EasyUI
Home
Demo
Tutorial
Documentation
Download
Extension
Contact
Forum
EasyUI Forum
January 24, 2026, 02:31:10 PM
Welcome,
Guest
. Please
login
or
register
.
1 Hour
1 Day
1 Week
1 Month
Forever
Login with username, password and session length
News
:
Home
Help
Search
Login
Register
EasyUI Forum
>
General Category
>
EasyUI for jQuery
>
[SOLVED]How can I get field with datagrid onRowContextMenu
Pages: [
1
]
« previous
next »
Print
Author
Topic: [SOLVED]How can I get field with datagrid onRowContextMenu (Read 11577 times)
aiit
Newbie
Posts: 12
[SOLVED]How can I get field with datagrid onRowContextMenu
«
on:
June 26, 2019, 04:48:52 PM »
How can I get field with datagrid onRowContextMenu(e, index, row)
thanks
«
Last Edit: June 27, 2019, 10:46:30 PM by aiit
»
Logged
stworthy
Administrator
Hero Member
Posts: 3581
Re: How can I get field with datagrid onRowContextMenu
«
Reply #1 on:
June 27, 2019, 07:15:54 AM »
Please try this code to get the field name.
Code:
onRowContextMenu: function(e){
var td = $(e.target).closest('td[field]');
var field = td.attr('field');
console.log(field)
}
Logged
aiit
Newbie
Posts: 12
Re: How can I get field with datagrid onRowContextMenu
«
Reply #2 on:
June 27, 2019, 10:45:43 PM »
Thank you very much Mr. stworthy.
Quote from: stworthy on June 27, 2019, 07:15:54 AM
Please try this code to get the field name.
Code:
onRowContextMenu: function(e){
var td = $(e.target).closest('td[field]');
var field = td.attr('field');
console.log(field)
}
Logged
Pages: [
1
]
Print
« previous
next »
Jump to:
Please select a destination:
-----------------------------
General Category
-----------------------------
=> News
=> General Discussion
=> EasyUI for jQuery
=> EasyUI for Angular
=> EasyUI for Vue
=> EasyUI for React
=> Bug Report
Loading...