Sponsored Content
Full Discussion: Help with Dynamic variable
Top Forums Shell Programming and Scripting Help with Dynamic variable Post 302484563 by m.d.ludwig on Saturday 1st of January 2011 12:14:25 PM
Old 01-01-2011
You need an quoted "$" in front of "USER" for the eval to work:
Code:
eval USER=\$USER${X}

This User Gave Thanks to m.d.ludwig For This Post:
 

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Dynamic Variable Declatation

Evening all, Have been trying to create the following environment variable: ${MD_SYSTEM}_ZZ_EMAIL_SUPPORT="myname@domain.com" However when the script that contains the above is executed it returns: ksh: MDQA_ZZ_EMAIL_SUPPORT=myname@domain.com: not found Is what I'm trying to do... (2 Replies)
Discussion started by: Cameron
2 Replies

2. UNIX for Dummies Questions & Answers

Dynamic variable values

Bit of a newbie :D with regard to unix scripting and need some advice. Hopefully someone can help with the following: I have a predefined set of variables as follows: AAA_IP_ADD=1.1.1.1 BBB_IP_ADD=2.2.2.2 I have a funnction call which retrieves a value into $SUPPLIER which would be... (3 Replies)
Discussion started by: ronnie_uk
3 Replies

3. Shell Programming and Scripting

Dynamic variable assignment

Hi All, I have the below scenario: A file test.cfg with three fields>> DATA1 DATA2 DATA3 In the script I need to assign each of the fields to variables. The number of fields will not be constant (this case we have three). Im trying to do something like this: NUM=1 OUT_DAT_NO=3 ... (4 Replies)
Discussion started by: deepakgang
4 Replies

4. Shell Programming and Scripting

dynamic variable name

I found one post in another site with a solution for my problem the below solution should explain what I want. #!/bin/sh first="one" second="two" third="three" myvar="first" echo ${!myvar} But this gives error 'bad substitution' System info SunOS sundev2 5.9... (3 Replies)
Discussion started by: johnbach
3 Replies

5. Shell Programming and Scripting

Dynamic variable assignment

Hi all, I’m very new to UNIX programming. I have a question on dynamic variable 1. I’m having delimited file (only one row). First of all, I want to count number of columns based on delimiter. Then I want to create number of variables equal to number of fields. Say number of... (5 Replies)
Discussion started by: mkarthykeyan
5 Replies

6. Shell Programming and Scripting

dynamic variable value assignmnet

QUERY IN BRIEF Listing the query in short #! /bin/csh -f #say i have invoked the script with two arguments : a1 and 2 set arg = $1 # that means arg = a1 echo "$arg" #it prints a1 #now what i want is: echo "$a1" #it will give error message :a1 undefined. #however what i need is that the... (2 Replies)
Discussion started by: animesharma
2 Replies

7. Shell Programming and Scripting

Dynamic file name in variable

Hi guys. i have a requirment as below. I have a scripts which perform for loop for i in /backup/logs -- it will give all the logs file SC_RIO_RWVM_20120413064217303.LOG SC_RIO_RWXM_20120413064225493.LOG SC_RIO_RXXM_20120413064233273.LOG ... do open script.sh ---- in this file... (3 Replies)
Discussion started by: guddu_12
3 Replies

8. Shell Programming and Scripting

Passing dynamic variable within another variable.

I have a small program which needs to pass variable dynamically to form the name of a second variable whose value wil be passed on to a third variable. ***************** Program Start ****************** LOC1=/loc1 PAT1IN=/loc2 PAT2IN=/loc3 if ; then for fpattern in `cat... (5 Replies)
Discussion started by: Cyril Jos
5 Replies

9. Shell Programming and Scripting

Dynamic variable assignment

My Code : -------------------------------------------- #!/bin/bash for i in `echo server1 server2` do eval ${i}_name = "apache" echo ${i}_name done -------------------------------------------- Current output : >./test.sh ./test.sh: line 5: server1_name: command not found... (3 Replies)
Discussion started by: sameermohite
3 Replies

10. Shell Programming and Scripting

Dynamic variable name in bash

Hello, I'm trying to build a small script to store a command into a dynamic variable, and I have trouble assigning the variable. #!/bin/bash declare -a var1array=("value1" "value2" "value3") var1arraylength=${#var1array} for (( i=1; i<${var1arraylength}+1; i++ )); do mkdir... (7 Replies)
Discussion started by: alex2005
7 Replies
ALTER 
GROUP(7) PostgreSQL 9.2.7 Documentation ALTER GROUP(7) NAME
ALTER_GROUP - change role name or membership SYNOPSIS
ALTER GROUP group_name ADD USER user_name [, ... ] ALTER GROUP group_name DROP USER user_name [, ... ] ALTER GROUP group_name RENAME TO new_name DESCRIPTION
ALTER GROUP changes the attributes of a user group. This is an obsolete command, though still accepted for backwards compatibility, because groups (and users too) have been superseded by the more general concept of roles. The first two variants add users to a group or remove them from a group. (Any role can play the part of either a "user" or a "group" for this purpose.) These variants are effectively equivalent to granting or revoking membership in the role named as the "group"; so the preferred way to do this is to use GRANT(7) or REVOKE(7). The third variant changes the name of the group. This is exactly equivalent to renaming the role with ALTER ROLE (ALTER_ROLE(7)). PARAMETERS
group_name The name of the group (role) to modify. user_name Users (roles) that are to be added to or removed from the group. The users must already exist; ALTER GROUP does not create or drop users. new_name The new name of the group. EXAMPLES
Add users to a group: ALTER GROUP staff ADD USER karl, john; Remove a user from a group: ALTER GROUP workers DROP USER beth; COMPATIBILITY
There is no ALTER GROUP statement in the SQL standard. SEE ALSO
GRANT(7), REVOKE(7), ALTER ROLE (ALTER_ROLE(7)) PostgreSQL 9.2.7 2014-02-17 ALTER GROUP(7)
All times are GMT -4. The time now is 04:43 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy