Having two cases in the same switch statement or branches in the same if structure with the same implementation is at best duplicate code, and at worst a coding error. If the same logic is truly needed for both instances, then in an if structure they should be combined, or for a switch, one should fall through to the other.

Moreover when the second and third operands of a ternary operator are the same, the operator will always return the same value regardless of the condition. Either the operator itself is pointless, or a mistake was made in coding it.

Noncompliant Code Example

switch ($i) {
  case 1:
    doSomething();
    break;
  case 2:
    doSomethingDifferent();
    break;
  case 3:  // Noncompliant; duplicates case 1's implementation
    doSomething();
    break;
  default:
    doTheRest();
}

if ($a >= 0 && $a < 10) {
  doTheThing();
else if ($a >= 10 && $a < 20) {
  doTheOtherThing();
}
else if ($a >= 20 && $a < 50) {
  doTheThing();  // Noncompliant; duplicates first condition
}
else {
  doTheRest();
}

if ($b == 0) {
  doOneMoreThing();
}
else {
  doOneMoreThing(); // Noncompliant; duplicates then-branch
}

var b = a ? 12 > 4 : 4;  // Noncompliant; always results in the same value

Compliant Solution

switch ($i) {
  case 1:
  case 3:
    doSomething();
    break;
  case 2:
    doSomethingDifferent();
    break;
  default:
    doTheRest();
}

if (($a >= 0 && $a < 10) || ($a >= 20 && $a < 50)) {
  doTheThing();
else if ($a >= 10 && $a < 20) {
  doTheOtherThing();
}
else {
  doTheRest();
}

doOneMoreThing();

b = 4;

or

switch ($i) {
  case 1:
    doSomething();
    break;
  case 2:
    doSomethingDifferent();
    break;
  case 3:
    doThirdThing();
    break;
  default:
    doTheRest();
}

if ($a >= 0 && $a < 10) {
  doTheThing();
else if ($a >= 10 && $a < 20) {
  doTheOtherThing();
}
else if ($a >= 20 && $a < 50) {
  doTheThirdThing();
}
else {
  doTheRest();
}

if ($b == 0) {
  doOneMoreThing();
}
else {
  doTheRest();
}

int b = a ? 12 > 4 : 8;