linux - Shell script substring extraction and manipulation -

-

i able extract substring string; no matter syntax changes try (and i've tried many), unable enter if block in snippet though print out suggests client_name string matches expected 1 (attached output first echo). first echo prints anything. doing wrong here? ideas appreciated!

the idea if client named aa_nnnn, need extract aa , nnnn , check if aa matches known string (say "xx") , if does, then, calculate version nnnn , if version nnnn exceeds known version mmmm.

#! /bin/sh client=$1 ... client_name="${client:0:2}" client_version=2015 echo "before compare; client: $client_name; version: $client_version" if [ "$client_name" == "xx" ];    client_version="${client:3:4}"    echo "inside compare; client: $client_name; version: $client_version"    if [ $client_version -ge 2016 ];       ...    fi fi

first echo output: 

before compare; client: xx; version: 2015

/bin/sh --version returns: 

gnu bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu) copyright (c) 2009 free software foundation, inc. license gplv3+: gnu gpl version 3 or later <http://gnu.org/licenses/gpl.html>  free software; free change , redistribute it. there no warranty, extent permitted law.

i got same result running code, removing quotes substring extraction (and replacing , : in second one) fixed it...

client_name=${client:0:2}

...and...

client_version=${client:3:4}

Comments

Post a Comment

Popular posts from this blog

ruby - Trying to change last to "x"s to 23 -

-
i have been playing around irb , have string looks this: irb(main):072:0> puts ["[\"4354 5432 5432 xxxxx\", \"6547 6547 8543 xxxxx\", \"2344 6543 6674 xxxxx\", \"2346 6236 7543 xxxxx\", \"1273 5585 5587 xxxxx\"]"] => nil irb(main):073:0> what want gsub last 2 "x" s of each 17 digit combination, 23 . what i've tried far (i've been using rubular): a.gsub(/\d[x]/,"23") <= grabs "x" s i'm close a.gsub(/\w[x{2}]/,"23") <= grabs "x" s grabs two digits each combination. is there easier way i'm not understanding? a.gsub(/\d{4} \d{4} \d{4} xxx\kxx/, '23') \d{4} = 4 digits what \k drop matched far. can use don't replace entire number, still able validate have in desired format.
Read more

jquery - Clone last and append item to closest class -

-
this bare-bones code on page: <div class="list-container"> <div class="single-list-item"> <button>remove</button> </div> <div class="single-list-item"> <button>remove</button> </div> <div class="single-list-item"> <button>remove</button> </div> </div> <p><button class="add-list-item">add</button></p> <div class="list-container"> <div class="single-list-item"> <button>remove</button> </div> <div class="single-list-item"> <button>remove</button> </div> <div class="single-list-item"> <button>remove</button> </div> </div> <p><button class="add-list-item">add</button></p> <di...
Read more

css - Can I use the :after pseudo-element on an input field? -

-
i trying use :after css pseudo-element on input field, not work. if use span , works ok. <style type="text/css"> .mystyle:after {content:url(smiley.gif);} .mystyle {color:red;} </style> this works (puts smiley after "buu!" , before "some more") <span class="mystyle">buuu!</span>a more this not work - colors somevalue in red, there no smiley. <input class="mystyle" type="text" value="somevalue"> what doing wrong? should use pseudo-selector? note: cannot add span around input , because being generated third-party control. :after , :before not supported in internet explorer 7 , under, on elements. it's not meant used on replaced elements such form elements (inputs) , image elements . in other words it's impossible pure css. however if using jquery can use $(".mystyle").after("add smiley here"); api docs on .after to ...
Read more