根据rand为变量赋值给变量一个字符串（1,9）

Question

我试图为变量赋值，我想知道是否有更方便和更简单的东西，通常的if / else if方式。

$typeofvariable = rand (1,9);
if ($typeofvariable == 1) {
$stringvalue = "type a";
}
else if ($typeofvariable == 2) {
$stringe value == "type b";
}
.......

提前谢谢。

Answer 1

您可以使用switch声明

$typeofvariable = rand (1,9);
switch ($typeofvariable) {
    case 1:
        $stringvalue = "type a";
        break;
    case 2:
        $stringvalue = "type b";
        break;
    case 3:
        // etc..
        break;
    default:
        // Code to be executed if non of above.
        break;
}

Answer 2

您可以像这样使用switch()：

<?php
$typeofvariable = rand (1,9);
switch ($typeofvariable) {
    case 1:
        //code to be executed if $typeofvariable=1;
        break;
    case 2:
        //code to be executed if $typeofvariable=2;
        break;
    case 3:
        //code to be executed if $typeofvariable=3;
        break;
    ...
    default:
        code to be executed if $typeofvariable is different from all labels;
} 
?>

Answer 3

因为这就是堆栈的工作原理。

这应该适合你：

<?php

    $map = array("type a", "type b", ...);
    $string = $map[array_rand($map)];

?>