Hi Guys
i have created on child table there is two fields one name as data type and other one is view as button type so when i do refresh the child table button is disappearing
and whenever i clicked on child table suddenly button appears
thanks and regards
Eliyas
I was wondering the same as I’m facing the same issue. Any idea on how to fix button and make it visible with no need of clicking on the row?
U can show button by using jquery
refresh(frm) {
frm.fields_dict["field_name"].$wrapper.find('.grid-body .rows').find(".grid-row").each(function (i, item) {
$(item).find('[data-fieldname="field_name"]').empty().append("<button class='btn btn-xs btn-success'>Finish</button>").click(function () {
let cdn = $(item).attr('data-name')
let cdt = frm.fields_dict["field_name"].grid.doctype
frm.script_manager.trigger('field_name', cdt, cdn)
})
});
};
@Thiago_Henry yes,
i have created new data field ‘action_button’ as readonly in child table
In refresh function i am changing field type to button using below script
let btn_class;
let status;
let html_tag;
let html_attribute;
let button_label = 'Select'
frm.fields_dict[child_table_fieldname].$wrapper.find('.grid-body .rows').find(".grid-row").each(function (i, item) {
btn_class = 'btn btn-primary';
html_tag = 'input';
html_attribute = 'type="button"';
$(item).find('[data-fieldname="action_button"]').replaceWith('<'+ html_tag +' class="col col-xs-1 ' + btn_class + '"' +
'data-fieldname="action_button" data-fieldtype="Data" ' + html_attribute +
' value="'+ button_label +'" style="margin: 2px;">' +
'</div><div class="static-area ellipsis"></div></' + html_tag + '>');
});
@niyaz_razak your script is add button only in one row , when we have add the new row this is not working what to to . please give me the suggection.
your script is add button only in one row , when we have add the new row this is not working what to to . please give me the suggection.