Version: 20 R7 BETA

OB Copy

OB Copy ( object {; resolvePtrs | {; option {; groupWith}}} ) : Object

Parameter	Type		Description
object	Object, Object	→	Structured object
resolvePtrs	Boolean	→	True = resolve pointers, False or omitted = do not resolve pointers
option	Integer	→	ck shared: return a shared object,ck resolve pointers: resolve pointers before copying
groupWith	Collection, Object	→	Shared collection or object to be grouped with the resulting object
Function result	Object	←	Deep copy of object

Description

The OB Copy command returns an object containing a complete (deep) copy of the properties, sub-objects and values for the object.

First syntax: OB Copy(object{; resolvePtrs})

If object contains pointer type values, by default the copy also contains the pointers. However, you can resolve pointers when copying by passing True in the resolvePtrs parameter. In this case, each pointer present as a value in object is evaluated when copying and its dereferenced value is used.

Note: If properties of object are shared objects or shared collections, they become regular (not shared) objects or collections in the returned copy. Use the second syntax if you want to return shared elements (see below).

Second syntax: OB Copy(object{; option{; groupWith}})

If passed, the option parameter can contain one or both of the following constants:

option	Description
ck resolve pointers	If the original object contains pointer type values, by default the copy also contains the pointers. However, you can resolve pointers when copying by passing the ck resolve pointers constant. In this case, each pointer present in the object is evaluated when copying and its dereferenced value is used.
ck shared	By default, OB Copy returns a regular (not shared) object, even if the command is applied to a shared object. Pass the ck shared constant to create a shared object. In this case, you can use the groupWith parameter to associate the shared object with another collection or object (see below).

The groupWith parameter allows you to designate a collection or an object with which the resulting object should be associated.

Notes:

OB Copy can be used with an entity selection object. By default if ck shared is omitted, a non-shareable entity selection is returned. If ck shared is passed, a shareable entity selection is returned. In this context, the groupWith option is useless since an entity selection does not have a locking identifier.
The same principle applies to entity selections stored inside properties of object. However, the command is optimized when ck shared is passed and a nested entity selection is shareable: the same entity selection reference is returned in the resulting object.
Datastore, dataclass, and entity objects are not copiable. If OB Copy command is called with them, a Null value is returned

Example 1

You want to duplicate an object containing simple values:

 var $Object : Object
 var $JsonString : Text
 
 ARRAY OBJECT($arraySel;0)
 ALL RECORDS([Product])
 While(Not(End selection([Product])))
    OB SET($Object;"id";[Product]ID_Product)
    OB SET($Object;"Product Name";[Product]Product_Name)
    OB SET($Object;"Price";[Product]Price)
    OB SET($Object;"Tax rate";[Product]Tax_rate)
    $ref_value:=OB Copy($Object) //direct copy
    APPEND TO ARRAY($arraySel;$ref_value)
  //the contents of $ref_value are identical to those of $Object
    NEXT RECORD([Product])
 End while
  //the contents of $ref_value
 $JsonString:=JSON Stringify array($arraySel)

Example 2

You duplicate an object containing pointers (first syntax):

 var $ref : Object
 
 OB SET($ref;"name";->[Company]name) //object with pointers
 OB SET($ref;"country";->[Company]country)
 ARRAY OBJECT($sel;0)
 ARRAY OBJECT($sel2;0)
 
 ALL RECORDS([Company])
 
 While(Not(End selection([Company])))
    $ref_comp:=OB Copy($ref) // copy without evaluating pointers
  // $ref_comp={"name":"->[Company]name","country":"->[Company]Country"}
    $ref_comp2:=OB Copy($ref;True) //copy with evaluation of pointers
  // $ref_comp={"name":"4D SAS","country":"France"}
    APPEND TO ARRAY($sel;$ref_comp)
    APPEND TO ARRAY($sel2;$ref_comp2)
    NEXT RECORD([Company])
 End while
 
 $Object:=JSON Stringify array($sel)
 $Object2:=JSON Stringify array($sel2)
 
  // $Object = [{"name":"","country":""},{"name":"","country":""},...]
  // $Object2 = [{"name":"4D SAS","country":"France"},{"name":"4D, Inc","country":"USA"},{"name":"Catalan","country":"France"}...]

Example 3

We want to copy the regular (non shared) $person object into the $sharedObject shared object. To do this, we must create a shared copy of the object ($sharedObject).

 var $person;$copy;$sharedObject : Object
 var $text : Text
 
 $text:=Document to text(Get 4D folder(Current resources folder)+"person.txt")
 $person:=JSON Parse($text) //$person is a standard object
 $sharedObject:=New shared object()
 $copy:=OB Copy($person;ck shared) //$copy is a shared object
 
  //So it can be put in $sharedObject
 Use($sharedObject)
    $sharedObject.person:=$copy
 End use

Example 4

$obj contains a pointer ("name" property) on the current record "name" field.

 var $obj;$objWithPtr;$sharedObjWithPtr : Object
 $obj:=New object()
 
  //$obj is an object with a pointer
 OB SET($obj;"name";->[Persons]name)
 
 ALL RECORDS([Persons])
  //Now there is a current record on [Persons] table so [Persons]name is filled
  //
  // If we want to copy $obj as a standard object with evaluation of pointers
  // We do this:
 $objWithPtr:=OB Copy($obj;True)
  //
  // If we want to copy $obj as a shared object with evaluation of pointers
  // We do this:
 $sharedObjWithPtr:=OB Copy($obj;ck resolve pointers+ck shared)

Example 5

We want to copy $sharedObj in $sharedColl but since they belong to different shared groups, a direct copy would result in an error. We must make a copy of $sharedObj and designate $sharedColl as shared group for the copy.

 var $sharedObj;$objCopy : Object
 var $sharedColl : Collection
 
  //$sharedObj belongs to a shared group
 $sharedObj:=New shared object("lastname";"Smith";"address";New shared object("city";"New York"))
  //$sharedColl belongs to another shared group
 $sharedColl:=New shared collection(New shared object("lastname";"Brown"))
 
 $objCopy:=OB Copy($sharedObj;ck shared;$sharedColl)
  //$objCopy is now in the same shared group as $sharedColl
 
  //So $objCopy can be put in $sharedColl without error
 Use($sharedColl)
    $sharedColl.push($objCopy)
 End use

Description​

Example 1​

Example 2​

Example 3​

Example 4​

Example 5​

See also​