Sponsored Content
Top Forums Shell Programming and Scripting Passing space seprated variables in awk Post 302556705 by manubatham20 on Monday 19th of September 2011 10:34:16 AM
Old 09-19-2011
Passing space seprated variables in awk

Hi,

How to pass space seprated variables in awk. (HP-UX, sh shell)

I have the problem like below:

Code:
var="Hello"
var2="Manu Batham"
echo $var2 | awk -v variable=${var} '{ printf "%s %s", variable, $1}'

and its output is:
Hello Manu

while output should be:
Hello Manu Batham

Please help me to accept it spaces as well.

Thanks,
Manu
 

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Passing Variables to AWK

Does anybody have an explanation for the following: The following scripts runs fine on IRIX64 6.5 but has bugs on Solaris 8. #! /bin/sh echo run only on an SGI machine echo type in linenumber read j echo value read value awk -f rmspass2 level=$value $j'step1.mlf' When the script is... (5 Replies)
Discussion started by: AreaMan
5 Replies

2. Shell Programming and Scripting

Passing Variables to Awk

Hi I have a unix shell script with an awk statement. I would like to print some of the fields of an input file. However, I would like to print them dynamically, ie by passing the literal $1 $3 into the script to define the output. I have tried the following: variable1='$1' awk... (2 Replies)
Discussion started by: Bab00shka
2 Replies

3. Shell Programming and Scripting

Passing awk Variables

I am trying to pass the results from a variable gathered from awk, however when I echo the 'PARSE' and 'SUB', the response is blank. This is my command. awk -F= '/Unit/''{ PARSE=substr($2,1,5) ; SUB=substr($2,1,1) }' inputfile.lst Is this a kind of valid attempt or am I obligated to declare... (3 Replies)
Discussion started by: gozer13
3 Replies

4. Shell Programming and Scripting

Passing awk variables to shell

Hi. I need to parse file and assign some values to variables, right now i do like below MYHOMEDIR=`awk '/Home/ {print $NF}' output.txt` MYSHELL=`awk '/Shell/ {print $NF}' output.txt` PRGRP=`awk '/Primary/ {print $NF}' output.txt` SECGRP=`awk '/Second/ {print $NF}' output.txt` In this... (10 Replies)
Discussion started by: urello
10 Replies

5. Shell Programming and Scripting

ksh passing to awk multiple dyanamic variables awk -v

Using ksh to call a function which has awk script embedded. It parses a long two element list file, filled with text numbers (I want column 2, beginning no sooner than line 45, that's the only known thing) . It's unknown where to start or end the data collection, dynamic variables will be used. ... (1 Reply)
Discussion started by: highnthemnts
1 Replies

6. Shell Programming and Scripting

Passing variables into AWK

I'm trying to use awk to write new entries to a hosts file if they don't exist. I need to do so depending on the type of system I have. Below is what I have, but it isn't working. awk -v myip1=$IP1 myip2=$IP2 myhost1=$HOST1 myhost2=$HOST2' BEGIN { mqhost1=0; mqhost2=0; stap1=0; stap2=0; } ... (4 Replies)
Discussion started by: Boomn4x4
4 Replies

7. Shell Programming and Scripting

PAssing variables to awk arithmetic

Hi all, I am wanting to pass variables from a file to an awk arithmetic formula. When I use the formula with the value it works well. As soon as I make these variables I get an inf (infinity) response. I can certainly echo the variables back and they look correct. My googling for answers has... (3 Replies)
Discussion started by: gafoleyo73
3 Replies

8. Shell Programming and Scripting

Passing variables to awk

Hi guys, I need to fetch data from logfile between two given dates,i got the below code from our forum.It works perfect,but i need to enter the value dynamically to awk while running. awk '/2012 Jun/{p=1}!/2012 Jul/ && prev~/2012 Jul/ && p{p=0}{prev=$0}p' file i tried the below code,but... (4 Replies)
Discussion started by: mohanalakshmi
4 Replies

9. Shell Programming and Scripting

Passing awk variables to bash variables

Trying to do so echo "111:222:333" |awk -F: '{system("export TESTO=" $2)}'But it doesn't work (2 Replies)
Discussion started by: urello
2 Replies

10. Shell Programming and Scripting

awk - passing variables in and out

Am looking to pass some Linux environment variables into AWK , can I simply use the -v option ? awk -F: -v AHOME=$HOME '{ if {rm AHOME/file.txt a=2 } }' config.txt ... (4 Replies)
Discussion started by: alldbest
4 Replies
DEBUG_ZVAL_DUMP(3)							 1							DEBUG_ZVAL_DUMP(3)

debug_zval_dump - Dumps a string representation of an internal zend value to output

SYNOPSIS
void debug_zval_dump (mixed $variable, [mixed $...]) DESCRIPTION
Dumps a string representation of an internal zend value to output. PARAMETERS
o $variable - The variable being evaluated. RETURN VALUES
No value is returned. EXAMPLES
Example #1 debug_zval_dump(3) example <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump(&$var1); ?> The above example will output: &string(11) "Hello World" refcount(3) Note Beware the refcount The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump(3). This behavior is further compounded when a variable is not passed to debug_zval_dump(3) by reference. To illustrate, consider a slightly modified version of the above example: Example #2 <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump($var1); // not passed by reference, this time ?> The above example will output: string(11) "Hello World" refcount(1) Why refcount(1)? Because a copy of $var1 is being made, when the function is called. This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value): Example #3 <?php $var1 = 'Hello World'; debug_zval_dump($var1); ?> The above example will output: string(11) "Hello World" refcount(2) A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening? When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump(3)), PHP's engine opti- mizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made, but only at the moment of writing. This is known as "copy on write." So, if debug_zval_dump(3) happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call. SEE ALSO
var_dump(3), debug_backtrace(3), References Explained, References Explained (by Derick Rethans). PHP Documentation Group DEBUG_ZVAL_DUMP(3)
All times are GMT -4. The time now is 01:30 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy