I have the following error:
ls -lt | awk 'BEGIN NR > 1 { print $2, $9 }'
Syntax Error The source line is 1.
The error context is
BEGIN >>> NR <<< > 1 { print $2, $9 }
awk: 0602-500 Quitting The source line is 1.
What I want to do is ls a directory, skip the first... (3 Replies)
Hello,
I have the following command that does 2 searches.
awk '{if ($0 ~ /STRING1/) {c++} }{if ( c == 2 ) {sub(/STRING1/,"NEWSTRING") } } { print }' FILE
How do I search up after the first search?
thanks (4 Replies)
i have a little awk script that I use looks this:
awk '{if (FNR==1){print FILENAME; print $0}else print $0}' file1...file2....fi... > bundled.
i have completely forgotten how to unbundle this. I have tried several different approaches and still can not remember how to unbundle the file bundled.... (2 Replies)
I am trying to read through a file, gather the states in that file and change it from an abbreviation to the ful text.
Can anyone provide some assistance.
Thanks!! (4 Replies)
How I can rid of the following presentation du -sk /u*/oradata/TEST/*.dbf |awk '{print total+=$1} 1.28003e+06
4.35109e+06
4.36134e+06
4.4535e+06
5.47752e+06
5.48777e+06
7.52554e+06
7.73036e+06
9.06158e+06
:confused: thank you (3 Replies)
Can anyone help with this this one liner:
nawk -v RS='' '$1=$1' InputFile
What I have in the file:
0.0013985457223116
-0.0002338180925628
0.0
0.0003709430584958
-0.0005763523138347
0.0
And the output I want:
0.0013985457223116 -0.0002338180925628 0.0
0.0003709430584958... (1 Reply)
I have a script problem that I am not able to solve due my very limited understanding of unix/awk.
This is the contents of test.sh
awk '{print $1}'
From the prompt if I enter:
./test.sh Hello World
I would expect to see "Hello" but all I get is a blank line. Only then if I enter "Hello... (2 Replies)
Use and complete the template provided. The entire template must be completed. If you don't, your post may be deleted!
1. The problem statement, all variables and given/known data:
im using ls -l | xargs | awk '{what ever files here}'
im trying to get something that looks like this... (7 Replies)
Hi Experts,
I am trying to get system output to capture inside awk , but not working:
Please advise if this is possible :
I am trying something like this but not working, the output is coming wrong:
echo "" | awk '{d=system ("date") ; print "Current date is:" , d }'
Thanks, (5 Replies)
Discussion started by: rveri
5 Replies
LEARN ABOUT PHP
datetime.setisodate
DATETIME.SETISODATE(3) 1 DATETIME.SETISODATE(3)DateTime::setISODate - Sets the ISO date
Object oriented style
SYNOPSIS
public DateTime DateTime::setISODate (int $year, int $week, [int $day = 1])
DESCRIPTION
Procedural style
DateTime date_isodate_set (DateTime $object, int $year, int $week, [int $day = 1])
Set a date according to the ISO 8601 standard - using weeks and day offsets rather than specific dates.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $year
- Year of the date.
o $week
- Week of the date.
o $day
- Offset from the first day of the week.
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.0 | |
| | |
| | Changed the return value on success from NULL to |
| | DateTime. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
DateTime.setISODate(3) example
Object oriented style
<?php
$date = new DateTime();
$date->setISODate(2008, 2);
echo $date->format('Y-m-d') . "
";
$date->setISODate(2008, 2, 7);
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create();
date_isodate_set($date, 2008, 2);
echo date_format($date, 'Y-m-d') . "
";
date_isodate_set($date, 2008, 2, 7);
echo date_format($date, 'Y-m-d') . "
";
?>
The above examples will output:
2008-01-07
2008-01-13
Example #2
Values exceeding ranges are added to their parent values
<?php
$date = new DateTime();
$date->setISODate(2008, 2, 7);
echo $date->format('Y-m-d') . "
";
$date->setISODate(2008, 2, 8);
echo $date->format('Y-m-d') . "
";
$date->setISODate(2008, 53, 7);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2008-01-13
2008-01-14
2009-01-04
Example #3
Finding the month a week is in
<?php
$date = new DateTime();
$date->setISODate(2008, 14);
echo $date->format('n');
?>
The above examples will output:
3
SEE ALSO DateTime.setDate(3), DateTime.setTime(3).
PHP Documentation Group DATETIME.SETISODATE(3)