10-25-2006
Compare the first column and get the second column from file.
Hi All ,
I have an output value with two columns like this...
Days User
10 A
500 B
1 C
How i can compare the first column value and passing the user name as parameter?
For example :
while read -r days
If (days<=30) ;
then
value=days/30 x100
update tableA set value = ${value} where user=A
else
update tableA set value=100 where user=B
Anyone can help me?
Thank you!!
Last edited by EDBGSK; 10-25-2006 at 10:21 AM..
10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
Hi All,
Wish you a Happy New year...
I have to find the difference between two dates, the result should be the number of days. I have seen the "datecalc" function. Its good, can I have any other alternative.
Thanks in Advance
Raju (4 Replies)
Discussion started by: rajus19
4 Replies
2. Shell Programming and Scripting
Hi All
How to get the difference between two dates in no of days ??? My date format is like this YYYY/MM/DD. I have to get the no of days between two dates in the given format.
I tried to search the forum but nothing came up similar to my requitement. Your help will be appreciated.
... (1 Reply)
Discussion started by: csaha
1 Replies
3. Shell Programming and Scripting
Hi all.
My question may seems to be similar to one that already been here. But i need a little other solution.
I have two dates in format dd/mm/yyyy. I need to find number of days between them. I need to do it in bash script.
I am running on Solaris machine and have cutted 'date' command version... (1 Reply)
Discussion started by: kukuruku
1 Replies
4. Shell Programming and Scripting
Hi,
Is there any way I can get the difference between two dates in terms of days?
I have used this method so far, but I cant format it in terms of days.
@a=&DateCalc($date1,$date2,0);
The o/p that I am getting is sort of like this:
+0:0:0:4:0:0:0
I just want to get 4 days as an o/p.... (1 Reply)
Discussion started by: King Nothing
1 Replies
5. Shell Programming and Scripting
Hi!
I have two parameters like this: YYYY-MM-DD YYYY-MM-DD
My question is, there is a direct command for get the elapsed time between the 2 dates, or I have to find another way?
Thx! (1 Reply)
Discussion started by: MalaTomi
1 Replies
6. Shell Programming and Scripting
hi all,
I need a help for below requirement.
Difference between two dates"12-11-2009" and "03-25-2012" (mm-dd-yy format") in weeks and days and hours
Please help me for this. Thanks in adv....
I am working in AIX, so dont have below command:-
date --version (2 Replies)
Discussion started by: gani_85
2 Replies
7. Shell Programming and Scripting
I want the difference between two following date using scripts in terms of no.of days. How I can accomplish this.
lastdate=Tue Nov 13 10:30:56 2012
currdate=Wed Dec 15 15:58:21 PAKST 2012
Ouput should be like this:
Your Password will expire after = 32 Days on Wed Dec 15 15:58:21 PAKST... (1 Reply)
Discussion started by: m_raheelahmed
1 Replies
8. Shell Programming and Scripting
Hi Friends,
I have a file that has the contents like below:
file1.txt
5,13/07/2013 23:25:25,14/07/2013 19:40:21
5,13/07/2013 23:25:25,14/07/2013 19:40:43
5,12/07/2013 23:50:50,13/07/2013 20:30:26
5,12/07/2013 23:20:24,13/07/2013 19:40:53
60,14/07/2013 00:00:00,14/07/2013 23:00:39... (5 Replies)
Discussion started by: vsachan
5 Replies
9. Fedora
I have a script which is printing date in below format while writing the logs.
theDate=`date +"%m%d%Y"`
theTime=`date +"%H%M%S"`
echo $theDate $theTime
How can i find out difference current time and above format. Appreciate your help. (6 Replies)
Discussion started by: srikanth38
6 Replies
10. UNIX for Beginners Questions & Answers
Hi There
I am trying to find the difference between two dates in seconds, by taking the first 10 digits of the file name itself, which I have done as shown below:
current_time=`date +%s`
last_login_of_tim=`date -d @1489662376 +%s`
diff_sec=$(($current_time-$last_login_of_tim))
... (5 Replies)
Discussion started by: simpsa27
5 Replies
SHADOW(5) File Formats Manual SHADOW(5)
NAME
shadow - encrypted password file
DESCRIPTION
shadow contains the encrypted password information for user's accounts and optional the password aging information. Included is
Login name
Encrypted password
Days since Jan 1, 1970 that password was last changed
Days before password may be changed
Days after which password must be changed
Days before password is to expire that user is warned
Days after password expires that account is disabled
Days since Jan 1, 1970 that account is disabled
A reserved field
The password field must be filled. The encryped password consists of 13 to 24 characters from the 64 characters alphabet a thru z, A thru
Z, 0 thru 9, . and /. Optionally it can start with a "$" character. This means the encrypted password was generated using another (not DES)
algorithm. For example if it starts with "$1$" it means the MD5-based algorithm was used.
Refer to crypt(3) for details on how this string is interpreted.
The date of the last password change is given as the number of days since Jan 1, 1970. The password may not be changed again until the
proper number of days have passed, and must be changed after the maximum number of days. If the minimum number of days required is greater
than the maximum number of day allowed, this password may not be changed by the user.
An account is considered to be inactive and is disabled if the password is not changed within the specified number of days after the pass-
word expires. An account will also be disabled on the specified day regardless of other password expiration information.
This information supercedes any password or password age information present in /etc/passwd.
This file must not be readable by regular users if password security is to be maintained.
FILES
/etc/passwd - user account information
/etc/shadow - encrypted user passwords
SEE ALSO
chage(1), login(1), passwd(1), su(1), passwd(5), pwconv(8), pwunconv(8), sulogin(8)
AUTHOR
Julianne Frances Haugh (jockgrrl@ix.netcom.com)
SHADOW(5)