I would like to extract the following fields from the text file attached. I copied the contents from a pdf file and pasted them into the text file so I can use awk to extract them. The layout is as listed below.
name1,name2,name3,name4,Title,designation,nationality,dob,
national ... (1 Reply)
Hi,
When i run the script ./script.sh sun, this give me no output, it should give me the list of file.
If i run the script without the argument it should send me echo inside usage().
What is the problem?
please help
-Adsi
#!/bin/sh
ROOT_PATH=/net/icebox/vol/local_images/spins... (2 Replies)
Hi,
I have this script, searches and sets variables, then searches and sets more variables from multiple files.
I'd need to debug it a bit.
#!/bin/bash
egrep $1 `find | grep MAGT` >/tmp/resRA-$$
thread=`sed -n '/{0x/ {s/^.*{0x\(*\).*/\1/p;q}' /tmp/resRA-$$`
tag=`sed -n '/Tag=/... (5 Replies)
Hi I am relatively new in shell scripting Below is the code which i developed but for some reason, it keeps giving me error:
/apps/bss/BatchProg/at1/batch/scripts/ksh/TBATLC02.ksh: syntax error at line 41 : `then' unmatched
#!/usr/bin/ksh... (4 Replies)
Hi all,
I am using the below given sqlplus command in my unix script to invoke a stored procedure which returns a value .It works fine.
RET_CODE=$(/opt/oracle/product/10.2.0.4.CL/bin/sqlplus -S $USER/$PASSWD@$DB_NAME <<EOF
EXEC MY_PKG.MY_SP (:COUNT);
PRINT COUNT;
commit;
... (6 Replies)
Hi,
I am trying to get squid up and running using a redirector process, and every time I try to load a web page, squid fails miserably.
Can some one with perl and squid knowledge take a look at these codes and tell if something is wrong here.
#!/usr/bin/perl
#
$| = 1;
@endings = qw/... (0 Replies)
I want to make a script that check for the argument passed to it and generates an error in case any character/string argument passed to it.
I am using below code, but its not working. can anyone help.
#!/bin/bash
if ]; then
echo 'An integer argument is passed to the script hence... (3 Replies)
I'm sorry if the title is really criptic, but I don't know how to phrase my problem.
I know I can't really ask for a solution, and I normally wouldn't but this is really escaping my abilities.
Antefacts.
I developed a program using the zeromq messaging library.
I got to a point where the... (11 Replies)
How can I debug this script?
I want to know what it is doing or not doing?
#!/bin/bash
#
#
if ; then
# Do the thing you want before suspend here
echo "we are suspending." > /tmp/systemd_suspend_test.txt
elif ; then
# Do the thing you want after resume here
echo "and we are... (21 Replies)
Discussion started by: drew77
21 Replies
LEARN ABOUT PHP
datetime.add
DATETIME.ADD(3) 1 DATETIME.ADD(3)DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::add (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_add (DateTime $object, DateInterval $interval)
Adds the specified DateInterval object to the specified DateTime object.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $interval
- A DateInterval object
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
EXAMPLES
Example #1
DateTime.add(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-11
Example #2
Further DateTime.add(3) examples
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-01 10:00:30
2007-06-05 04:03:02
Example #3
Beware when adding months
<?php
$date = new DateTime('2000-12-31');
$interval = new DateInterval('P1M');
$date->add($interval);
echo $date->format('Y-m-d') . "
";
$date->add($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.sub(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.ADD(3)