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Old 12-25-2016, 05:51 PM   #1
slimcharles
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Regular expression on apt-cache output


Hi,
I want to easily read apt-cache search output.
So I apply regular expression inside a function for apt-cache search inside .bashrc.

Code:
searchgrep() {
    sudo apt-cache search $1 | grep -E '.*?\ -\ '
}
alias search=searchgrep
If you type
Code:
search finance
then you get this

Sample Output:
erlang-percept - Erlang/OTP concurrency profiling tool
golang - Go programming language compiler - metapackage
golang-1.6 - Go programming language compiler - metapackage
golang-1.6-doc - Go programming language - documentation
golang-1.6-go - Go programming language compiler, linker, compiled stdlib

However all I want is that grep only marks the first part, software name part.
How should I edit the regular expression to look the example below?
Code:
grep -E '.*?\ -\ '
This is what I want to achieve:
erlang-percept - Erlang/OTP concurrency profiling tool
golang - Go programming language compiler - metapackage
golang-1.6 - Go programming language compiler - metapackage
golang-1.6-doc - Go programming language - documentation
golang-1.6-go - Go programming language compiler, linker, compiled stdlib

This is not a big problem but I am curious.
Thanks for your time.
 
Old 12-25-2016, 06:12 PM   #2
syg00
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regex is generally greedy, as you discovered. One (simple) way to circumvent that is to more specifically specify what you are searching for - "non whitespace" for example rather than "any character" (dot).
 
Old 12-27-2016, 03:37 PM   #3
slimcharles
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Quote:
Originally Posted by syg00 View Post
regex is generally greedy, as you discovered. One (simple) way to circumvent that is to more specifically specify what you are searching for - "non whitespace" for example rather than "any character" (dot).
Thanks for your reply. Can you give me more specific details, please? Whatever I did, I couldn't find a way.
 
Old 12-27-2016, 06:53 PM   #4
grail
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You want to look for the first thing you do not want, in your case it would be whitespace. Have a look into the [^...] construct.
 
Old 12-27-2016, 07:54 PM   #5
syg00
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I would like to see a few of those attempts so we can assist.
 
Old 12-27-2016, 11:39 PM   #6
nodir
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You can simply do
apt-cache search --names-only searchpattern
If that ain't enough i do what grail proposed ( "the [^...] construct" )
Most of the times one gets away with --names-only though.
 
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Old 12-28-2016, 03:21 PM   #7
MadeInGermany
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*? is actually perl RE for minimum match.
Try
Code:
grep -P '.*? - '
 
Old 12-28-2016, 05:03 PM   #8
Habitual
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Quote:
Originally Posted by nodir View Post
You can simply do
apt-cache search --names-only searchpattern
Man, I've been chewing on "extra" output in apt search for literally years, and now I know why. Thanks!
If it were me? I'd
Code:
alias <myalias>="sudo apt-cache search --names-only"
in ~/.bashrc or equivalent.

and it spews:
Code:
mysearch finance
[sudo] password for <user>:
libfinance-bank-ie-permanenttsb-perl - perl interface to the PermanentTSB Open24 homebanking
libfinance-qif-perl - Parse and create Quicken Interchange Format files
libfinance-quote-perl - Perl module for retrieving stock quotes from a variety of sources
libfinance-quotehist-perl - Perl modules for fetching historical stock quotes from the web
libfinance-streamer-perl - Perl5 module with interface to Datek Streamer
libfinance-yahooquote-perl - Perl module for retrieving stock quotes from Yahoo! Finance
Have fun!
 
Old 01-03-2017, 05:45 AM   #9
MadeInGermany
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Quote:
Originally Posted by MadeInGermany View Post
*? is actually perl RE for minimum match.
Try
Code:
grep -P '.*? - '
Hmm, still does not work. Well, the man page sais "experimental".
Here is an original perl that returns the matching portion (like grep -o ).
Code:
sudo apt-cache search "$1" | perl -lne 'print $1 if /(.*?) - /'
Or, maybe you can use grep and match from the end?
Code:
sudo apt-cache search "$1" | grep ' - .*'
 
  


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