Sponsored Content
Top Forums Shell Programming and Scripting Variable name change in a loop Post 302436873 by canduc17 on Tuesday 13th of July 2010 10:44:58 AM
Old 07-13-2010
Variable name change in a loop

I need to do something like this:
Code:
for I in var1 var2 var3 ; do
  $I = "Something calculated inside the loop"
done

Obviously it doesn't work...but is it possible doing that in other ways?

Thanks in advance.
 

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

change if-else loop to while-do

Hello friends, i wrote a script which includes a couple of if-else loops but i have to change it to while-do loop as it is not allowed to use "break" in if-else loop in bash. #!/bin/bash -x NUM=`find . -name ufsdump_output1.txt | xargs egrep "End-of-tape detected"` if ; then echo "OK!"... (6 Replies)
Discussion started by: EAGL€
6 Replies

2. Shell Programming and Scripting

Variable value dosent change after the loop changes it

i am new to shell scripting. this is similar to the code which i was writing for some other work. i want the variable 'x' to have the value which it will finally have at the end of the loop ( the number of directories ). but the value of 'x' only changes inside the loop and it remains '0' out-side... (3 Replies)
Discussion started by: chathura666
3 Replies

3. Shell Programming and Scripting

Change Variable Value from Multiple Scripts and Use these Variable

Hi to All, Please find below details. file_config.config export file1_status="SUCCESS" export file2_status="SUCCESS" file_one.sh I am calling another two shell script from these script. I need to pass individual two script status (If it's "FAILED") to file_main.sh. file_main.sh I... (2 Replies)
Discussion started by: div_Neev
2 Replies

4. Shell Programming and Scripting

[SHELL: /bin/sh] For loop using variable variable names

Simple enough problem I think, I just can't seem to get it right. The below doesn't work as intended, it's just a function defined in a much larger script: CheckValues() { for field in \ Group_ID \ Group_Title \ Rule_ID \ Rule_Severity \ ... (2 Replies)
Discussion started by: Vryali
2 Replies

5. Shell Programming and Scripting

printing variable with variable suffix through loop

I have a group of variables myLINEcnt1 - myLINEcnt10. I'm trying to printout the values using a for loop. I am at the head banging stage since i'm sure it has to be a basic syntax issue that i can't figure out. For myIPgrp in 1 2 3 4 5 6 7 8 9 10; do here i want to output the value of... (4 Replies)
Discussion started by: oly_r
4 Replies

6. Shell Programming and Scripting

Change a field value in a loop

Hi guys, i have an executable file that contains several records and fields. One of the records has a variable filed that must be changed each time i want to execute the file. Would it be possible that i can use a loop to change the value of that field? Suppose that the field address is: Record... (5 Replies)
Discussion started by: saeed.soltani
5 Replies

7. Shell Programming and Scripting

Array Variable being Assigned Values in Loop, But Gone when Loop Completes???

Hello All, Maybe I'm Missing something here but I have NOOO idea what the heck is going on with this....? I have a Variable that contains a PATTERN of what I'm considering "Illegal Characters". So what I'm doing is looping through a string containing some of these "Illegal Characters". Now... (5 Replies)
Discussion started by: mrm5102
5 Replies

8. Shell Programming and Scripting

Doing a loop to change a value in a file

Hi, I have an N number of files in a directory. I like to write a shell script that would make identical plots for each one of these files. The files have names such as: t00001.dat t00002.dat t00003.dat t00004.dat t00005.dat . . . t00040.dat i.e. the... (4 Replies)
Discussion started by: lost.identity
4 Replies

9. Shell Programming and Scripting

[Solved] How to increment and add variable length numbers to a variable in a loop?

Hi All, I have a file which has hundred of records with fixed number of fields. In each record there is set of 8 characters which represent the duration of that activity. I want to sum up the duration present in all the records for a report. The problem is the duration changes per record so I... (5 Replies)
Discussion started by: danish0909
5 Replies

10. UNIX for Dummies Questions & Answers

Change argument in a for loop

In a "for i in *FD.CPY do" loop, I need to change .CPY to .layout so the executed command would be reclay blahFD.CPY >blahFD.layout What do I need to do to modify a copy of i to use after the > symbol? TIA (5 Replies)
Discussion started by: wbport
5 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 11:25 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy