I prefer to use perl. it works on mostly all platforms.
with the limited information you provided, I was able to generate a program that resets the password for any given user ID.
note!(edit) I forgot to mention that this code will maintain a file called pcount in the local directory.
the purpose of this file is to maintain a count of how many times the password has been set. At the moment the file only maintains
a password for one user. However, it can be easily modified to maintain password for all possible users.
the usage is as follows:
I should note that this program is missing the actual command that would be used to reset the password. Mostly because I did not know if you were reseting the password on a unix, windows, or some specific application.
Such commands should be added the code some where around line 43 below.
the actual code is as follows:
Last edited by busi386; 07-18-2014 at 05:12 PM..
Reason: provided more information.
hi,
i want to increment a Variable but it doesnt work.
here my codé
COUNT=1
COUNT= 'expr $COUNT + 1'
i've tried it in the prompt but it print me:
expr: syntaxerror
What does I make wrong? (4 Replies)
Hi All,
I have created a script...
#!/bin/sh
datafile=ABC2008101601.OUT
indfile=ABCIND20081016.1.OUT
waittime=600
starttime=0
while
do
if
then
echo "Indicator file has arrived."
break
else
sleep 10;
((starttime=$starttime+10))
echo $starttime (2 Replies)
Hello
Does anyone know how to auto-increment the value of a variable, preferably using awk or sed?
I need to read values from a file and auto-increment those values to use them as line numbers
I'd be doing:
while read line do
# auto-increment
sed -n${line}p file> file1
done... (6 Replies)
Hi All,
I have a variable n that stores a number.
Eg. echo $n comes out to be 120.
I need to print 121 using echo command on n.
Please advice.
Thanks in advance !! (4 Replies)
Hi,
I have a variable lets say DATA_DATE.
I have to pass some value to this variable in YYYYMMDD format.
lets say today I have passed this variable as :
DATA_DATE=20100107
Then pls help me how to calculate another variable DATA_DATE1 (which is DATA_DATE+1).
The code should work... (3 Replies)
hi Friends,
Today_Dt=`date "+%Y-%m-%d"`
So the Today date is 2010-05-03
I have a file which has date values as below
2010-04-27
2010-04-02
2010-04-18
2010-04-28
2010-04-29
.. (1 Reply)
Hi,
I have a small query with gawk which i'm unsure how to solve. My csv input data is as follows:
1 58352.9 34549 -469.323 LINE_149
2 58352.9 34499 -469.323 LINE_149
3 58352.9 34549 -469.323 LINE_151
4 58352.9 34503.4 -489.841 LINE_151
5 58352.9 34549 -469.323 LINE_152
6 58352.9... (1 Reply)
I have to increment time ... by sec but i am getting the output like this.
for m in {2..3}
> do
> for (( i = 1; i <= 13; i++ ))
> do
> echo "$m:$i"
> done
> done
2:1
2:2
2:3
2:4
2:5
2:6
2:7
2:8 (2 Replies)
Discussion started by: kalyankalyan
2 Replies
10. Forum Support Area for Unregistered Users & Account Problems
I was unable to login and so used the "Forgotten Password' process. I was sent a NEWLY-PROVIDED password and a link through which my password could be changed. The NEWLY-PROVIDED password allowed me to login.
Following the provided link I attempted to update my password to one of my own... (1 Reply)