从含有阵列

Question

阵列删除重复我有一个JSON数组这样

[{"Email":"someone@some.com","Name":"ACO","Groups":["MOD_SW","MOD_PI","MOD_GE"],"Id":63,"Url":"aco"}, 
{"Email":"someone@some.com","Name":"Agpo","Groups":["MOD_PI"],"Id":22,"Url":"agpo"}, 
{"Email":"someone@some.com","Name":"Akatherm","Groups":["MOD_SW"],"Id":64,"Url":"akatherm"}, 
{"Email":"someone@some.com","Name":"Albrand","Groups":["MOD_PI,"MOD_GE"],"Id":23,"Url":"albrand"}] 

我想创建一个新的数组（选择标记）具有鲜明Groups。

这个Groups是一个数组。

我想要的选择框有以下值：

MOD_SW 
MOD_PI 
MOD_GE 

我的JS：

UpdateSelectMenu: function (selectId, data) { 
     $(selectId).empty(); 
     $(selectId).html("<option value='all' selected='selected'>All groups</option>"); 
     var array_unique_values = []; 

     for (var i = 0; i < data.Groups.length; i++) { 
     for (var j = i+1; j < data.Groups.length; j++) { 
      if (data.Groups[i] === data.Groups[j]) { 
       j = ++i; 
      } 
     } 
     array_unique_values.push(data.Groups[i]); 
     } 

     array_unique_values = array_unique_values.sort(); 

     $.each(array_unique_values, function (k, v) { 
     $(selectId).append("<option value='" + v + "'>" + v + "</option>"); 
     }); 
    } 

我也试过

for (var i = 0; i < data.length; i++) { //browse whole data 
     for (var j = 0; j < data[i].Groups.length; j++) { //browse Groups array 
      for (var k = j + 1; j < data[i].Groups.length; k++) { 
       if (data[i].Groups[j] === data[i].Groups[k]) { 
        continue; 
       } 
      } 

      array_unique_values.push(data[i].Groups[j]); 
     } 
     } 

但由于出现错误：Groups.length为空或不是物体

此代码附加到select标记组值，但它显示为重复项，因为Groups是一个数组。

我必须创建一个新的for语句来浏览Groups数组？

或者还有另外一种方法避免嵌套for陈述？

谢谢

Answer 1

2个嵌套循环可以做的工作：

var data = [... your data ...]; 
var groups = []; 
$.each(data, function(i, item) { 
    $.each(item.Groups, function(j, group) { 
     if ($.inArray(group, groups) == -1) { 
      groups.push(group); 
     } 
    });  
}); 

// at this stage groups = ["MOD_SW", "MOD_PI", "MOD_GE"] 

，如果你想直接附加选项，您的下拉列表：

var groups = []; 
$.each(data, function(i, item) { 
    $.each(item.Groups, function(j, group) { 
     if ($.inArray(group, groups) == -1) { 
      groups.push(group); 
      $(selectId).append(
       $('<option/>', { 
        value: group, 
        text: group 
       }) 
      ); 
     } 
    });  
}); 

---

UPDATE：

又使这个效率更高，你可以定义一个静态的方法：

$.extend({ 
    distinct : function(arr) { 
     var result = []; 
     $.each(arr, function(index, value) { 
      if ($.inArray(value, result) == -1) { 
       result.push(value); 
      } 
     }); 
     return result; 
    } 
}); 

，然后使用.map方法：

var data = [... your data ...]; 
var groups = $.distinct($(data).map(function() { 
    return this.Groups;  
})); 
$.each(groups, function(index, group) { 
    $(selectId).append(
     $('<option/>', { 
      value: group, 
      text: group 
     }) 
    ); 
}); 

Answer 2

在最后一组首先看看："Groups":["MOD_PI,MOD_GE"] 你不需要逗号之后MOD_PI后关闭的报价并打开它们或这是一套班子？

如果用引号是问题的脚本可以是这个样子：

var obj = [{"Email":"someone@some.com","Name":"ACO", "Groups":["MOD_SW","MOD_PI","MOD_GE"],"Id":63,"Url":"aco"}, 
{"Email":"someone@some.com","Name":"Agpo", "Groups":["MOD_PI"],"Id":22,"Url":"agpo"}, 
{"Email":"someone@some.com","Name":"Akatherm", "Groups":["MOD_SW"],"Id":64,"Url":"akatherm"}, 
{"Email":"someone@some.com","Name":"Albrand", "Groups":["MOD_PI","MOD_GE"],"Id":23,"Url":"albrand"}] 

var temp = {}, 
    result = []; 

for (var i = 0; i < obj.length; i+=1) { 
    for (var j = 0; j < obj[i]['Groups'].length; j+=1) { 
     if (typeof temp[obj[i]['Groups'][j]] === 'undefined') { 
      result.push(obj[i]['Groups'][j]); 
      temp[obj[i]['Groups'][j]] = true; 
     } 
    } 
} 

如果这是一组集：

for (var i = 0; i < obj.length; i+=1) { 
    for (var j = 0; j < obj[i]['Groups'].length; j+=1) { 
     currentGroups = obj[i]['Groups'][j].split(','); 
     for (var k = 0; k < currentGroups.length; k += 1) { 
      currentGroups[k] = currentGroups[k].replace(/ /g,""); 
      if (typeof temp[currentGroups[k]] === 'undefined') { 
       result.push(currentGroups[k]); 
       temp[currentGroups[k]] = true; 
      }  
     } 
    } 
} 

我认为这是最有效的方式，因为你正在用O（1）检查重复项，你不必做额外的工作（例如排序任何数组）。