Being new to shell scripting I hope this question isn't too elementary but here goes.
I run a grep statement - grep 'sqr' sqrmodule.par and the grep statement returns correctly the information that I'm looking for. Now I want to take the output from the grep statement and load it into a... (2 Replies)
i have such awk working fine but how to use variable instead of strings
awk '/asdasd.*asda.*asdasd/' file2.txt
This is not working:
awk '/${a}.*${b}.*${c}/' file2.txt
Thanks & regards
Peter (7 Replies)
Hi,
I have a reqmt as i have some values in array and I want to search each value in a file by grep command.
Here goes my scripting:
#!/bin/ksh
set -A ArrayA CENTER LEFT RIGHT
echo "ArrayA contains: ${ArrayAİ*¨}"
grep -e "${ArrayAİ*¨}" filename.txt
The above grep is working for... (4 Replies)
Help,
I have a ksh script that has some variables within the grep command, I am then setting another variable that greps the variables that have greps within them.
cat $WORKINGDIR/meter_list.txt | while read meter
do
serialnum=$(cat masterlogfile.txt | grep "$meter" | awk '{ print $19 }'... (7 Replies)
Hello,
I want to only print lines where variables occur more than once using grep.
For eg:
Input:
$this is a comment
int a,b,c,b;
int b,c;
int d,e;
int f,g,f;
x=y+5;
For the above input, the output would be
int a,b,c,b;
int f,g,f;
I have done grep... (3 Replies)
I would like to know if grep can extract the following requirement.
I have the folllowing piece of SQL in a file and need to grep the FROM part.
mp db-ter-fast-export C100_Input_Target_Table__table_ "${DB}"'/teradata.dbc' -select 'SELECT UPPER(trim(proj_id)) as proj_id,
latest_job_id,... (9 Replies)
Good day Geeks,
Am having an issue with using variables in a rather simple script, the script is as follows:
#!/bin/bash
### Script written by Adigun Gbenga
### Date: April 28, 2012
array=( 1 2 3 4 5 29 7 8 9... (6 Replies)
Hello, this will be my first post. I've been browsing around for a bit and have found a lot of useful information on here, hopefully a solution can be provided to me.
Issue:
Alright, I am looking to search for strings within a file using variables.
I have a script that will accept 3 or 4... (2 Replies)
Use and complete the template provided. The entire template must be completed. If you don't, your post may be deleted!
1. The problem statement, all variables and given/known data:
The issue I am having is part of a validation problem. My script will validate 3 or 4 parameters entered by the... (4 Replies)
Hi, I'm currently trying to use variables in grep on my script. Printing the variable via echo works fine. Also, if I hard coded the date of the appointment it works just fine. But, if I try to use the $DATE as an argument in grep it doesn't do anything.
#!/bin/bash
DATE=${2}/${3}/${1} ... (6 Replies)
Discussion started by: nuclearpenguin
6 Replies
LEARN ABOUT PHP
debug_zval_dump
DEBUG_ZVAL_DUMP(3) 1 DEBUG_ZVAL_DUMP(3)debug_zval_dump - Dumps a string representation of an internal zend value to outputSYNOPSIS
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)