05-07-2008
Snytax error on If Statement--help
year=`date '+%Y'`
month=`date '+%m'`
day=`date '+%d'`
day=`expr $day - 1`
case $month in
1 | 3 | 5 | 7 | 8 | 10 | 12);;
if($day =7 ); then
$day=6
fi
4 | 6 | 9 | 11);;
if [ $day = 0 ] ; then
$day=31
fi
2);;
if [ $day = 0 ] ; then
if [ `expr $year % 4` -eq 0]; then
$day = 28
else
$day = 29
fi
fi
*) echo error: too many arguments 1>&2 ;;
esac
DATE=$year$month$day
cp $HOME/*$DATE*.log $HOME/QVGLog
cd $HOME/QVGLog
gzip *
the error message is
./call: line 9: syntax error near unexpected token `('
./call: line 9: `if($day =7 ); then'
Pls help, i've checked the syntax for if else statement on web, but cant fix it.
10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
hello,
I am trying to parse an error returned by a command inside the if statement but it is just displays the full error instead and then stops.
if ; then
echo "no such package"
else
echo "similar version found will use pkgrm"
fi
the above code just displays
please let me know... (2 Replies)
Discussion started by: rakeshou
2 Replies
2. Shell Programming and Scripting
:b:hi,
I have a script as given below:
pr_det="1"
if
then
awk ' BEGIN {printf("%23s","session")}' >> report.txt
awk ' BEGIN {printf "\n"} ' >> report.txt
else
awk ' BEGIN {printf("%55s","file_dsc")} ' >> report.txt
awk ' BEGIN {printf("%101s","no_recs")} '... (1 Reply)
Discussion started by: jisha
1 Replies
3. Linux
Hi ,
I am getting an error when I run the script for checking word "view" in a file . I am using if statement. like this
if
then
VW_VAR=` cat $TN.${ecmdate}.sql1 | grep -i view | awk '{print $3}' | cut -d '.' -f2 `
echo " VW_$VW_VAR "
sed -e... (16 Replies)
Discussion started by: capri_drm
16 Replies
4. Shell Programming and Scripting
HI i am getting error while executing the given statement
for filename in `cat a/file.lst`
do
if then
echo "Exit Code Description :File $filename - is missing in Input Directory" >a.log
exit
else
count1=`awk 'END {print NR}' $filename`
echo "$count1">>a.log
count2=`awk 'END {print... (4 Replies)
Discussion started by: ravi214u
4 Replies
5. Shell Programming and Scripting
hi,
When i try to run the code below, i get the following error
"ksh: syntax error: `(' unexpected"
i am not able to figure it out. Can anyone help me?
Code: (2 Replies)
Discussion started by: ragavhere
2 Replies
6. UNIX for Dummies Questions & Answers
Hi,
This is my script to catch any oracle errors.
In this, the $sqlerr returns
ORA-01017: invalid username/password; logon denied
when i specify wrong username/password
the if condition is failing. how can i resolve the issue.
the if statement gives error
sqloutput=`sqlplus -s -L... (1 Reply)
Discussion started by: Swapna173
1 Replies
7. Shell Programming and Scripting
I am working on script for stale nfs.
the file consists of
cat data01stale.log
- - - - /abcd/backup
- - - - /abcd/data
Script (16 Replies)
Discussion started by: nareshkumar522
16 Replies
8. Shell Programming and Scripting
Hi
Can you please tell me what is wrong with this line:
if && ]; then
basically i want to check if x = 12 and F (Filename) end with 'g'. But it is throwing syntax error. (7 Replies)
Discussion started by: rtagarra
7 Replies
9. Shell Programming and Scripting
Can anybody tell the correct way to use the following awk pattern check within an if statement?
When I run this command outside of the if statement
cat /tmp/test2.out | awk -v T=$TIME -v G=$GROUP -v C=$CDATE '$0 ~ T && $0 ~ G && $0 ~ C' | grep -i "Starting the group"
I get the following... (0 Replies)
Discussion started by: kieranfoley
0 Replies
10. UNIX for Beginners Questions & Answers
hello all
im new to unix and when i use below script i get an error :
#! /bin/bash
Echo -e "enter the name of the file : \c"
read file_name
if
then
echo "$file_name found"
else
echo "$file_name not found"
fi
running the script i get below error :
$ ./hello (26 Replies)
Discussion started by: Ibrahims1
26 Replies
LEARN ABOUT PHP
gregoriantojd
GREGORIANTOJD(3) 1 GREGORIANTOJD(3)
gregoriantojd - Converts a Gregorian date to Julian Day Count
SYNOPSIS
int gregoriantojd (int $month, int $day, int $year)
DESCRIPTION
Valid Range for Gregorian Calendar 4714 B.C. to 9999 A.D.
Although this function can handle dates all the way back to 4714 B.C., such use may not be meaningful. The Gregorian calendar was not
instituted until October 15, 1582 (or October 5, 1582 in the Julian calendar). Some countries did not accept it until much later. For exam-
ple, Britain converted in 1752, The USSR in 1918 and Greece in 1923. Most European countries used the Julian calendar prior to the Grego-
rian.
PARAMETERS
o $month
- The month as a number from 1 (for January) to 12 (for December)
o $day
- The day as a number from 1 to 31
o $year
- The year as a number between -4714 and 9999
RETURN VALUES
The julian day for the given gregorian date as an integer.
EXAMPLES
Example #1
Calendar functions
<?php
$jd = GregorianToJD(10, 11, 1970);
echo "$jd
";
$gregorian = JDToGregorian($jd);
echo "$gregorian
";
?>
SEE ALSO
jdtogregorian(3), cal_to_jd(3).
PHP Documentation Group GREGORIANTOJD(3)