想用逗号分隔绑定字符串 [英] want to bind string using comma separated

问题描述

的字符串JSON = QUOT; [

{

"的ParentId"：" a9764da3147845c184bd272cef6a5937" ;,

"路径"：" ; / LMS / Cabinet / bdc2cd8e1da3451c84e332d1aa74f605"，

" CreatedBY"：" admin"，

" IsActive"：" Y"，

CabinetName：LMS，

" FolderTag"：IT mobile computing，Comm Skills，

" Name"：" ; JAVA"，

" UpdatedBY"：""，

" Type"：" Folder"，

" IsDelete"：" N"，

" UpdatedON&qu​​ot;：""，

" Id"：" bdc2cd8e1da3451c84e332d1aa74f605"，

" CreatedON&qu​​ot;：" 2015_09_08-11：19：50"，

" TemplateId"：" dd42c8a71a954c1d948ef35492ee1242"

}，

{

&q uot; ParentId"：" a9764da3147845c184bd272cef6a5937"，

" Path"：" / LMS / Cabinet / 3aae020c256f4dc1ab705af67bede2c7"，

" CreatedBY"：" admin" ，

IsActive：Y，

" CabinetName"：LMS，

" FolderTag"： IT移动计算，通信技能，

"名称"：Spring，

" UpdatedBY"：""，

" Type"：" Folder"，

" IsDelete"：" N"，

" UpdatedON&qu​​ot;：""，

"标识"：" 3aae020c256f4dc1ab705af67bede2c7" ;,

" CreatedON"：" 2015_09_04-16：58：05" ;,

TemplateId：" dd42c8a71a954c1d948ef35492ee1242"

}，

{

" ParentId"：" a9764da3147845c184bd272cef6a5937 " ;,

"路径"：" / LMS /柜/ c139b33a22a94a25bf624b94450aee3e" ;,

" CreatedBY"："管理员" ;,

" IsActive"：" Y" ;,

" CabinetName"：" LMS" ;,

" FolderTag"："社会技能"，

" Name"：" SQL"，

" UpdatedBY"：""，

" Type"："文件夹，

" IsDelete"：" N"，

" UpdatedON&qu​​ot;：""，

" Id" ;：c139b33a22a94a25bf624b94450aee3e，

" CreatedON&qu​​ot;：2015_09_04-16：54：44"，

" TemplateId"：dd42c8a71a954c1d948ef35492ee1242"

}

]"



这是我的json字符串，我只希望名字逗号分隔成另一个变量的字段...



例如：字符串res = {" JAVA"，" SPRING"，" SQL"，... }



就像那个plz给我解决方案

解决方案

var objdata =

.parseJSON（json）;



在objdata中，您将获得使用for循环的所有值，获取索引6中的值，这是名称

string json="[
{
"ParentId":"a9764da3147845c184bd272cef6a5937",
"Path":"/LMS/Cabinet/bdc2cd8e1da3451c84e332d1aa74f605",
"CreatedBY":"admin",
"IsActive":"Y",
"CabinetName":"LMS",
"FolderTag":"IT mobile computing,Comm Skills",
"Name":"JAVA",
"UpdatedBY":"",
"Type":"Folder",
"IsDelete":"N",
"UpdatedON":"",
"Id":"bdc2cd8e1da3451c84e332d1aa74f605",
"CreatedON":"2015_09_08-11:19:50",
"TemplateId":"dd42c8a71a954c1d948ef35492ee1242"
},
{
"ParentId":"a9764da3147845c184bd272cef6a5937",
"Path":"/LMS/Cabinet/3aae020c256f4dc1ab705af67bede2c7",
"CreatedBY":"admin",
"IsActive":"Y",
"CabinetName":"LMS",
"FolderTag":"IT mobile computing,Comm Skills",
"Name":"Spring",
"UpdatedBY":"",
"Type":"Folder",
"IsDelete":"N",
"UpdatedON":"",
"Id":"3aae020c256f4dc1ab705af67bede2c7",
"CreatedON":"2015_09_04-16:58:05",
"TemplateId":"dd42c8a71a954c1d948ef35492ee1242"
},
{
"ParentId":"a9764da3147845c184bd272cef6a5937",
"Path":"/LMS/Cabinet/c139b33a22a94a25bf624b94450aee3e",
"CreatedBY":"admin",
"IsActive":"Y",
"CabinetName":"LMS",
"FolderTag":"Social Skills",
"Name":"SQL",
"UpdatedBY":"",
"Type":"Folder",
"IsDelete":"N",
"UpdatedON":"",
"Id":"c139b33a22a94a25bf624b94450aee3e",
"CreatedON":"2015_09_04-16:54:44",
"TemplateId":"dd42c8a71a954c1d948ef35492ee1242"
}
] "

this is my json string and i want to only "Name" field with comma separated into another variable...

example: string res={"JAVA","SPRING","SQL",...}

like that plz give me solution

解决方案

var objdata =

.parseJSON(json);

In objdata you will get all the values in which using for loop, get the values in the index 6, which is name

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