GridView Layout/Output - asp.net

I have an Website using ASP.Net 2.0 with SQL Server as Database and C# 2005 as programming language. In one of the pages I have a GridView with following layout.
Date -> Time -> QtyUsed
The sample values are as follows: (Since this GridView/Report is generated for a specific month only, I have extracted and displaying only the Day part of the date ignoring the month and year part.
01 -> 09:00 AM -> 05
01 -> 09:30 AM -> 03
01 -> 10:00 AM -> 09
02 -> 09:00 AM -> 10
02 -> 09:30 AM -> 09
02 -> 10:00 AM -> 11
03 -> 09:00 AM -> 08
03 -> 09:30 AM -> 09
03 -> 10:00 AM -> 12
Now the user wants the layout to be like:
Time 01 02 03 04 05 06 07 08 09
-------------------------------------------------------------------------
09:00 AM -> 05 10 08
09:30 AM -> 03 09 09
10:00 AM -> 09 11 12
The main requirement is that the days should be in the column header from 01 to the last date (the reason why I extracted only the day part from the date). The Timeslots should be down as rows.
From my experience with Excel, the idea of Transpose comes to my mind to solve this, but I am not sure.
Please help me in solving this problem.
Thank you.
Lalit Kumar Barik

You will have to generate the dataset accordingly. I am guessing you are doing some kind of grouping based on the hour so generate a column for each hour of the day and populate the dataset accordingly.

In SQL Server, there is a PIVOT function that may be of use.
The MSDN article specifies usage and gives an example.
The example is as follows
Table DailyIncome looks like
VendorId IncomeDay IncomeAmount
---------- ---------- ------------
SPIKE FRI 100
SPIKE MON 300
FREDS SUN 400
SPIKE WED 500
...
To show
VendorId MON TUE WED THU FRI SAT SUN
---------- ----------- ----------- ----------- ----------- ----------- ----------- -----------
FREDS 500 350 500 800 900 500 400
JOHNS 300 600 900 800 300 800 600
SPIKE 600 150 500 300 200 100 400
Use this select
SELECT * FROM DailyIncome
PIVOT( AVG( IncomeAmount )
FOR IncomeDay IN
([MON],[TUE],[WED],[THU],[FRI],[SAT],[SUN])) AS AvgIncomePerDay
Alternatively, you could select all of the data from DailyIncome and build a DataTable with the data pivoted. Here is an example.

Related

Append list of logged in users to a log file using crontab?

I need to create a basic log file through the use of a crontab job that appends a timestamp, followed by a list of logged in users. It must be at 23:59 each night.
(I have used 18 18 * * * as an example to make sure the job works for now)
So far, I have;
!#/bin/bash
59 23 * * * (date ; who) >> /root/userlogfile.txt
for my crontab script, the output;
Fri Dec 9 18:18:01 UTC 2022
root console 00:00 Dec 9 18:15:15
My required output is something similar to;
Fri 09 Dec 23:59:00 GMT 2022
user1 tty2 2017-11-30 22:00 (:0)
user5 pts/1 2017-11-30 20:35 (192.168.1.1)
How would I go about this?

How can I write a shell scripts that calls another script for each day starting from start date upto current date [duplicate]

This question already has answers here:
How to loop through dates using Bash?
(10 answers)
Closed 1 year ago.
How can I write a shell scripts that calls another script for each day
i.e currently have
#!/bin/sh
./update.sh 2020 02 01
./update.sh 2020 02 02
./update.sh 2020 02 03
but I want to just specify start date (2020 02 01) and let is run update.sh for every day upto current date, but don't know how to manipulate date in shell script.
I made a stab at it, but rather messy, would prefer if it could process date itself.
#!/bin/bash
for j in {4..9}
do
for k in {1..9}
do
echo "update.sh" 2020 0$j 0$k
./update.sh 2020 0$j 0$k
done
done
for j in {10..12}
do
for k in {10..31}
do
echo "update.sh" 2020 $j $k
./update.sh 2020 $j $k
done
done
for j in {1..9}
do
for k in {1..9}
do
echo "update.sh" 2021 0$j 0$k
./update.sh 2021 0$j 0$k
done
done
for j in {1..9}
do
for k in {10..31}
do
echo "update.sh" 2021 0$j $k
./update.sh 2021 0$j $k
done
done
You can use date to convert your input dates into seconds in order to compare. Also use date to add one day.
#!/bin/bash
start_date=$(date -I -d "$1") # Input in format yyyy-mm-dd
end_date=$(date -I) # Today in format yyyy-mm-dd
echo "Start: $start_date"
echo "Today: $end_date"
d=$start_date # In case you want start_date for later?
end_d=$(date -d "$end_date" +%s) # End date in seconds
while [ $(date -d "$d" +%s) -le $end_d ]; do # Check dates in seconds
# Replace `echo` in the below with your command/script
echo ${d//-/ } # Output the date but replace - with [space]
d=$(date -I -d "$d + 1 day") # Next day
done
In this example, I use echo but replace this with the path to your update.sh.
Sample output:
[user#server:~]$ ./dateloop.sh 2021-08-29
Start: 2021-08-29
End : 2021-09-20
2021 08 29
2021 08 30
2021 08 31
2021 09 01
2021 09 02
2021 09 03
2021 09 04
2021 09 05
2021 09 06
2021 09 07
2021 09 08
2021 09 09
2021 09 10
2021 09 11
2021 09 12
2021 09 13
2021 09 14
2021 09 15
2021 09 16
2021 09 17
2021 09 18
2021 09 19
2021 09 20

How to combine two files sequentially based on certain conditions in Unix

I am trying to format files in Unix (In this case RHEL).
File 1
AAAAA|AAA|1582|YNYY
BBBBB|BAV|1234|NYYY
File 1 has 1 sample record (row). There are 4 columns in each record. In Column 4 we have 4 status values.
File 2
20190103|W 2019 01
20190203|W 2019 02
20190303|W 2019 03
20190403|W 2019 04
Output has to be as follows:
AAAAA|1582|Y|20190103|W 2019 01
AAAAA|1582|N|20190203|W 2019 02
AAAAA|1582|Y|20190303|W 2019 03
AAAAA|1582|Y|20190403|W 2019 04
BBBBB|1234|N|20190103|W 2019 01
BBBBB|1234|Y|20190203|W 2019 02
BBBBB|1234|Y|20190303|W 2019 03
BBBBB|1234|Y|20190403|W 2019 04
I have tried AWK and Paste but am not able to get the required output.
Using awk
awk -F'|' '{split($4,a,""); b=$1"|"$2"|"$3} { getline < "file2"; for (i in a ) print b"|"a[i]"|"$0 }' < file1`
Demo:
$cat file1 file2
AAAAA|AAA|1582|YNYY
BBBBB|BAV|1234|NYYY
20190103|W 2019 01
20190203|W 2019 02
20190303|W 2019 03
20190403|W 2019 04
$awk -F'|' '{split($4,a,""); b=$1"|"$2"|"$3} { getline < "file2"; for (i in a ) print b"|"a[i]"|"$0 }' < file1
AAAAA|AAA|1582|Y|20190103|W 2019 01
AAAAA|AAA|1582|N|20190103|W 2019 01
AAAAA|AAA|1582|Y|20190103|W 2019 01
AAAAA|AAA|1582|Y|20190103|W 2019 01
BBBBB|BAV|1234|N|20190203|W 2019 02
BBBBB|BAV|1234|Y|20190203|W 2019 02
BBBBB|BAV|1234|Y|20190203|W 2019 02
BBBBB|BAV|1234|Y|20190203|W 2019 02
$
Explanation:
awk -F'|' <-- Set field seprator as |
'{split($4,a,""); <-- Split 4th field and store in array a
b=$1"|"$2"|"$3} <-- Store Column 1-2 in variable b
getline < "file2"; <-- read input from file2 row by row
for (i in a ) print b"|"a[i]"|"$0 <-- Loop through array a and append variable b and input record from file2
Note: When you use getline value of internal variables $0, NF, NR get changed

