我在数组中得到这样的格式 这是我的数组=
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我想要这样的格式
var array= [{"address":"Jaipur"},{"address":"Mumbai"},{"address":"Mumbai"}]
在JSON中应该做些什么,这样我才能获得所需的数组。
答案 0 :(得分:7)
您可以使用Set获取唯一值,并将spread再次添加到数组中。
var array= [{"address":"Jaipur"},{"address":"Mumbai"},{"address":"Mumbai"}];
var res = [...new Set(array.map(x => x.address))];
console.log(res)
答案 1 :(得分:2)
var array= [{"address":"Jaipur"},{"address":"Mumbai"},{"address":"Mumbai"}];
var address=[];
$.each(array,function(add,val){
address.push(val.address);
});
var address = Array.from(new Set(address));
console.log(address);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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答案 2 :(得分:1)
你可以reduce
数组:
var array= [{"address":"Jaipur"},{"address":"Mumbai"},{"address":"Mumbai"}];
var formatedArray = array.reduce((output, item) => {
if(!output.includes(item.address)) {
output.push(item.address)
}
return output
}, [])
console.log(formatedArray)
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答案 3 :(得分:-1)
string json = DataTableToJSON(dt_main);
jsonnew = @"{""Data"":" + json + "}";
private static string DataTableToJSON(DataTable table)
{
List<string[]> result = table.Rows
.Cast<DataRow>()
.Select(row => row.ItemArray
.Select(x => x.ToString())
.ToArray())
.ToList();
JavaScriptSerializer serializer = new JavaScriptSerializer();
return serializer.Serialize(result);
//------another method----------------
//ArrayList arr = new ArrayList();
//List<ArrayList> list = new List<ArrayList>();
//foreach (DataRow row in table.Rows)
//{
// arr = new ArrayList();
// foreach (DataColumn col in table.Columns)
// {
// arr.Insert(col.Ordinal, row[col]);
// }
// list.Add(arr);
//}
// JavaScriptSerializer serializer = new JavaScriptSerializer();
// return serializer.Serialize(list);
}
答案 4 :(得分:-2)
试试这个
var array= [{"address":"Jaipur"},{"address":"Mumbai"},{"address":"Mumbai"}];
var array_new=[];
for(var i=0;i<array.length;i++){
array_new.push(array[i].address);
}