I guess you have GNU/Linux, then it is -d or --date= - or you get an error message instead of an output.
Further, it should be a dash not an underbar, and %y not %Y, as was already addressed by R.Singh.
Hi,
I'd like to grep a variable that I saved in the program.
Like
grep '0\$variable1' file1
Does someone know what's wrong with this command?
Thanks a lot! (2 Replies)
Hi, I can't get this script to work (returns 0, should return 3):
$ cat A.lst | \
while read LINE
do
echo "$LINE"
grep -c "$LINE" B.tmp
done> > > > >
Socket
0
$
but in contrast this one works fine (returns 3 as expected):
$ LINE=Socket
$ grep -c $LINE B.tmp
3
$ (5 Replies)
Hi all,
I am trying to do a simple thing in my mind. However I am fairly new to bash. What I need to do is create a folder for each partition on each CD, and each partition has a unique name (with spaces in it, do not ask why, it is already done :confused: ) . All CD's will show up... (2 Replies)
can i grep a variable
say i have a variable var=`hostname` and I want to make an if statement like
if grep "esp-ueh" $var;then......
how can i do this
I dont want to store this variable in a file and the grep it because my script will be used at the same time on multiple stations and then that... (9 Replies)
how can I use grep with a variable to find a value?
cat data.out
Hello World
grep "Hello World" data.out
Hello World ## Value found
I want to do something like this but can't seem to get it to work any
suggestions would be appreciated.
var="Hello World"
grep $var data.out (3 Replies)
I want to search a text in file but that file is pointing to variable.
ex:
file=there was nothing special
grep "there was nothing" $file
but its not working . Can u let me know that how we can use variable($file) in grep command.
Please use code tags (6 Replies)
I have a pattern like:
column "5" is missing
PS: the no is in double quotes.
The number usally changes, so we use a loop to grep.
grep 'column "$number" is missing' filename.txt
But it is not working....
How to solve this? (2 Replies)
Hello,
I usually search extensively and have to date found what I've needed. However, this one's got me stumped. I need to create a variable as follow. The issue however is that upon execution, it freezes. $var1 isn't always present in usage.log and this is fine but I'd like it to continue with... (6 Replies)
I've got a file that I'm trying to grep through that looks like this:
alpha1
alpha2
alpha3
beta1
beta2
gamma5
gamma6
gamma7
gamma8
gamma9
and I want the output to only contain the line with the highest value for each, so the output I want is:
alpha3
beta2
gamma9
I also need... (11 Replies)
Hi,
I am trying to grep one variable over the other variable
Example:
i=abc
j=ab
grep $j $i
I am getting this error:
The error is due to $i being variable and not file. I know I could do it by putting the value of abc in a file
and then greping it. (1 Reply)
Discussion started by: pinnacle
1 Replies
LEARN ABOUT PHP
datetime.sub
DATETIME.SUB(3) 1 DATETIME.SUB(3)DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_sub (DateTime $object, DateInterval $interval)
Subtracts the specified DateInterval object from 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.sub(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-10
Example #2
Further DateTime.sub(3) examples
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-19 13:59:30
1992-08-15 19:56:58
Example #3
Beware when subtracting months
<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-03-30
2001-03-02
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.add(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.SUB(3)