My JSON object looks like this:
{
"destination_addresses" : [ "Paris, France" ],
"origin_addresses" : [ "Amsterdam, Nederland" ],
"rows" : [
{
"elements" : [
{
"distance" : {
"text" : "504 km",
"value" : 504203
},
"duration" : {
"text" : "4 uur 54 min.",
"value" : 17638
},
"status" : "OK"
}
]
}
],
"status" : "OK"
}
I will need the "504 km" value from distance. How can i do this?
You can use the DBXJSON
unit, included since Delphi 2010.
Try this sample
uses
DBXJSON;
{$R *.fmx}
Const
StrJson=
'{ '+
' "destination_addresses" : [ "Paris, France" ], '+
' "origin_addresses" : [ "Amsterdam, Nederland" ], '+
' "rows" : [ '+
' { '+
' "elements" : [ '+
' { '+
' "distance" : { '+
' "text" : "504 km", '+
' "value" : 504203 '+
' }, '+
' "duration" : { '+
' "text" : "4 uur 54 min.", '+
' "value" : 17638 '+
' }, '+
' "status" : "OK" '+
' } '+
' ] '+
' } '+
' ], '+
' "status" : "OK" '+
'}';
procedure TForm6.Button1Click(Sender: TObject);
var
LJsonObj : TJSONObject;
LRows, LElements, LItem : TJSONValue;
begin
LJsonObj := TJSONObject.ParseJSONValue(TEncoding.ASCII.GetBytes(StrJson),0) as TJSONObject;
try
LRows:=LJsonObj.Get('rows').JsonValue;
LElements:=TJSONObject(TJSONArray(LRows).Get(0)).Get('elements').JsonValue;
LItem :=TJSONObject(TJSONArray(LElements).Get(0)).Get('distance').JsonValue;
ShowMessage(TJSONObject(LItem).Get('text').JsonValue.Value);
finally
LJsonObj.Free;
end;
end;