这是jqgrid - json looping issue的延续，区别在于我已将 colNames 和 colModel 移至JSON。
JSON
{
"colModel": [
{
"name": "linkimg",
"index": "linkimg",
"width": 60,
"align": "left",
"jsonmap": "cells.0.links"
}
],
"colNames": [
"linkimg"
],
"mypage": {
"outerwrapper": {
"page": "1",
"total": "1",
"records": "1",
"innerwrapper": {
"rows": [
{
"id": "1",
"read": true,
"cells": [
{
"label": "linkimg",
"value": "Link-A",
"links": [
{
"name": "link1"
},
{
"name": "link2"
},
{
"name": "link3"
}
]
}
]
},
{
"id": "2",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-B",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "3",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-C",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "4",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-D",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "5",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-E",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "6",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-F",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "7",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-G",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "8",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-H",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "9",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-I",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "10",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-J",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "11",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-K",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "12",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-L",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "13",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-M",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "14",
"read": false,
"cells": [
{
"label": "linkimg",
"value": "Link-N",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
},
{
"id": "15",
"read": true,
"cells": [
{
"label": "linkimg",
"value": "Link-O",
"links": [
{
"name": "link1"
},
{
"name": "link2"
}
]
}
]
}
]
}
}
}
}
JQGrid定义
$(document).ready(function () {
$.ajax({
type: "GET",
url: "myjqgrid.json",
data: "",
dataType: "json",
success: function(response){
var columnData = response.mypage.outerwrapper,
columnNames = response.colNames,
columnModel = response.colModel;
$("#myjqgrid").jqGrid({
datatype: 'jsonstring',
datastr: columnData,
colNames: columnNames,
colModel: columnModel,
jsonReader: {
root: "innerwrapper.rows",
repeatitems: false
},
gridview: true,
pager: "#Pager",
rowNum: 2,
rowList: [2, 4, 6, 8],
viewrecords: true,
recordpos: 'left',
multiboxonly: true,
multiselect: true,
sortname: 'id',
sortorder: "desc",
sorttype: "text",
sortable: true,
caption: "<h2>My JQGrid</h2>",
width: "1406",
height: "100%",
scrolloffset: 0,
loadonce: true,
cache: true,
loadComplete: function(data){
}
});
}
});
$("#myjqgrid").jqGrid('navGrid','#Pager');
});
问题
如何从formatter:中取出colModel？如果我将它保存在JSON中，则jqgrid不会显示。
答案 0 :(得分：2)
如果您不需要unformatter，可以使用格式化程序扩展预定义格式化程序列表：
$.fn.fmatter.myLinkFormatter = function (val, options, rawObject) {
....
};
然后在列定义中使用"formatter": "myLinkFormatter"
{
"colModel": [
{
"name": "linkimg",
"index": "linkimg",
"width": 60,
"align": "left",
"formatter": "myLinkFormatter",
"jsonmap": "cells.0.links"
}
],
"colNames": [
"linkimg"
],
"mypage": {
"outerwrapper": {
...
}
}
请参阅the demo。
还有一句话：你应该像我在演示中所做的那样在$("#myjqgrid").jqGrid('navGrid','#Pager');处理程序中移动success。