将相关模型的方法放到jqgrid的单独列中 [英] Taking method of the related model into a separate column of jqgrid

问题描述

我正在使用 django 1.4 和 django_gems 包中的 jqgrid以下代码试图通过连接名字和姓氏将 virtual 列带到 grid 上.

I am using django 1.4 together with jqgrid from django_gems package The following code is trying to bring the virtual column onto the grid by concatenating first and last name.

但是它失败了

无法将关键字 client__get_fullname 解析为字段.

Cannot resolve keyword client__get_fullname into field.

有什么可以接受的方法来达到这个目的吗?

Is there any acceptable way to achieve this?

 class Car(models.Model):
     number = models.CharField(max_length = 50)
 client = models.ForeignKey('Client')

 class Client(models.Model):
     first_name = models.CharField(max_length = 50)
 last_name = models.CharField(max_length = 50)
 def get_fullname(self):
     return '%s %s' % (self.first_name, self.last_name)

 from django_gems.jqgrid
 import JqGrid
 class CarGrid(JqGrid):
     queryset = Car.objects.all()
 fields = ['number', 'client__get_fullname']

 jqgrid config = {
     "altRows": true,
     "rowList": [10, 25, 50, 100],
     "sortname": "id",
     "viewrecords": true,
     "autowidth": false,
     "forcefit": false,
     "shrinkToFit": false,
     "height": "auto",
     "colModel": [{
         "index": "id",
         "editable": false,
         "name": "id",
         "label": "ID"
     }, {
         "index": "number",
         "editable": false,
         "name": "number",
         "label": "number"
     }, {
         "index": "first_name",
         "editable": false,
         "name": "client__first_name",
         "label": "first name"
     }],
     "caption": "Cars",
     "datatype": "json",
     "gridview": true,
     "sortorder": "asc",
     "viewsortcols": true,
     "url": "main/examplegrid",
     "rowNum": 10,
     "pager": "#pager",
     "jsonReader": {
         "repeatitems": false
     }
 }

 sample data = {
     "total": 1,
     "records": 1,
     "rows": [{
         "client__first_name": "Bill",
         "client__last_name": "Clinton",
         "id": 1,
         "number": "111222"
     }],
     "page": 1
 }

推荐答案

好的！让我们获取 JSON 数据

OK! Let us you get the JSON data

{
    "total": 1,
    "records": 1,
    "rows": [
        {
            "client__first_name": "Bill",
            "client__last_name": "Clinton",
            "id": 1,
            "number": "111222"
        }
    ],
    "page": 1
}

jqGrid 包含一个额外的列

and the jqGrid contains an additional column

{name: "client__full_name", label: "full name"}

应该由 client__first_name 和 client__last_name 构成.在这种情况下，最简单的方法是使用 beforeProcessing 回调函数:

which should be constructed from client__first_name and client__last_name. In the case the most simplest way would be to use beforeProcessing callback function:

$("#list").jqGrid({
    url: "main/examplegrid",
    datatype: "json",
    colModel: [
        {name: "id", label: "ID"},
        {name: "client__first_name", label: "first name"},
        {name: "client__last_name", label: "last name"},
        {name: "client__full_name", label: "full name"}
    ],
    gridview: true,
    jsonReader: { repeatitems: false },
    //... other parameters
    beforeProcessing: function (data) {
        var items = data.rows, n = items.length, i, item;
        for (i = 0; i < n; i++) {
            item = items[i];
            item.client__full_name = item.client__first_name + ' ' +
                item.client__last_name;
        }
    }
});

回调函数beforeProcessing会在从服务器接收到数据之后，在数据处理之前被jqGrid调用.因此，我们可以通过简单的方式实现任何虚拟"列.

The callback function beforeProcessing will be called by jqGrid after the data are received from the server and before the data will be processed. So in the simple way we can implement any "virtual" column.

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