I'm using the Scheduler Timeline view of Fullcalendar limited to one slot per month (12 slots for the current year), but I need a resource column (or a custom column) appended to the right side of the scheduler. Is possible to append a custom column to the view? I saw that currently is possible to append resource columns to the left side of the Calendar, but not to the right.
| | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sept | Oct | Nov | Dec | Custom Column |
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| Content | | | | | | | | | | | | | Text |
Related
I have a data set, each line representing a "service visit" for customers. A customer might have between 0 and 5 service calls. If there isn't a service call for someone, the columns associated with a service call would all be empty.
+--------------+-------------------+-------------------+------------------------+---------------------+
| Project Name | Customer Name | Service Call.Name | Service Call Date Time | Service Call Status |
+--------------+-------------------+-------------------+------------------------+---------------------+
| OO-99999 | A | SC-001762 | 3/21/2022 7:00:00 PM | Completed |
| OO-99999 | A | SC-002323 | null | Completed |
| OO-99999 | A | SC-002357 | 10/3/2022 7:00:00 PM | 2nd Visit Scheduled |
| OO-88888 | B | SC-001260 | 2/1/2022 8:00:00 PM | Completed |
| OO-88888 | B | SC-002938 | 8/25/2022 7:00:00 PM | Scheduled |
| OO-55555 | C | SC-000957 | 12/27/2021 8:00:00 PM | Completed |
| OO-55555 | C | SC-001418 | 2/7/2022 4:30:00 PM | Completed |
| OO-55555 | C | SC-003007 | null | null |
| OO-66666 | D | SC-001626 | null | No Longer Required |
| OO-66666 | D | SC-002329 | 6/9/2022 7:00:00 PM | Completed |
| OO-66666 | D | SC-002538 | null | Completed |
| OO-66666 | D | SC-002932 | null | Call Reviewed |
| OO-66666 | D | SC-003350 | 9/29/2022 7:00:00 PM | Scheduled |
| OO-11111 | F | null | null | null |
+--------------+-------------------+-------------------+------------------------+---------------------+
My goal is to filter out duplicates. I only want one row per customer, but I want to keep a specific row. A duplicate only appears if someone has multiple service calls.
If someone has a service call (Service Call.Name not equal to null), and one of those has a service call status of something OTHER than "Completed" or "Not required", I want to keep that row. So for Customer A, I want the third row since the service call status is not "completed" or "Not required".
If someone has multiple service calls, like customer , and they are all "completed" or "Not required". I don't care which one I keep, as long as I only keep one.
If someone has one service call or no service call, there will be no duplicate of that person, so I want to keep that row.
EDIT
There were cases of duplicates I didn't realize I had, I've edited the data to show them.
For someone with more than one open service call like customer E, I only want to keep one of them. If there is a date for both, I want the latest date of the two. If one has a date and the other doesn't, i want the one with a date. If neither have a date, i don't care which is kept, but i only want one.
I am working in Power BI, but I have access to R and think that might be easier.
Here is a solution. duplicated will give what rows to keep by customer name and another logical index, created with %in%, the rows to keep by status.
dat <- read.table(text = '+--------------+---------------+-------------------+------------------------+---------------------+
| Project Name | Customer Name | Service Call.Name | Service Call Date Time | Service Call Status |
+--------------+---------------+-------------------+------------------------+---------------------+
| OO-99999 | A | SC-001762 | 3/21/2022 7:00:00 PM | Completed |
| OO-99999 | A | SC-002323 | null | Completed |
| OO-99999 | A | SC-002357 | 10/3/2022 7:00:00 PM | 2nd Visit Scheduled |
| OO-88888 | B | SC-001260 | 2/1/2022 8:00:00 PM | Completed |
| OO-88888 | B | SC-002938 | 8/25/2022 7:00:00 PM | Scheduled |
| OO-55555 | C | SC-000957 | 12/27/2021 8:00:00 PM | Completed |
| OO-55555 | C | SC-001418 | 2/7/2022 4:30:00 PM | Completed |
| OO-55555 | C | SC-003007 | null | null |
| OO-11111 | D | null | null | null |
+--------------+---------------+-------------------+------------------------+---------------------+
', header = TRUE, sep = "|", comment.char = "+", strip.white = TRUE, check.names = FALSE)
dat <- dat[-c(1, ncol(dat))]
not_wanted <- c("Completed", "Not required")
i <- dat[['Service Call Status']] %in% not_wanted
i <- ave(i, dat[['Customer Name']], FUN = \(k) {
if(all(k)) k[1] <- FALSE
!k
})
result <- dat[i,]
j <- ave(result[['Service Call Status']], result[['Customer Name']], FUN = duplicated)
result <- result[!as.logical(j), ]
result
#> Project Name Customer Name Service Call.Name Service Call Date Time Service Call Status
#> 3 OO-99999 A SC-002357 10/3/2022 7:00:00 PM 2nd Visit Scheduled
#> 5 OO-88888 B SC-002938 8/25/2022 7:00:00 PM Scheduled
#> 8 OO-55555 C SC-003007 null null
#> 9 OO-11111 D null null null
Created on 2022-10-26 with reprex v2.0.2
i need to have a datagrid with the following layout:
------------------------------------------------
| | | player |
| date | time |-------------------------------|
| | | first name | last name | age |
|----------------------------------------------|
| jan | 14:02 | roy | batty | 3 |
|----------------------------------------------|
| mar | 17:12 | pika | chu | 1 |
|----------------------------------------------|
| dec | 05:31 | louie | dickens | 33 |
------------------------------------------------
Preliminary inquiries seem to reveal that dojo does not support this kind of behavior, am i right?
thank you very much
This is possible in dojo, you need to use CompoundColumns feature for this and then you will achieve this.