Convert date with Time Zone formats in R

I have my dates in the following format :- Wed Apr 25 2018 00:00:00 GMT-0700 (Pacific Standard Time) or 43167 or Fri May 18 2018 00:00:00 GMT-0700 (PDT) all mixed in 1 column. What would be the easiest way to convert all of these in a simple YYYY-mm-dd (2018-04-13) format? Here is the column:
dates <- c('Fri May 18 2018 00:00:00 GMT-0700 (PDT)',
'43203',
'Wed Apr 25 2018 00:00:00 GMT-0700 (Pacific Standard Time)',
'43167','43201',
'Fri May 18 2018 00:00:00 GMT-0700 (PDT)',
'Tue May 29 2018 00:00:00 GMT-0700 (Pacific Standard Time)',
'Tue May 01 2018 00:00:00 GMT-0700 (PDT)',
'Fri May 25 2018 00:00:00 GMT-0700 (Pacific Standard Time)',
'Fri Apr 06 2018 00:00:00 GMT-0700 (PDT)','43173')
Expected format:2018-05-18, 2018-04-13, 2018-04-25, ...
I believe similar questions have been asked several times before. However, there
is a crucial point which needs special attention:
What is the origin for the dates given as integer (or as character string which can be converted to integer to be exact)?
If the data is imported from the Windows version of Excel, origin = "1899-12-30" has to be used. For details, see the Example section in help(as.Date) and the Other Applications section of the R Help Desk article by Gabor Grothendieck and Thomas Petzoldt.
For conversion of the date time strings, the mdy_hms() function from the lubridate package is used. In addition, I am using data.table syntax for its conciseness:
library(data.table)
data.table(dates)[!dates %like% "^\\d+$", new_date := as.Date(lubridate::mdy_hms(dates))][
is.na(new_date), new_date := as.Date(as.integer(dates), origin = "1899-12-30")][]
dates new_date
1: Fri May 18 2018 00:00:00 GMT-0700 (PDT) 2018-05-18
2: 43203 2018-04-13
3: Wed Apr 25 2018 00:00:00 GMT-0700 (Pacific Standard Time) 2018-04-25
4: 43167 2018-03-08
5: 43201 2018-04-11
6: Fri May 18 2018 00:00:00 GMT-0700 (PDT) 2018-05-18
7: Tue May 29 2018 00:00:00 GMT-0700 (Pacific Standard Time) 2018-05-29
8: Tue May 01 2018 00:00:00 GMT-0700 (PDT) 2018-05-01
9: Fri May 25 2018 00:00:00 GMT-0700 (Pacific Standard Time) 2018-05-25
10: Fri Apr 06 2018 00:00:00 GMT-0700 (PDT) 2018-04-06
11: 43173 2018-03-14
Apparently, the assumption to choose the origin which belongs to the Windows version of Excel seems to hold.
If only a vector of Date values is required:
data.table(dates)[!dates %like% "^\\d+$", new_date := as.Date(lubridate::mdy_hms(dates))][
is.na(new_date), new_date := as.Date(as.integer(dates), origin = "1899-12-30")][, new_date]
[1] "2018-05-18" "2018-04-13" "2018-04-25" "2018-03-08" "2018-04-11" "2018-05-18"
[7] "2018-05-29" "2018-05-01" "2018-05-25" "2018-04-06" "2018-03-14"

Converting UTC Time to Local Time with Days of Week and Date Included

I have the following 2 columns as part of a larger data frame. The Timezone_Offset is the difference in hours for the local time (US West Coast in the data I'm looking at). In other words, UTC + Offset = Local Time.
I'm looking to convert the UTC time to the local time, while also correctly changing the day of the week and date, if necessary. For instance, here are the first 5 rows of the two columns.
UTC Timezone_Offset
Sun Apr 08 02:42:03 +0000 2012 -7
Sun Jul 01 03:27:20 +0000 2012 -7
Wed Jul 11 04:40:18 +0000 2012 -7
Sat Nov 17 01:31:36 +0000 2012 -8
Sun Apr 08 20:50:30 +0000 2012 -7
Things get tricky when the day of the week and date also have to be changed. For instance, looking at the first row, the local time should be Sat Apr 07 19:42:03 +0000 2012. In the second row, the month also has to be changed.
Sorry, I'm fairly new to R. Could someone possibly explain how to do this? Thank you so much in advance.
Parse as UTC, then apply the offset in seconds, ie times 60*60 :
data <- read.csv(text="UTC, Timezone_Offset
Sun Apr 08 02:42:03 +0000 2012, -7
Sun Jul 01 03:27:20 +0000 2012, -7
Wed Jul 11 04:40:18 +0000 2012, -7
Sat Nov 17 01:31:36 +0000 2012, -8
Sun Apr 08 20:50:30 +0000 2012, -7", stringsAsFactors=FALSE)
data$pt <- as.POSIXct(strptime(data$UTC, "%a %b %d %H:%M:%S %z %Y", tz="UTC"))
data$local <- data$pt + data$Timezone_Offset*60*60
Result:
> data[,3:4]
pt local
1 2012-04-08 02:42:03 2012-04-07 19:42:03
2 2012-07-01 03:27:20 2012-06-30 20:27:20
3 2012-07-11 04:40:18 2012-07-10 21:40:18
4 2012-11-17 01:31:36 2012-11-16 17:31:36
5 2012-04-08 20:50:30 2012-04-08 13:50:30
>

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