Question

我创建了一个json对象，想要读取内容并将值放入dataTable.But我无法读取Json对象的内容。

以下是我的javascript函数：

finalGrid : function(data){
            console.log("final Grid");
            var strJson = JSON.stringify(data);
            var dataJson = JSON.parse(strJson);
            console.log(dataJson);
            var table = $('#' + Dcn.tableId).dataTable();
            var tableObj = table.api();
            var rows = [];
            var field = Dcn.tableField;
            var count=0;
            $.each(dataJson, function(index, obj) {
                var row = [];
                row[field.dcnRefNo.index] = obj[field.dcnRefNo.name];
                row[field.name.index] = obj[field.name.name]";
                row[field.package.index] = obj[field.package.name];
                row[field.priority.index] = obj[field.priority.name];

                row.push("<button class='btn btn-labeled btn-success project-edit' " +
                        "onclick=\"Dcn.List.bindDcnStatusUpdate('"+obj["_id"]+"')\"> " +
                        "<span class='btn-label'><i class='fa fa-edit'></i></span>View Dcn</button>");
            rows.push(row);
            count = count+1;
            console.log(count);
        });

        tableObj.rows.add(rows).draw();
        $('#' + Dcn.tableId).show();

    }

怎么办？

Answer 1

如果你的json看起来像这样：

{"package": "package-a", "priority": "1", "dcnRefNo": "DCN/Duraline/001", "name": "abc"}

您可以使用此代码示例：

var data = [{"package": "package-a", "priority": "1", "dcnRefNo": "DCN/Duraline/001", "name": "abc"}];
function printJson(data){
            var strJson = JSON.stringify(data);
            var dataJson = JSON.parse(strJson);
            $.each(dataJson, function(index, obj) {
              console.log(obj.dcnRefNo);
              console.log(obj.name);
              console.log(obj.package);
              console.log(obj.priority);
        });
}

无法从Json Object读取字段的值

1 个答案: