php if 和 elseif 语句具有多个/超过 3 个条件

Question

我想在用户注册网站时为其分配随机个人资料图片。我的代码有问题。如果我将“if 和 elseif”语句限制为 2 个条件，则效果很好。但是当我超过 2 个条件时(在本例中我有 10 个条件)，代码就不起作用了。如果应该使用“switch”语句来代替，您将如何编写代码？

这是我的代码

$rand = rand(1,10); //random number between 1 and 10

if($rand == 1)
$profile_pic = "/defaults/profile_pic1.png";

else if($rand == 2)
$profile_pic = "/defaults/profile_pic2.png";

.
.
.

else if($rand == 9)
$profile_pic = "/defaults/profile_pic9.png";

else
$profile_pic = "/defaults/profile_pic10.png";

Answer 1

我希望这就是你想要的

<?php
$rand = rand(1,10);
switch ($rand) {
    case "1":
        $profile_pic = "/defaults/profile_pic1.png";
        echo "1";
        break;
    case "2":
        $profile_pic = "/defaults/profile_pic2.png";
        echo "2";
        break;
    case "3":
        $profile_pic = "/defaults/profile_pic3.png";
        echo "3";
        break;
    case "4":
        $profile_pic = "/defaults/profile_pic4.png";
        echo "4";
        break;
    case "5":
        $profile_pic = "/defaults/profile_pic5.png";
        echo "5";
        break;
    case "6":
        $profile_pic = "/defaults/profile_pic6.png";
        echo "6";
        break;
    case "7":
        $profile_pic = "/defaults/profile_pic7.png";
        echo "7";
        break;
    case "8":
        $profile_pic = "/defaults/profile_pic8.png";
        echo "8";
        break;
    case "9":
        $profile_pic = "/defaults/profile_pic9.png";
        echo "9";
        break;
    case "10":
        $profile_pic = "/defaults/profile_pic10.png";
        echo "10";
        break;
    default:
        $profile_pic = "/defaults/profile_picDEFAULT.png";
        echo "default PHOTO";
}
?>