jqGrid has a column chooser functionality as like described here:
http://www.trirand.com/jqgridwiki/doku.php?id=wiki:jquery_ui_methods
and a demonstration from #Oleg:
http://www.ok-soft-gmbh.com/jqGrid/SimpleLocalGridWithColumnChooser.htm
However its too complex for my needs. Is there any simple way as like FlexGrid's show-hide columns choosing functionality:
http://flexigrid.info/
You can check Example 1. Columns can be shown and hide by clicking columns. I don't want to same thing but it would be great just clicking somewhere at or near columns and choosing just columns to show or hide.
Any ideas?
PS: I want a basic show-hide column field not a complex screen that opens and people selects many options on them. It should be build on table as like Flexigirid. users click checkboxes and thats all.
Sorry, but you can't have the same look in one software product like you know in another one.
Do you can configure flexigrid so that it looks like jqGrid? Do you can that iPhone looks like Windows Phone or Mac OS X looks like Windows? Every software product has its own design and the set of features.
The columnCooser are based on the multiselect widget and allows you to show or hide any columns or reorder columns with respect of drag & drop. flexigrid has another implementation of the features.
If you want to use some free open soure product you should understand the features and the possibilities which provide the product and just use there.
Related
I need to have a BI Dashboard where the columns(eg, Brand) or group of columns ( eg, Company, Company site) have to hide/show in all analysis of dashboard based on check box selections with Brand and Company respectively. I am able to pass the column header and formula based on selection through Presentation Variable in Prompts, but stuck with hiding columns when the respective check box in unchecked. Note than I'm using OACv5.8.
Thanks in advance for any help on how to achieve/proceed further.
Why don't you use standard, out-of-the-box functionality like column selectors or show/hide sections?
Instead pf playing with presentation variables and trying to fiddle around.
I'm a FM beginner and would like to know how I can show a searchable dropdown. When I create a product, I need to select a supplier from a large list. I would like to tyoe some characters, from where I find a value in the dropdown. As far as I see now, that isn't possible and the only way to do this is through a popover.
But my question is then, how to show a full table in the popover, from which I can select a value, which is filled in my form.
I would like to tyoe some characters, from where I find a value in the dropdown. As far as I see now, that isn't possible and the only way to do this is through a popover.
Actually, it's the other way around: with a drop-down list you get the option to Auto-complete using value list, which is unavailable with a pop-up menu.
how to show a full table in the popover, from which I can select a value,
You cannot show a full table in a popover. You can show a portal in a popover, and make it show all records in a table by using the x operator when defining the relationship.
Or - preferably, IMHO - use a card window to show all the records you want to select from; this could be the full table or a reduced set as the result of a find.
Selecting products or other related records using popovers always feels slow when you try to search/filter the list of records, with the introduction of card windows doing such searches/selects became a lot faster and easier to implement, if you have 2 different layouts that need to search the same list of records you can use the same card window for both, that was also a nice thing.
Using extjs 4.1.1
I have a grid with lots of columns >20.
Initially, most of these columns are hidden.
If a user wants to unhide the column, they select the menu on any column, then select the "columns" choice, this expands another dropdown/dropout which shows all of the columns. Those with checkboxes next to them are shown.
My issue is this:
The columns in the dropdown are shown in the order in which they are defined/displayed in the grid. The order in which they are displayed in the grid has been chosen for a good reason( e.g. id as the first column). However, when a user wants to display one of the hidden columns, it is hard for them to find it in the list. This is because the list is sorted in the order the columns are defined. I want to sort the column dropdown/dropout list in alphabetical order, without effecting the order of the columns in the grid.
How can this done?
I think I found the solution to your question.
First of all I don't have the Ext JS 4.1.1. framework on my current PC. So I tried to figure out you problem reading the Ext JS 4.1.3. documentation available on Sencha's site. But I don't think that they have made drastical changes in this part of the framework between the two minor releases so my solution should work in your case too.
I have tried out my solution using JSFiddle. Unfortunately they did not have the 4.1.1. ext-all.css file, so I have linked manually the 4.0.2 file available at Sencha, so the menu is looking a bit missplaced.
The header menu and it's submenus are managed by the Ext.grid.header.Container class. The column submenu is constructed by the getColumnMenu method. The whole menu is purged and reconstructed on every drag and drop or other event which should affect the grid view. As a result it is enough to overwrite this method in order to solve the problem. Because the headercontainer class is too deep in the framework it is hard to extend it, so you have to make use of the Ext.base.override method.
The column submenu's menu items are created from the result of the
items = headerContainer.query('>gridcolumn[hideable]')
query. So you have to first sort alphabetically the result, before creating the menu items. I have added to the class the sortColumns method which does all the sorting stuff.
So here is what I did: link to my solution.
I hope that this is what you were looking for.
I want to change the default behavior of the configurable products option. By default it shows options in a dropdown box; I want to show that on a grid in place of the dropdown.
Creating a new product selector is a pretty big undertaking. I cannot tell you how to turn it into a grid because you have to handle multiple selections, etc. But what I can tell you is that the file you are going to spend a lot of time in is:
/app/design/frontend/base/default/template/catalog/product/view/options.phtml
I have done something similar.
The main Problem you will face is to make it work with the magento javascript components, so you can still use all functions which are provided by Magento.
As this is nearly impossible not to breake, i used a script, which places an own html control over the selectbox.
https://github.com/claviska/jQuery-SelectBox
This is mainly for customice Design of a DropDown box, but you can "misuse" it to have arbitary structures, as you have usual html elements you can place how you wish with CSS.
You could try looking at using a plugin such as this one by Magento Mechanics which will do all the hard work for you:
http://www.magemechanics.com/product-grid-options.html
I have used and recommend there plugin.
I'm currently using jqGrid and ASP.Net MVC. With my current project, my goal is to provide a grid of data to the end user, and they can then edit this. The data is machine-generated, and the users will be confirming if the machine is correct or not.
I think ideally for speed, I'd like to provide a row per item, with a radio button group as the editable. The users could then pick from the values 'Unknown', 'Correct', 'Incorrect'.
As there will be a lot of data, I'd also like to provide a control of some type that can set all rows in the grid to one of the available radio button choices, for the user experience.
Given that there seems to be no native support for this in jqGrid, I wanted to ask if anyone has had any experience writing something like this, and whether this is achievable and reliable, or whether I should stick with the drop-down editable approach that is native to jqGrid.
To implement radio button as the editable instead of the standard drop-down editable approach you can use so named custom editing feature of jqGrid (see http://www.trirand.com/jqgridwiki/doku.php?id=wiki:common_rules#custom). This allows you to create any custom control to edit the cell value. An example of the implementation you can find here: Add multiple input elements in a custom edit type field.
To set all rows in the grid to one of the available radio button choices you can use either a control outside of jqGrid or add an additional custom button in the navigation bar (see http://www.trirand.com/jqgridwiki/doku.php?id=wiki:custom_buttons). If you search for navButtonAdd you will find a lot of examples how to implement this, for example, Jqgrid: navigation based on the selected row. Because you use server based data, you can just call a method on the server to make the changes which you need and then call trigger("reloadGrid") to refresh jqGrid data.