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Find data by name
Source position: fpjson.pp line 114
public function TJSONData.FindPath( |
const APath: TJSONStringType |
):TJSONData; |
APath |
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Path to search for |
The found data element, or nil
FindPath finds a value based on its path. If none is found, Nil is returned. The path elements are separated by dots and square brackets, as in object member notation or array notation. The path is case sensitive.
The following code will return the value itself, i.e. E will contain the same element as D:
Var D,E : TJSONData; begin D:=TJSONIntegerNumber.Create(123); E:=D.FindPath(''); end.
The following code will not return anything:
Var D,E : TJSONData; begin D:=TJSONIntegerNumber.Create(123); E:=D.FindPath('a'); end.
The following code will return the third element from the array:
Var D,E : TJSONData; begin D:=TJSONArray.Create([1,2,3,4,5]); E:=D.FindPath('[2]'); Writeln(E.AsJSON); end.
The output of this program is 3.
The following code returns the element Age from the object:
Var D,E : TJSONData; begin D:=TJSONObject.Create(['Age',23, 'Lastame','Rodriguez', 'FirstName','Roberto']); E:=D.FindPath('Age'); Writeln(E.AsJSON); end.
The code will print 23.
Obviously, this can be combined:
Var D,E : TJSONData; begin D:=TJSONObject.Create(['Age',23, 'Names', TJSONObject.Create([ 'LastName','Rodriguez', 'FirstName','Roberto'])]); E:=D.FindPath('Names.LastName'); Writeln(E.AsJSON); end.
And mixed:
var D,E : TJSONData; begin D:=TJSONObject.Create(['Children', TJSONArray.Create([ TJSONObject.Create(['Age',23, 'Names', TJSONObject.Create([ 'LastName','Rodriguez', 'FirstName','Roberto']) ]), TJSONObject.Create(['Age',20, 'Names', TJSONObject.Create([ 'LastName','Rodriguez', 'FirstName','Maria']) ]) ]) ]); E:=D.FindPath('Children[1].Names.FirstName'); Writeln(E.AsJSON); end.
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Class of TJSONArray |
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Class of TJSONObject |
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GetPath |