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[Bash] Remove everything after the first space in a string

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Gau36o Posted messages 14 Status Member -  
Gau36o Posted messages 14 Status Member -
Hello,

Beginner in Bash, I have a string stored in a variable.

For example: maVar="ceci 001 est un exemple 5600"

To remove a strict occurrence such as the ceci (including the space following the word ceci) from the string above, no problem, I do:
line1=${maVar#*ceci }

After this, I would now like to remove the space and everything that follows this space, after the pattern 001. Knowing that 001 is necessarily a group of digits but with a variable length (between 2 and 6 characters).

I tried by doing:
line2=${line1#* }

As well as other attempts, but I can’t get the desired result...

Any help is welcome :-)

Thanks in advance!

Configuration: Windows / Chrome 58.0.3029.110

3 answers

Answer 1 / 3
dubcek Posted messages 18627 Registration date   Status Contributor Last intervention   5 659
 
hello
or with a table 
$ maVar="ceci 001 est un exemple 5600" 
$ t=($maVar)
$ echo ${t[0]}
ceci
$ echo ${t[1]}
001
2
Gau36o Posted messages 14 Status Member
 
Hi,
Thanks for this answer!
This method suits me much better :)
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Answer 2 / 3
Gau36o Posted messages 14 Status Member
 
Problem solved after all...

If it can help someone... 
line2=$(echo $line1 | cut -d " " -f 1)


So the truncated result of the variable is stored in $line2
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Answer 3 / 3
zipe31 Posted messages 34620 Registration date   Status Contributor Last intervention   6 501
 
Hi,

We can do this as well: 

$ maVar="ceci 001 est un exemple 5600"

$ echo "${maVar}"
ceci 001 est un exemple 5600

$ echo "${maVar#* }"
001 est un exemple 5600

$ echo "${maVar##* }"
5600

$ echo "${maVar% *}"
ceci 001 est un exemple

$ echo "${maVar%% *}"
ceci


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