How to grep the required string from the result of alternatives –display java using start and ends with?

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英文:

How to grep the required string from the result of alternatives --display java using start and ends with?

问题

I wanted to fetch two outputs from the alternatives --display java list, one is /usr/lib/jvm/jdk-1.8-oracle-x64/bin/java and the other is /usr/java/jdk-11.0.17/bin/java I use the below command

alternatives --display java | grep -o '^/usr.*bin/java>'

and I get output

/usr/lib/jvm/jdk-1.8-oracle-x64/bin/java
/usr/java/jdk-11.0.17/bin/java

To be specific, this server has jdk 8 and jdk 11 installed. What additional regex can be added to the above command to get the output separately?

I need help with the above command to get the single output as

/usr/lib/jvm/jdk-1.8-oracle-x64/bin/java

when trying to fetch jdk-1.8 and

/usr/java/jdk-11.0.17/bin/java

when trying to fetch jdk-11.0.17

For example, when I run alternatives --display java | grep -o '^/usr.*1[.]8*bin/java>', the output should be /usr/lib/jvm/jdk-1.8-oracle-x64/bin/java, and when I run alternatives --display java | grep -o '^/usr.*11[.]0[.]17*bin/java>', the output should be /usr/java/jdk-11.0.17/bin/java. Unfortunately, I get no result when I run the command in the example.

英文:

I wanted to fetch two outputs from the alternatives --display java list
, one is /usr/lib/jvm/jdk-1.8-oracle-x64/bin/java and the other is /usr/java/jdk-11.0.17/bin/java I use the below command

alternatives --display java | grep -o '^/usr.*bin/java\>'

and I get output

/usr/lib/jvm/jdk-1.8-oracle-x64/bin/java
/usr/java/jdk-11.0.17/bin/java

To be specific, this server has jdk 8 and jdk 11 installed. What additional regex can be added to the above command to get the output seperately ?
I need help with the above command to get the single output as

/usr/lib/jvm/jdk-1.8-oracle-x64/bin/java

when trying to fetch jdk-1.8
and

/usr/java/jdk-11.0.17/bin/java

when trying to fetch jdk-11.0.17

eg when i run alternatives --display java | grep -o '^/usr.*1[.]8*bin/java\>' , the output should be /usr/lib/jvm/jdk-1.8-oracle-x64/bin/java and when i run alternatives --display java | grep -o '^/usr.*11[.]0[.]17*bin/java\>' , the output should be /usr/java/jdk-11.0.17/bin/java . Unfortunately , I get no result when i run the command in the example

答案1

得分: 0

From OP's comment the following do not work:

grep -o '^/usr.*1[.]8*bin/java>'
                   ^^^^^
grep -o '^/usr.*11[.]0[.]17*bin/java>'
                         ^^^^^

问题在于 8*bin7*bin,它们表示匹配0个或多个字符 87 后跟字符串 bin;当然,在输入中不存在字符串 .bin.8bin.8888bin.1bin.17bin.17777bin,因此不会生成任何输出。

尝试将 * 应用于单字符通配符 (.),就像在模式的前面应用的那样 (user.*);所以,对 OP 当前代码稍作修改:

grep -o '^/usr.*1[.]8.*bin/java>'
                     ^
grep -o '^/usr.*11[.]0[.]17.*bin/java>'
                           ^
英文:

From OP's comment the following do not work:

grep -o '^/usr.*1[.]8*bin/java\>'
                    ^^^^^
grep -o '^/usr.*11[.]0[.]17*bin/java\>'
                          ^^^^^

The problem here is the 8*bin and 7*bin which says to match on 0 or more of the characters 8 or 7 followed by the string bin; of course, the strings .bin, .8bin, .8888bin, .1bin, .17bin and .17777bin do not occur in the input so no output is generated.

Try applying the * to the single character wildcard character (.), as is done at the front of the pattern (user.*); so, slight modification to OP's current code:

grep -o '^/usr.*1[.]8.*bin/java\>'
                     ^
grep -o '^/usr.*11[.]0[.]17.*bin/java\>'
                           ^

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  • 本文由 发表于 2023年6月15日 01:04:00
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