Bash script read specific value from files of an entire folder
Hello,
I heva a problem creating a script that read specifc value from all the files of an entire folder
I have a number of email files into a directory and i need to extrect from each file 2 specific values.
After that i have to put them into a new file that looks like that:
Code:
To: value1
value2
This is what i want to do, but i don't know how to create the script:
Code:
ls -l | awk '{print $9 }' >tmpfile - i am putting the name of the files into a temp file
date=`date +"%T"` - used for the name of a file
var1=`cat tmpfile | grep "To: 0" | awk '{print $2 }' | cut -b -10 ` - the first specific value from file(phone number)
var2=cat file | grep Subject | awk '{print $2$3$4$5$6$7$8$9$10 }' - the second specific value from file(subject)
echo "To: 4"$var1"" > sms-$date - put the first value in a new file on the first row
echo ""$var2"" >>sms-$date put the second value in the same file on the second row
and do the same for every file in the directory
I tried using while and for functions but i couldn't finalize the script
Thank You
Last edited by Scott; 03-28-2014 at 11:25 AM..
Reason: Please use code tags
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#!/bin/sh
cd workspace
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echo ${array}
But it doesn't change to workspace
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Discussion started by: valdez
13 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)