Chisum asked 5 years ago
is there any requirejs integration magento2 uses it and I really dislike everything about it but I have to use it for this client. I just keep phpgrid out of the requirejs, the only problem is that there’s no way to delay jQuery being called on load with phpgrid, and any hooks I write to make jquery global aren’t ready yet because requirejs is slow. I’d need a way to delay the jQuery calls of phpgrid at all until I choose to fire them. Simply having jQuery load from a script tag breaks other modules that use requirejs and jquery on the page because requirejs should never be used by anyone for any reason.
Abu answered 5 years ago
Hello,
I’ve not tested it with requriejs. Give me some time and i’ll update you back.
Abu answered 5 years ago
One solution could be to move all initialization JS code inside a function and call it when you need.
Edit lib/inc/jqgrid_dist.php
Find this code segment:
< script >
var phpgrid = jQuery(“#<?php echo $grid_id?>”);
var phpgrid_pager = jQuery(“#<?php echo $grid_id.”_pager”?>”);
var fx_ajax_file_upload;
var fx_replace_upload;
…..
jQuery(document).ready(function(){
<?php echo $this->render_js($grid_id,$out);?>
});
< /script >
and then wrap it in a funciton, e.g.
< script >
function phpgrid_init()
{
var phpgrid = jQuery(“#<?php echo $grid_id?>”);
var phpgrid_pager = jQuery(“#<?php echo $grid_id.”_pager”?>”);
var fx_ajax_file_upload;
var fx_replace_upload;
…..
jQuery(document).ready(function(){
<?php echo $this->render_js($grid_id,$out);?>
});
}
< /script >
This way, grid will not call any jQuery code until you call phpgrid_init() function.
Hope it helps.
Abu answered 5 years ago
A little correction, to make phpgrid global JS available, you need to move function vars outside phpgrid_init() declaration.
— script start —
var fx_ajax_file_upload;
var fx_replace_upload;
var fx_bulk_update;
var fx_bulk_unrequire;
var fx_get_dropdown;
var fx_reload_dropdown;
var fx_grid_resize;
var fx_show_form;
var fx_tooltip_init;
var fx_show_view_dialog;
function phpgrid_init()
{
var phpgrid = jQuery(“#<?php echo $grid_id?>”);
var phpgrid_pager = jQuery(“#<?php echo $grid_id.”_pager”?>”);
…
jQuery(document).ready(function(){
<?php echo $this->render_js($grid_id,$out);?>
});
}
— script end —
Also declare this var outside phpgrid_init:
var fx_tooltip_init;
and replace:
function fx_tooltip_init()
with:
fx_tooltip_init = function ()
Abu GhufranChisum answered 5 years ago
I keep getting some return not in function error I\’ll have to try again soon
Chisum answered 5 years ago
Your code worked I just had to deal with caching.
I got it to load error free but it was a nightmare for something that would have been braindead simple and no extra work otherwise. Screw requirejs.
require.config({
paths: {
//’jqjquery’: ‘/phpgrid/lib/js/jquery.min’,
‘grid-en’: ‘/phpgrid/lib/js/jqgrid/js/i18n/grid.locale-en’,
‘jqgrid’: ‘/phpgrid/lib/js/jqgrid/js/jquery.jqGrid.min’,
‘jq-ui’: ‘/phpgrid/lib/js/themes/jquery-ui.custom.min’,
},
shim: {
/*’jqjquery’: {
exports:’jquery’
},*/
‘grid-en’: {
deps:[‘jquery’]
},
‘jqgrid’: {
deps:[‘jquery’]
},
‘jq-ui’: {
deps:[‘jquery’]
}
}
});
require([‘jquery’,’grid-en’,’jqgrid’,’jq-ui’], function(jQuery) {
window.$ = jQuery;
window.jQuery = jQuery;
phpgrid_init();
jQuery(document).ready(dobinds);
});
