I am using symfony 3.4.8 version. When I am updating form data I want to see the differences before submit and after submit.
After $formEdit->handleRequest($request); line old value is changing to the new value but I don’t want it.
but I could not do this.
Here is my controller.
/**
 * @Route("/contract/{contractId}/Edit", name="contract_edit")
 */
public function editContract(Request $request, $contractId)
{
    $log[]='';
    $contractInfo = $this->getDoctrine()->getRepository('AppBundle:myContractContract')->find($contractId);
    $formEdit = $this->createForm(ContractEditType::class, $contractInfo);
    $log[0]= $contractInfo;
    dump($log[0]); // Old value is coming here
    $formEdit->handleRequest($request);
    if ($formEdit->isSubmitted() && $formEdit->isValid()) {
        $log[1]= $contractInfo; //new value is comming
        dump($log[0]);
        dump($log[1]);
        // I want to compare the contractInfo before submitting and after submitting.
        // I want to see the differences before and after to take a log.
        // when I submit log[0] and log[1] are coming with same value.
        //Updating Contract table..
        $this->getDoctrine()->getManager()->flush();
        dump($this->getDoctrine()->getManager()->flush());
        $this->addFlash(
            'notice', 'Data is Updated.'
        );
        return $this->redirectToRoute('staff_index');
    }
    return $this->render('contract/contract_edit.html.twig',
        array(
            'contractInfo' => $contractInfo,
            'form' => $formEdit->createView(),
        )
    );
}
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Answer
In php, objects are passed by reference. Have a look at this small example.
<?php
class A {
    public $b = 1;
}
$a = new A();
$b = $a;
$bStill = clone($a);
// when we change $a it will affect $b, but not $bStill, because theres cloned object from previous state of object $a
$a->b = 2;
$c = $a;
// ["b"] => 2
var_dump($b);
// ["b"] => 2
var_dump($c);
// ["b"] => 1
var_dump($bStill);
So for your case, try using $log[0] = clone($contractInfo);