In the bash below the variable date displays in the echo. However when I use it in the for loop it does not. Basically, the user inputs a date then that date is converted to the desired format of (month-day-year, no leading 0). That input is used in the for loop to return every file that matches the .bam extension. I can not seem to figure out why the $date variable is not being passed to the path to search in? Thank you .
Code:
printf "\n\n"
printf "What is the date to analyze : "; read id
date -d "$id" +%-m-%-d-%Y ; read date #convert to month-date-year
echo "$date"
[ -z "$id" ] && printf "\n No date supplied. Leaving match function." && sleep 2 && return
for f in /home/cmccabe/Desktop/NGS/bam/$date/*.bam ; do
bname=`basename $f`
pref=${bname%%.bam}
echo "The bam files found in directory at $f are ${pref}"
done
Anybody know what's wrong with this syntax?
awk -v job="$job" 'BEGIN { FS="|"}
{print $1,$2," ",$4," ",$3\n,$5,"\n"}' list
It's keeping give me this message:
awk: syntax error near line 1
awk: bailing out near line 1
It seems awk has problem with my BEGIN command.
Any... (8 Replies)
Does anybody know how to print a variable passed to awk command?
awk -F"|" 'BEGIN {print $job,"\n","Question \n"} {print $1,$2$4," ",$3}' "job=$job1" file1
I am trying to pass job the variable job1.
the output is blank.
?? (3 Replies)
Hi,
I have a unix script which can accept n number of parameters .
I can get the parameter count using the following command and assign it to a variable
file_count=$#
Is there a similar command through which i can assign a variable all the values that i have passed as a parameter
... (2 Replies)
Hi all..
Does anyone know have an example of passing the contents of a bash array to oracle?
basically I am looking to loop through the contents of a file(not csv!) and store each line into a bash array. Once i have this I can then pass the array into an oracle procedure that accepts an array... (1 Reply)
Hi,
Can you please help. I am scripting in sh and I am trying to simply copy one directory to another but for some reason my variables are not recognised?
echo "The latest version of the program is being found......."
cd $SOFTWARE/src/$progname
version=`ls $SOFTWARE/src/$progname | grep... (13 Replies)
I am having difficulties with the fllowing script:
!/bin/sh
voicemaildir=/var/spool/asterisk/voicemail/$1/$2/INBOX/
echo `date` ':' $voicemaildir >> /var/log/voicemail-notify.log
for audiofile in `ls $voicemaildir/*.wav`; do
transcriptfile=${audiofile/wav/transcript}
... (4 Replies)
I have a script.
#!/bin/sh
cur_$1_modify_time=Hello
echo "cur_$1_modify_time"
When I run like
sh /root/script1 jj
I expect value "Hello" being assigned to variable "cur_jj_modify_time" and output being "Hello" ie echoing $cur_jj_modify_time
But the output comes as
# sh... (3 Replies)
EDIT: -- SOLVED --
Heyas,
Getting used to optargs, but by far not understanding it.
So i have that script that shall be 'changeable', trying to use the passed arguments to modify the script visuals.
Passing:
browser -t test -d sect $HOME
Where -t should change the title,
and -d... (0 Replies)
I have a file that has 2 fields called b_file:
11977 DAR.V3.20150209.1.CSV
3295 DAR.V3.20150209.1.CSV
1721 DAR.V2.20150210.1.CSV
I need to search a sftplog using the field 1, but want to maintain the relationship between field 1 and 2. I am passing field 1 as a parameter in a bash loop.
... (14 Replies)
Hello,
I am writing a script which is not giving the desired result. When I check the content of the 'InputFile_009_0.sh', it shows following with missing Index in this command
sed -i "s/L1ITMBLT.root/L1ITMBLT_"".root/g" run_DttfFromCombinedPrimitives_cfg.py
of .
Any help?
... (13 Replies)
Discussion started by: emily
13 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)
12-17-2015
.