I have an IG in my app which returns 138 rows and most of my users are used to “CTRL+F” to find something on the page and they are not able to find the result, as only 40 rows are fetched at a time. I know this is a natural drawback of any kind of pagination. Any suggestions on how to increase the number of rows fetched on page load?
Tell most of your users to use Apex' search capabilities:
Alternatively, modify number of rows displayed per page; as there are 138 rows, set it to 1000 (don't forget to save the report, then!). Though, what will you do if report contains more than 1000 rows? I'd do what I suggested first: search using the "Search" bar.
Related
I am working on a form in which I need to take numeric value from one field and based on that number(n) I need to create n fields for another column. How can I do so?
Probably you can use APEX_ITEM API if I understand your task right:
https://docs.oracle.com/en/database/oracle/application-express/19.2/aeapi/APEX_ITEM.html
Something like this:
SELECT APEX_ITEM.TEXT(rownum)
FROM DUAL
CONNECT BY rownum < :YOUR_ITEM;
You will not be able to create n page items in a declarative way. APEX simply wasn't made for this.
I can think of two workarounds:
Use a plsql dynamic content region to generate html with the apex_item package just like Ivan Dubashinskii suggested. You will need to submit the page in order to have the region re-generate your page items when the amount n is changed. There are some downsides to this approach, as for example it won't be that simple to use your page items in a page process.
Just limit your page items to some number, like 10. Then, create 10 page items in a declarative way and use dynamic actions to show/hide them when n changes. This way, it will also be much easier to use your items in a page procress.
I am using django-endless-pagination with its twitter-style pagination. Now i want to paginate over shuffled query set. I have tried to add
return Fact.objects.all().order_by('?')
But then objects can appear more than 1 time.
How can I change this behavior?
I think using pagination is misleading in this case. When the user clicks on page 2 it's not actually the second page, it's just another 20 items.
A better option is having a single button (called Fetch for example) and fetch 20 items (or whatever the page size is) each time the user clicks on it.
To avoid the same item showing up twice, you can keep a list of viewed ids in session and exclude them from the subsequent queries.
Another approach that you can take try is something that I used and I found on another StackOverflow post here.
import random
items = sorted(Fact.objects.all().order_by('nr'), key=lamda x: random.random())
return items
i'm a little bit struggling regarding how can i add more column in the costumer tabs in the admin page OSCommerce. As we can i only can display 8 customer in one page. How can i increase it ?
Thank you so much
I can't say for sure without seeing your catalog/admin/customers.php file... Is the following code in your customers.php file?
$customers_split = new splitPageResults($HTTP_GET_VARS['page'], MAX_DISPLAY_SEARCH_RESULTS, $customers_query_raw, $customers_query_numrows);
If so, then check the value of MAX_DISPLAY_SEARCH_RESULTS in your database. MAX_DISPLAY_SEARCH_RESULTS is in the 'configuration" table. If it is set to 8 then adjust it to the number of customer rows you prefer to see in the customers table.
Keep in mind that MAX_DISPLAY_SEARCH_RESULTS is used elsewhere and whatever value you adjust it too will be used everywhere MAX_DISPLAY_SEARCH_RESULTS is used.
We are querying a large SP 2007 document library with over 12,000 documents using the Lists web service, for document comparison purposes.
All queries are built using CAML, to limit the results returned by one of the fields on the list.
In general, the CAML query will return no more than 200 records.
Unfortunately, we are finding that one query will return 20 documents, and the exact same query will return 23 documents 15 minutes later.
As this crawl occurs after hours, it is not possible that documents have been added during that time.
Has anyone experienced similar issues ?
If you're using the Lists.GetListItems method, try setting the RowLimit parameter to something larger.
rowLimit A string that specifies the number of items, or rows, to
display on a page before paging begins. If supplied, the value of this
parameter overrides the row limit set in the view specified by the
viewName parameter or the row limit set in the default view for the
list.
If you don't specify, it will use the limit for the default view which is probably 200 judging by your question.
I don't understand the second part of your question. The search index uses a completely separate web service and you'll never use CAML to query the search index.
Turns out that the issue was related to hardware errors on one of our front end web servers.
This caused a validation failure for some of the List Items.
In one of the models overview panel, after I filter the items by month, I have to select them all and then create a document with information regarding them (kind of like a monthly report). This is a problem when one month has more than 100 items as Django paginates the filtering results.
Is there a way to increase the number of items shown from 100 to 400 or select all the items from the filtering result?
Selecting all the items from one page, creating a document, going to the next, creating another, etc, then merging the documents isn't an option.
See http://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_per_page
In your ModelAdmin definition, set list_per_page:
class MyModelAdmin(admin.ModelAdmin):
list_per_page = 400
I believe you can also add the all GET parameter to your query (ie, add ?all to the end of your url), but that only works if you have less than 200 items, and this limit is hardcoded in the admin. Therefore it's only useful if you have more than list_per_page (100) and less than 200 items, but the admin will offer you a link for this anyway when this is the case.
If you're talking about admin, see this.
ModelAdmin.list_per_page