Please follow this link,
CompoundColumns for multilevel row in dojo grid
Please do try this and do let me know if have any issue/concern.
Normally, I can do show partitions <table> in hive. But when it is a parquet table, hive does not understand it. I can go to hdfs and check the dir structure, but that is not ideal. Is there any better way to do that?
I am using Impala 1.4.0 and I can see partitions.
From the impala-shell give the command:
show partitions <mytablename>
I have something looking like this:
+-------+-------+-----+-------+--------+---------+--------------+---------+
| year | month | day | #Rows | #Files | Size | Bytes Cached | Format |
+-------+-------+-----+-------+--------+---------+--------------+---------+
| 2013 | 11 | 1 | -1 | 3 | 25.87MB | NOT CACHED | PARQUET |
| 2013 | 11 | 2 | -1 | 3 | 24.84MB | NOT CACHED | PARQUET |
| 2013 | 11 | 3 | -1 | 2 | 19.05MB | NOT CACHED | PARQUET |
| 2013 | 11 | 4 | -1 | 3 | 23.63MB | NOT CACHED | PARQUET |
| 2013 | 11 | 5 | -1 | 3 | 26.56MB | NOT CACHED | PARQUET |
Alternatively you can go to your table in HDFS . They are normally seen in this path:
/user/hivestore/warehouse/<mytablename> or
/user/hive/warehouse/<mytablename>
Unfortunately no. Issue is open though. So checking it manually seems to be the only option right now.
I'm trying to figure out how to to use org-mode to calculate the duration between two time points, however, whilst I figured out how to do it for two separate dates, when I add in the time component, it gives an answer, but I'd rather have the answer in
XX days, xx hours, xx minutes
| Start | End | Duration |
|------------------------+------------------------+----------|
| <2013-07-16 Tue 15:15> | <2013-07-17 Wed 11:15> | 0.833333 |
| | | 0 |
#+TBLFM: $3=(date(<$2>)-date(<$1>))
You may use the T flag to use the form HH:MM[:SS]. Example:
| Start | End | Days | HH:MM:SS |
|------------------------+------------------------+----------+----------|
| <2013-07-15 Tue 10:15> | <2013-07-17 Wed 11:15> | 2.041667 | 49:00:00 |
| | | 0 | 00:00:00 |
#+TBLFM: $3=date(<$2>)-date(<$1>)::$4=60*60*24*$3;T
I'm trying to created a report for my asp.net application which will show the quantity of each item in combination with unit that was ordered for each day of the week. The days of the week are columns.
To be more specific:
I have two table, one is the Orders table with order id, customer name, date etc...
The second table is OrderItems, this table has order id as a foreign key, order Item id, item name, unit (exp: each, box , case), quantity, and price.
When a user picks a date range for the report, for example from 3/2/12 to 4/2/12, on my asp page, the report will group order items by week and will look as follows:
**week (1) starting from sunday of such date to saturday of such date**
item | unit | Sun | Mon | Tues | Wedn | Thur | Fri | Sat | Total Price for week
item1 | bag | 3 | 0 | 12 | 8 | 45 | 1 | 4 | $1234
item4 | box | 2 | 4 | 5 | 0 | 5 | 2 | 6 | $1234
**week (2) starting from sunday of such date to saturday of such date**
item | unit | Sun | Mon | Tues | Wedn | Thur | Fri | Sat | Total Price for week
item1 | bag | 3 | 0 | 12 | 8 | 45 | 1 | 4 | $1234
item4 | box | 2 | 4 | 5 | 0 | 5 | 2 | 6 | $2354
**week (2) starting from sunday of such date to saturday of such date**
item | unit | Sun | Mon | Tues | Wedn | Thur | Fri | Sat | Total Price for week
item1 | bag | 3 | 0 | 12 | 8 | 45 | 1 | 4 | $1234
item4 | box | 2 | 4 | 5 | 0 | 5 | 2 | 6 | $2354
I wish I could have something to show that I have already started, but crystal isn't my strong point and I dont even know where start tackling this one. I do know how to pass parameters and a datatable that I myself pre-filtered before passing it to the report. For example filtering items by date range and customer or order id.
any help would be much appreciated
Create a formula for each day of the week that totals the order.
ie Sunday quantity:
if dayOfWeek(dateField) = 'Sun'
then order.quantity
else 0
Add each day formula to the detail section of the report and then summarize it for each group level. To group it by week, just group by the date field, then set the grouping option to by week. Suppress the detail, and you'll have what you are looking for.
I don't remember the exact name of the dayOfWeek function, but it's something like that.