changes 99998 to my stored $jobnum
starting with my j99998a.pst file (using $src as path location)
and outputting to new name with my $jobnum as the new filename
Just follow all of the double quotes to make sure variables all line up correctly.
I actually do several of these when setting up repetitive work - copying a global use file and customizing it for my current job/task. (By habit, I use 99998 as a dummy number - I found that using 99999 could run into issues since "99999" could be a mask for outputs or other uses.)
I have a 'sed' editor command trying to read contents of one file into another.
v=t1
PX=25
sed '/for/ r /$v/ext/tsfirmfiles.${PX}' /bb/bin/px${PX}.files.2b.deleted > result
I cannot come up with the syntax so 'sed' would do a variable substitution.
Any idea? Thanks a for help. -A (1 Reply)
Hi,
I am facing a strange problem. I have a script that used the following to search and replace text:
sed 's/'"${find_var_parm}"'/'"${find_var_filter}"'/g' $ParmFile > $TempFile
The values of $find_var_parm and $find_var_filter are set based on search criteria. The above seems to be working... (2 Replies)
Hi Chaps...
I have a log file as below:-
01 Oct 2009 12:57:56 DEBUG :
01 Oct 2009 12:14:21 DEBUG :.....
.
.
.
.05 Oct 2009 14:31:56 DEBUG :....
.
.
.05 Oct 2009 12:57:56 DEBUG :....
06 Oct 2009 01:23:11 DEBUG :....
.
.
.06 Oct 2009 12:53:46 DEBUG :.... (4 Replies)
Hi,
I have to insert a line at a particular line number in my file. But the line number is not fixed and it will vary every time. So, I have to use a variable to get the line number
I know we can use sed to insert lines at a particular line number but it does not work with variable... (3 Replies)
Give the code:
set line = 2
set year = `sed -n '2p' file
while ($line < 500)
echo $line > f.txt
@ line = $line + 1
end
How do I utilize the variable $line in the code instead of the number 2. I'm using this in a while loop and counter. I've tried quoting it, double/single... (1 Reply)
Hello, I searched the forum and unable to find a solution for my particular problem. I have a text file I'm trying to insert some text using sed after finding a pattern..
File contains in one line
Invoice date: xx/xx/xxxx Balance: $$$$ Name: xxxxxxxxxxxxxx
Trying to insert Invoice "Number:... (3 Replies)
Problem with the code below is that the value of the variable is not getting substituted in the sed expression.
#/bin/csh
set UNIX_ID="rajibd"
set X_ID="xrajibd"
sed -n 's/$UNIX_ID/$X_ID/g' passwd
When run , it is not giving expected output as shown below :
... (4 Replies)
Hi I am trying to do the following in a script find a string and add in a block of text two lines above on the command line this works fine
#/usr/bin/cat /usr/local/etc/dhcpd.conf_subnet | /usr/xpg4/bin/sed -n -e '1h;1\!H;${;g;s/}.*#END of 10.42.33.0/#START of RANGE $dstart\:option... (3 Replies)
I'm trying to change "F" to "G" in lines after the first one:
'FUE.SER' 5
1 1 F0501 F0401 F0502
2 1 F0301 E0501 F0201 E0502 F0302
3 1 F0503 E0503 E0301 E0201 E0302 E0504 F0504
4 1 F0402 F0202 E0202 F0101 E0203 F0203 F0403
5 1 F0505 E0505 E0303 E0204 E0304 E0506... (10 Replies)
Discussion started by: larrl
10 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)