces all occurrences. For instance:
$ echo ${var//123/}
asd
$ echo ${var//asd/}
123
Note that according with the syntax, the replacement part is empty string
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
add a comment |
In Bourne-like shells, such as dash ( /bin/sh on Ubuntu ), bash and ksh there is something known as parameter expansion, and is a feature specified by POSIX standard. Specifically, to quote dash manual:
${parameter%word} Remove Smallest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the suffix matched by the pattern deleted.
${parameter%%word} Remove Largest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the suffix matched by the pattern deleted.
${parameter#word} Remove Smallest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the prefix matched by the pattern deleted.
${parameter##word} Remove Largest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the prefix matched by the pattern deleted.
Thus, to replace asd part you can do:
$ var="asd 123"
$ echo ${var#asd*}
123
To remove 123 you can do:
$ echo ${var%*123}
asd
bash goes even further with this with another form: ${parameter/string/replacement} and ${parameter//string/replacement}. First one replaces first occurrence of string. Second form replaces all occurrences. For instance:
$ echo ${var//123/}
asd
$ echo ${var//asd/}
123
Note that according with the syntax, the replacement part is empty string
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
In Bourne-like shells, such as dash ( /bin/sh on Ubuntu ), bash and ksh there is something known as parameter expansion, and is a feature specified by POSIX standard. Specifically, to quote dash manual:
${parameter%word} Remove Smallest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the suffix matched by the pattern deleted.
${parameter%%word} Remove Largest Suffix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the suffix matched by the pattern deleted.
${parameter#word} Remove Smallest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest
portion of the prefix matched by the pattern deleted.
${parameter##word} Remove Largest Prefix Pattern. The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the largest
portion of the prefix matched by the pattern deleted.
Thus, to replace asd part you can do:
$ var="asd 123"
$ echo ${var#asd*}
123
To remove 123 you can do:
$ echo ${var%*123}
asd
bash goes even further with this with another form: ${parameter/string/replacement} and ${parameter//string/replacement}. First one replaces first occurrence of string. Second form replaces all occurrences. For instance:
$ echo ${var//123/}
asd
$ echo ${var//asd/}
123
Note that according with the syntax, the replacement part is empty string
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
answered Jan 4 at 6:46
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.6k9147310
70.6k9147310
add a comment |
add a comment |
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