Which among the below is the correct way to safely encode the URL "/admin/findUser.do?name=" + request.getParameter( "dangerousInput" )

technology security

  1. String safeURIToDisplay= "/admin/findUser.do?name=" + TCSSAPI.encoder().encodeForJavaScript(request.getParameter( "dangerousInput"));

  2. String safeURIToDisplay = TCSSAPI.encoder().encodeForURL( "/admin/findUser.do?name=" + request.getParameter( "dangerousInput" ) );

  3. String safeURIToDisplay= "/admin/findUser.do?name=" + com.tcs.sapi.io.ValidationUtil.encodeForURL(request.getParameter( "dangerousInput"));

  4. None of the above

Show answer
Correct Option: C

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