jquery easyui datagrid 嵌套对象绑定

jquery easyui datagrid 嵌套对象绑定，网上很多资料说是easyui本身不支持嵌套对象绑定，但很多时候我们恰恰需要这种嵌套对象绑定，而实际上是可以实现且支持的，只需调用formatter函数就可以实现。

json 格式：

{                                                     
 "total":239,                                                     
 "rows":[                                                         
  {"code":"001","user":{"id":1,"name":"name1"},"addr":"Address 11","col4":"col4 data"},        
  {"code":"002","user":{"id":1,"name":"name1"},"addr":"Address 13","col4":"col4 data"},        
  {"code":"003","user":{"id":1,"name":"name1"},"addr":"Address 87","col4":"col4 data"},        
  {"code":"004","user":{"id":1,"name":"name1"},"addr":"Address 63","col4":"col4 data"},        
  {"code":"005","user":{"id":1,"name":"name1"},"addr":"Address 45","col4":"col4 data"},        
  {"code":"006","user":{"id":1,"name":"name1"},"addr":"Address 16","col4":"col4 data"},         
  {"code":"007","user":{"id":1,"name":"name1"},"addr":"Address 27","col4":"col4 data"},         
  {"code":"008","user":{"id":1,"name":"name1"},"addr":"Address 81","col4":"col4 data"},         
  {"code":"009","user":{"id":1,"name":"name1"},"addr":"Address 69","col4":"col4 data"},         
  {"code":"010","user":{"id":1,"name":"name1"},"addr":"Address 78","col4":"col4 data"}    
 ]                                                         
}                                                          
js代码：

{field:'Id',title:'Id',width:120,
      formatter:function(value,rec){
       return rec.user.id;
      }
     },
     {field:'Name',title:'Name',width:120,
      formatter:function(value,rec){
       return rec.user.name;
      }
     }

注意，field命名是不能重复的，这样就可以实现嵌套对象绑定了

评论

3 楼 天助者自助 2014-07-07  

我想问一下   {"dataSize":2147483647,"diskSize":2147483647,"hosts":[{"health":true,"host":"localhost:"27017","id":0,"optime":0,"state":"PRIMARY"}],"id":1,"maxDataSize":0,"opLogSize":0,"percent":98,"syncOk":true,"writable":true}

数据是这样的 我想绑定hosts下面的host 应该怎么绑定  谢谢  

2 楼 zhenther 2014-01-17  

fenglingabc 写道

楼主说的都是对的，不过要注意一种情况：
如果你的结果集中某条记录 中嵌套的对象是null的话，数据就无法正常显示，比如：
{"code":"001","user":{"id":1,"name":"name1"},"addr":"Address 11","col4":"col4 data"},        
{"code":"001","user":null,"addr":"Address 11","col4":"col4 data"}
所以，我们还是在   formatter函数中加个判断，这样才是完整的：
formatter:function(value,rec){
  if(value != null){
     {
       return rec.user.name;
      }
  else{
       return "";
      }
}

是的，细节很重要

1 楼 fenglingabc 2013-12-26  

楼主说的都是对的，不过要注意一种情况：
如果你的结果集中某条记录 中嵌套的对象是null的话，数据就无法正常显示，比如：
{"code":"001","user":{"id":1,"name":"name1"},"addr":"Address 11","col4":"col4 data"},        
{"code":"001","user":null,"addr":"Address 11","col4":"col4 data"}
所以，我们还是在   formatter函数中加个判断，这样才是完整的：
formatter:function(value,rec){
  if(value != null){
     {
       return rec.user.name;
      }
  else{
       return "";
      }
}