Thanks for your input Franklin52. However, I need to pass it as a variable, since the program(s) I shall be executing subsequently on the input file will depend on the individual values I extract.
Thanks,
Guss.
You can use variables like this:
This User Gave Thanks to Franklin52 For This Post:
hi;
i have a file containing lines like:
1|1069108123|96393669788|00963215755711|2|0|941||;serv:Pps6aSyria;first:0;bear
i want to extract the second, third and fourth record of each line and store it in a file ";" seperated
this is what i wrote
while read line
do
... (3 Replies)
I am a newbie to awk and c programming, however am not a unix newbie. However, I do need help with a kshell script I am writing. It is almost complete, the last step is killing me. Any help would be greatly appreciated. What I am trying to do is cat a text file that has usernames. Then, using... (2 Replies)
Hi,
I'm new with this stuff, but I hope you can help me.
This is what I'm trying to do:
for id in $var; do
awk '{if ($1 == $id) print $2}' merg_data.dat > neigh.tmp
done
I need that for every "id", awk search the first column of the file merg_data.dat which contains "id" and... (3 Replies)
I am writing a script where I need awk to test if two columns are the same and shell to do something if they are or are not.
Here is the code I'm working with:
@ test = 0
...
test = `awk '{if($1!=$2) print 1; else print 0}' time_test.tmp`
#time_test.tmp holds two values separated by a space... (3 Replies)
Hi All,
I am new to AWK programming. I have the following for loop in my awk program.
cat printhtml.awk:
BEGIN
-------- <some code here>
END{
----------<some code here>
for(N=0; N<H; N++)
{
for(M=5; M<D; M++) print "\t" D "";
}
-----
}
... (2 Replies)
Hello All,
May i please why my shell variable is not getting passed into awk script.
#!/bin/bash -vx
i="1EB07C50"
/bin/awk -v ID="$i" '/ID/ {match($0,/ID/);print substr($0,RSTART,RLENGTH)}' /var/log/ScriptLogs/keys.13556.txt
Thank you. (1 Reply)
I have file called in in.txt contains with the below lines I want to display the lines between the value which I would be passing.
one
two
three
four
five
ten
six
seven
eight
Expected output if I have passed one and ten
two
three
four
five (8 Replies)
Discussion started by: mychbears
8 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)