![]() |
|
|
google unix.com
|
|||||||
| Forums | Register | Forum Rules | Links | Albums | FAQ | Members List | Calendar | Search | Today's Posts | Mark Forums Read |
| Shell Programming and Scripting Post questions about KSH, CSH, SH, BASH, PERL, PHP, SED, AWK and OTHER shell scripts and shell scripting languages here. |
More UNIX and Linux Forum Topics You Might Find Helpful
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Global variable declaration in shell scripting | susilgandhi | Shell Programming and Scripting | 1 | 02-06-2008 12:59 AM |
| Help with variable declaration | dsravan | Shell Programming and Scripting | 1 | 12-08-2006 03:15 PM |
| Double Quotes within a variable | burton_1080 | Shell Programming and Scripting | 4 | 12-01-2005 01:44 PM |
| double variable substitution | asal_email2 | UNIX for Dummies Questions & Answers | 1 | 06-14-2005 01:35 AM |
| Variable declaration | laila63 | High Level Programming | 2 | 07-07-2004 04:37 PM |
|
|
LinkBack | Thread Tools | Search this Thread | Rate Thread | Display Modes |
|
||||
|
double variable declaration
i have variables with different values as below in KSH
AU_Holiday=1 SG_Holiday=0 KR_Holiday=1 JP_Holiday=0 $country_cd is the second variable which be one of 'AU' 'SG' 'KR' 'JP' which comes in a for loop. form that i need to derive the variable like first one and check whether it is equal to 1 or not. if [ $"$country_cd"_Holiday = "1" ] then do something else do something fi the above logic did not work can someone please help ? |
| Bookmarks |
| Thread Tools | Search this Thread |
| Display Modes | Rate This Thread |
|
|