如果声明 - 要么是假的

Question

继承我的代码，我想拥有它，如果未检查无验证或验证，那么用户必须“选择购买选项”...我的方式下面的工作方式但只有一个更简单的方法如果声明？

if ($buyoption == "With Validation") {
} else if ($buyoption == "No Validation") {
} else {
    $buyoptionError = "Please select a buy option";
    echo "$buyoptionError <br>";
    $YesorNo = 0;
}

Answer 1

优化

if ($buyoption != 'With Validation' && $buyoption != "No Validation") {
     $buyoptionError = "Please select a buy option";
    echo "$buyoptionError <br>";
    $YesorNo = 0;
}

Answer 2

可能是你可以使用这个

if (((trim($buyoption) === "With Validation") || (trim($buyoption) === "No Validation")))===false) {
$buyoptionError = "Please select a buy option";
echo "$buyoptionError <br>";
$YesorNo = 0;}else{
// your code
}

Answer 3

您也可以使用开关：

switch($buyoption){

 case "With Validation":
  #code
  break;

 case "No Validation":
  #code
  break;

 default:
  $buyoptionError = "Please select a buy option";
  echo "$buyoptionError <br>";
  $YesorNo = 0;

}

Answer 4

echo ($YesorNo = ($buyoption == "With Validation" || $buyoption == "No Validation")) ? "": "Please select a buyoption";