Hello,
I want to rename multiple files at a time and I don't know how to do it.
I have various ".mp3" files, like "band name - music name.mp3" and I want to remove the "band name" from all files.
Anybody knows how to do it using shell script or sed or even perl?
Thanks (7 Replies)
I have several hundred files in one directory which I need to move to another directory with the new extension, for example:
/bb/data/rptmgr* are in the source directory need to be moved to
/bb/data55/rptmgr*.new
Is there an efficient way to do it? Thanks -A (4 Replies)
Hi all, I have some files like:
pickup.0000043200.t001.t001.data
pickup.0000043200.t001.t002.data
pickup.0000043200.t002.t001.data
pickup.0000043200.t002.t002.data
pickup.0000043200.t003.t001.data
pickup.0000043200.t003.t002.data
I need to rename these files to
... (4 Replies)
Hi everyone,
I'm very green in Linux.
Please help me to solve my problem.
I have thousands of files and I want to change their names.
They have naming convection: prefix_date_date+1_suffix.nc
prefix: ext-GLORY
date_date+1: 20020101_20020102
and two types of suffix: gridV_R20020130 and... (3 Replies)
Hi,
In my directory I have many files, for e.g.
file_123
file_124
file_125
file_126
file_127
Instead of renaming these files one by one, I would like to rename them at a same time using same command... they should appear like
123
124
125
126
127
What command(awk or ls or... (3 Replies)
Hi,
I have hundreds of files with XXX in their file name and I want to rename all of them with YYY in place of XXX.
for ex:
$ ls -1
123XXX789
345XXX678
Output
$ ls -1
123YYY789
345YYY678
I know we can loop in each file and sed to replace and rename each file but ren *XXX* *YYY*... (4 Replies)
I have multiple files in folder which i want to rename. hence I am using the below command in my script by I get an error:
export XXX_LOG_DIR="${LOG_DIR}/${XXX_HOST}/xxx/${REPORT_DATE}"
mv $XXX_LOG_DIR/*.audit.gz $XXX_LOG_DIR/*.audit.log.gz
But I get the below error:
mv: target... (5 Replies)
Hey guys,
I have wrote the following script to apply a module named "trinity" on my files. (it takes two input files and spit a trinity.fasta as output)
#!/bin/bash -l
#SBATCH -p node
#SBATCH -A <projectID>
#SBATCH -n 16
#SBATCH -t 7-00:00:00
#SBATCH --mem=128GB
#SBATCH --mail-type=ALL... (1 Reply)
OS : Oracle Linux 6.8
shell : bash
As shown below, I have multiple files like below (query1-extract_aa, query1-extract_ab, query1-extract_ac, ....)
$ ls -l
total 235680
-rw-rw-r-- 1 reportusr reportusr 30M May 3 11:25 query1-extract_aa
-rw-rw-r-- 1 reportusr reportusr 30M May 3 11:25... (5 Replies)
Discussion started by: kraljic
5 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)