hi,
i want to increment a Variable but it doesnt work.
here my codé
COUNT=1
COUNT= 'expr $COUNT + 1'
i've tried it in the prompt but it print me:
expr: syntaxerror
What does I make wrong? (4 Replies)
Hi All,
I have a variable n that stores a number.
Eg. echo $n comes out to be 120.
I need to print 121 using echo command on n.
Please advice.
Thanks in advance !! (4 Replies)
hi Friends,
Today_Dt=`date "+%Y-%m-%d"`
So the Today date is 2010-05-03
I have a file which has date values as below
2010-04-27
2010-04-02
2010-04-18
2010-04-28
2010-04-29
.. (1 Reply)
Hi,
I have a small query with gawk which i'm unsure how to solve. My csv input data is as follows:
1 58352.9 34549 -469.323 LINE_149
2 58352.9 34499 -469.323 LINE_149
3 58352.9 34549 -469.323 LINE_151
4 58352.9 34503.4 -489.841 LINE_151
5 58352.9 34549 -469.323 LINE_152
6 58352.9... (1 Reply)
Hello, this is a VI question more than anything...
I'm using: SunOS 5.10 Generic_150400-04 sun4v sparc SUNW,T5240
I'm trying to find a problematic line(s) in a script.
The only solution I can think of is to tag each repetative line, and increment it.
This is an oracle insert script.
... (14 Replies)
I have to increment time ... by sec but i am getting the output like this.
for m in {2..3}
> do
> for (( i = 1; i <= 13; i++ ))
> do
> echo "$m:$i"
> done
> done
2:1
2:2
2:3
2:4
2:5
2:6
2:7
2:8 (2 Replies)
I want a script which increments the count when the script runs. Basically I want to send an password reset email notification for an application, the password value should be keep on changing whenever the script is executed for example, first time i execute it should be password1, second time... (2 Replies)