Grid | Show/Hide Column Dynamically

[rajParticipant]

Hi,

I’ve a grid of 10 columns in which only 5 columns are important and the rest I made it as hidden in columns definition (hidden:true).

Also, I’ve built one custom listbox that will be like a filter menu i.e. this will be visible when cog wheel dropdown icon is clicked. I’ll show the custom listbox in a pop-up with “Apply” button. This listbox contains the list of all hidden column names with checkboxes.

So whenever the user wants to see any hidden columns, the respective column name checkbox is checked and click on “Apply” button. I’ll show the hidden column using following code along with the “getCheckedItems” list,

$('#userSummary').jqxGrid('hidecolumn', 'lastModifiedDate');

Let’s say, I’ve checked all 5 columns and clicked on “Apply” button. I can able to show all 5 hidden columns. Suppose If I uncheck all and clicked on “Apply“, I couldn’t able to restore the grid to the previous state i.e. grid with 5 important columns.

Hope I explained clearly. Can someone suggest me a solution?

Looking forward.

Regrads,
Raj

[VladimirParticipant]

Hello Raj,

You can call ‘showcolumn‘ for each checked item, and ‘hidecolumn‘ for each unchecked, when you press apply.
This way you will set the state for each column each time you hit apply.

Best Regards,
Vladimir

jQWidgets Team
http://www.jqwidgets.com

[rajParticipant]

Hi Vladimir,

I tried as you suggested and it is working fine :-). Thanks for your reply.

Regards,
Raj

[Author]

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