HI
I am trying to store the output of this awk command
awk -F, {(if NR==2) print $1} test.sr
in a variable when I am trying v= awk -F, {(if NR==2) print $1} test.sr
$v = awk -F, {(if NR==2) print $1} test.sr
but its not working out .
Any suggestions
Thanks
Arif (3 Replies)
Hi unix gurus,
I am trying to store the result of a command into a variable.
But it is not getting stored.
x='hello'
y=echo $x | wc -c
but it is giving the output as 0(zero)
Pls help me its very urgent (7 Replies)
Hi,
i have some files in one directory(say some sample dir) whose names will be like the following.
some_file1.txt
some_file2.txt.
i need to get the last modified file size based on file name pattern like some_
here i am able to get the value of the last modified file size using the... (5 Replies)
Hi All,
Hope someone can advise here as I have been struggling to find a syntax that works here. I have tried a stack of combination I have seed in the forums but I think because I have needed to use "" and `` in the statments another method is found.
I am reading in lines with the following... (1 Reply)
Hi all,
I am new to Linux/shell scripting having moderate knowledge.
In my script, I need to get execution time of a command (say 'ls') in mili seconds level. For this i tried using "time" command to retrieve the total execution time in milli seconds. But, the problem is that, how to save... (9 Replies)
My script below seems to be choking because I need the the output of the find command to be stored as a variable that can then be called by used lower in the script.
#!/bin/bash
cd "/resumes_to_be_completed"
var1=find . -mmin -1 -type f \( -name "*.doc" -o -name "*.docx" \)... (1 Reply)
I'm working on a script in which gives certain details in its output depending on user-specified options. So, what I'd like to do is something like:
if
then
awkcmd='some_awk_command'
else
awkcmd='some_other_awk_command'
fi
Then, later in the script, we'd do something like:
... (5 Replies)
I have a below syntax its working fine...
var12=$(ps -ef | grep apache | awk '{print $2,$4}')
Im getting expected output as below:
printf "%b\n" "${VAR12}"
dell 123
dell 456
dell 457
Now I wrote a while loop.. the output of VAR12 should be passed as input parameters to while loop and results... (5 Replies)
Hi All,
I want to run multiple sql queries and store the data in variable but i want to use sql command only once. Is there a way without running sql command twice and storing.Please advise.
Eg :
Select 'Query 1 output' from dual;
Select 'Query 2 output' from dual;
I want to... (3 Replies)
Hi,
I would like use the output of my cut command as a variable in my following awk command. Here's what I've written.
cut -f1 info.txt | awk -v i=xargs -F'' '{if($6 == $i) print $20}' summary.txt
Where obviously the 'xargs' doesn't do what I want. How can I pass my cut result to my awk... (3 Replies)
Discussion started by: heyooo
3 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)