php-如何将此 IF 条件放入开关内？

Question

请帮助我更正此代码，我想将 if 放在开关内：

switch ($urlcomecatid) {

 if ($urlcomeparentid == 1 || $urlcomeparentid == 2 || $urlcomeparentid == 3)
  break;

case "50":
case "51":
case "52":
case "109":
case "110":
    //do nothing and exit from switch
    break;
default:
    header ("Location:http://www.example.com/tech/tech.php"); exit();
    break;

}

Answer 1

正确的代码应该是这样的

switch ($urlcomecatid) {
    case "50":
    case "51":
    case "52":
    case "109":
    case "110":
        //do nothing and exit from switch
        break;
    default:
        if ($urlcomeparentid == 1 || $urlcomeparentid == 2 || $urlcomeparentid == 3)
            break;
        header ("Location:http://www.example.com/tech/tech.php"); exit();
        break;
 }

Answer 2

使用 if 而不是 switch。

<?php
if(!in_array($urlcomeparentid, array(1, 2, 3) && !in_array($urlcomecatid, array('50', '51', '52', '109', '110')){
    header ("Location:http://www.example.com/tech/tech.php");
    exit();
}

您应该避免 exit() 因为它会杀死脚本的其余部分，这在调试时是潜在的问题来源。