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1. UNIX for Advanced & Expert Users
I have a below syntax its working fine...
var12=$(ps -ef | grep apache | awk '{print $2,$4}')
Im getting expected output as below:
printf "%b\n" "${VAR12}"
dell 123
dell 456
dell 457
Now I wrote a while loop.. the output of VAR12 should be passed as input parameters to while loop and results... (5 Replies)
Discussion started by: sam@sam
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2. Shell Programming and Scripting
Hi All,
I have a file which has hundred of records with fixed number of fields. In each record there is set of 8 characters which represent the duration of that activity. I want to sum up the duration present in all the records for a report. The problem is the duration changes per record so I... (5 Replies)
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3. Red Hat
Hi all,
Hereby wish to have your advise for below:
Main concept is
I intend to get current directory of my script file.
This script file will be copied to /etc/init.d.
A string in this copy will be replaced with current directory value.
Below is original script file:
... (6 Replies)
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4. Shell Programming and Scripting
can anyone please help me with this:
i have written a shell script and a stored procedure which has one OUT parameter. now i want to use that out parameter as an input to the unix script but i am getting an error as variable not found. below are the unix scripts and stored procedure...
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5. Shell Programming and Scripting
Hi,
I am trying to do the following thing
var='date'
$var
Above command substitutes date for and in turn runs the date command and i am getting the todays date value.
I am trying to do the same thing as following, but facing some problems,
unique_host_pro="sed -e ' /#/d'... (3 Replies)
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6. Shell Programming and Scripting
Dear all,
I have basic knowledge of Unix script and her I am trying to process variable length and variable format CSV file.
The file length will depend on the numbers of Earnings/Deductions/Direct Deposits.
And
The format will depend on whether it is Earnings/Deductions or Direct Deposits... (2 Replies)
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7. Shell Programming and Scripting
Hi all,
I have a variable say var1 (output from somewhere, which I can't change)which store something like this:
echo $var1
name=fred
age=25
address="123 abc"
password=pass1234
how can I make the variable $name, $age, $address and $password contain the info?
I mean do this in a... (1 Reply)
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8. Shell Programming and Scripting
I want to instert Category:XXXXX into the 2. line
something like this should work, but I have somewhere the wrong sytanx. something with the linebreak goes wrong:
sed "2i\\${n}Category:$cat\n"
Sample:
Titel Blahh Blahh abllk sdhsd sjdhf
Blahh Blah Blahh
Blahh
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9. Shell Programming and Scripting
Hello,
i have another sed question.. I'm trying to do variable substition with sed and i'm running into a problem.
my var1 is a string constructed like this:
filename1 filerev1 filepath1
my var2 is another string constructed like this:
filename2 filerev2 filepath2
when i do... (2 Replies)
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10. Shell Programming and Scripting
This is what I tried:
vara=${varb}_count
(( vara += 1 ))
Thanks for help (4 Replies)
Discussion started by: pa3be
4 Replies
XCALPR(1) General Commands Manual XCALPR(1)
NAME
xcalpr - print xcal calendar entries
SYNTAX
xcalpr [ -c ][ -x ][ -f file ][ -d dir ][ -u user ][ date-spec ]
DESCRIPTION
Xcalpr prints the contents of the xcal files. It is intended to be used in situations when you have no access to an X screen. It can also
be used to generate entries for the standard UNIX calendar program.
With no arguments, it prints any entries that exist for the next seven days. The program also reads the contents of the seven daily files
and prints them at the appropriate point in the output stream. Each line in the output is preceded by the day of the week, the day of the
month, the month and the year.
Xcalpr can be given a date specification to select months and years. If the date spec consists of just a year number, then all the data
for that year is printed. For example:
xcalpr 1994
will print all the data for 1994. Several years can be specified.
If you give the name of a month, then the data for that month in the current year will be printed. If the month is in the past, then the
data for that month next year will be printed. For example, if
xcalpr oct jan
is typed in August, xcalpr will print October in the current year and January next year.
You can select a particular year by adding the number after any months that you need printing:
xcalpr oct nov 1994
will print October and November in 1994.
There are a couple of special `month' names. The name rest will print the data for the rest of the month, starting tomorrow. The rest
argument is not recognised if you give a year as a parameter. If tomorrow happens to be the first day of the next month, then all the data
for next month will be printed. The name next prints all the data for next month.
OPTIONS
The -c option causes xcalpr to output lines suitable for input to the standard UNIX calendar program.
The -d switch is followed by a directory name and specifies an alterative location for your Calendar directory. Your home directory is
prepended if the name doesn't start with a slash or a dot.
The -f option is followed by a file name and xcalpr will write it's output to that file, rather than standard output.
The -u option is followed by a user name and dumps their calendar files rather than yours.
The -x option makes xcalev operate with Calendar files that are compatible with the xcalendar program.
FILES
$HOME/Calendar/*
xc<dd><Mon><Year> A data file is day, Month in three letter format and the year.
xy<Year> A year directory.
xw<Day> A data file for the weekly code, one per day.
SEE ALSO
xcal(1), xcalev(1), xcal_cal(1)
AUTHOR
Copyright 1993 by Peter Collinson, Hillside Systems All rights reserved.
This product includes software developed by the University of California, Berkeley and its contributors.
X Version 11 R5 October 1993 XCALPR(1)