javascript - 在jqgrid条件中跳过多选复选框

Question

我看到了Oleg answer当单击 jqgrid 标题中的多选复选框时，它会在禁用复选框时删除检查。 (如果我错了请纠正我)。但在我的例子中，我想跳过行数据，或者如果行值 Status is approved，我不想选中复选框。

我试过这个

onSelectAll: function (aRowids, status) {
  $.each(aRowids, function (i, val) {
     var gridId = "#List";
     var rowData = jQuery(gridId).jqGrid('getRowData', val);
     var g = $("#List");
     var cbs = $("tr.jqgrow > td > " + rowData.Status == "Approved", g[0]);
     cbs.removeAttr("checked");
   }
}

但什么也没有发生。它仍然检查已批准的状态。

Answer 1

给您一个解决方案 http://jsfiddle.net/HJema/632/

var myData = [{
    id: 1,
    status: "Rejected"
}, {
    id: 2,
    status: "Approved"
}, {
    id: 3,
    status: "Rejected"
}, ];

$("#list").jqGrid({
    datatype: "local",
    colNames: ["Id", "Status"],
    colModel: [{
        name: "id",
        index: "id",
        sorttype: "int"
    }, {
        name: "status",
        index: "status"
    }],
    caption: "Viz Test",
    pager: '#pager',
    search: true,
    multiselect: true,
    data: myData,
    loadComplete: function(data) {
    	for (var i = 0; i < data.rows.length; i++) {
      	if(data.rows[i].status == "Approved"){   
        	$('#jqg_list_' + (i+1)).attr('disabled', true);
        }
      }
    }
});

<link href="https://code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css" rel="stylesheet"/>
<link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.14/themes/base/jquery-ui.css" rel="stylesheet"/>
<link href="http://trirand.com/blog/jqgrid/themes/ui.jqgrid.css" rel="stylesheet"/>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://code.jquery.com/ui/1.12.1/jquery-ui.js"></script>
<script src="http://trirand.com/blog/jqgrid/js/jquery.jqGrid.min.js"></script>
<script src="http://trirand.com/blog/jqgrid/js/i18n/grid.locale-en.js"></script>
<table id="list"></table>
<div id="pager"></div>

There is some problem with the Stackoverflow snippet, please refer to the jsfiddle.

希望对你有帮助