在$ .map中选择嵌套对象

如何在类似下面的代码中选择嵌套对象的键？

dominant = { 
    "owner1":{ 
     "sub1":{ 
      "super_sub1":"Amy", 
      "super_sub2":"Luara", 
      "super_sub3":"Daria", 
      "super_sub4":"Maria" 
     }, 
     "sub2":{ 
      "extra_sub1":"Puppy", 
      "extra_sub2":"Dog", 
      "extra_sub3":"Wolf", 
      "extra_sub4":"Pet" 
     } 
    }, 
    "owner2":{ 
     "sub1":{"super_sub1":"Issaballa"}, 
     "sub2":{"extra_sub1":"Puppy"} 
    }, 
    "owner3":{ 
     "sub1":{ 
      "super_sub1":"Mary", 
      "super_sub2":"Amy", 
      "super_sub3":"Laura", 
      "super_sub4":"Daria" 
     }, 
     "sub2":{ 
      "extra_sub1":"Wolf", 
      "extra_sub2":"Dog" 
     } 
    }, 
    "owner4":{ 
     "sub1":{ 
      "super_sub1":{"name":"Maria", "cords":[2,1,36,1,29,15,9,15]}, 
      "super_sub2":{"name":"Daria", "cords":[37,5,37,33,31,25,31,14]}, 
      "super_sub3":{"name":"Luara", "cords":[9,23,29,23,36,37,2,37]}, 
      "super_sub4":{"name":"Amy", "cords":[1,5,7,14,7,25,1,33]}, 
      "super_sub5":{"name":"Issaballa", "cords":[9,16,29,16,29,22,9,22]} 
     }, 
     "sub2":{ 
      "extra_sub1":{"name":"Puppy", "cords":[2,1,36,1,27,8,11,8]}, 
      "extra_sub2":{"name":"Dog", "cords":[37,1,37,37,29,30,29,8]}, 
      "extra_sub3":{"name":"Wolf", "cords":[11,30,27,30,36,37,2,37]}, 
      "extra_sub4":{"name":"Pet", "cords":[1,1,9,8,9,30,1,37]}, 
      "extra_sub5":{"name":"BarkBark", "cords":[10,10,28,10,28,28,10,28]} 
     } 
    }, 
}; 
$.map(dominant, function(dom, domi){ 
    console.log("DOM:");console.log(dom);//equal to owners {} 
    console.log("DOMI:");console.log(domi);//equal to owners 
    return (
     "<div class='tt_sect "+tvi+"'></div>" 
    ); 
}).join("")

我想选择业主键对象（SUB1 & SUB2）在$ .MAP内。我将如何做到这一点？我知道这不是太难。

来源

2016-08-11 Crystal Miller

据我理解你的问题，你需要访问SUB1和SUB2，所以这里顺便说一下，如果你只想地图做的是

$.map(dominant, function(dom, domi){ 
    console.log("DOM:");console.log(dom);//equal to owners {} 
    console.log("DOMI:");console.log(domi);//equal to owners 
    $.map(dom, function(sub, subi){ 
     console.log("DOM:");console.log(sub);//equal to subs {} 
     console.log("DOMI:");console.log(subi);//equal to subs  
    })  
    return ("<div class='tt_sect "+tvi+"'></div>"); 
}).join("")

，或者如果你只是想通过迭代sub1和sub2你可以用这个

$.map(dominant, function(dom, domi){ 
     console.log("DOM:");console.log(dom);//equal to owners {} 
     console.log("DOMI:");console.log(domi);//equal to owners 
     $.each(sub, function() { 
      console.log("DOM:");console.log(sub);//equal to subs {}   
     });      
    return ("<div class='tt_sect "+tvi+"'></div>"  
    }).join("")

来源

2016-08-11 19:50:28

在$ .map中选择嵌套对象

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