Question

我正在用PHP构建用户模块。它读取基于文本的用户类型，并分配一个数值来管理报表视图权限：

if ($usr_type == "Architect"){ $usr_rights = 0; }
if ($usr_type == "Director"){ $usr_rights  = 1; }
if ($usr_type == "Manager"){ $usr_rights  = 2; }
if ($usr_type == "Associate"){ $usr_rights  = 3; }
if ($usr_type == "Assistant"){ $usr_rights  = 4; }
if ($usr_type == "External"){ $usr_rights  = 5; }

有更简单，更优雅的方法吗？

Answer 1

您可以使用switch()执行if工作。

switch($usr_type) {
    case "Architect":
        $usr_rights = 0;
    break;
    case "Director":
        $usr_rights  = 1;
    break;
    case "Manager":
        $usr_rights  = 2;
    break;
    case "Associate":
        $usr_rights  = 3;
    break;
    case "Assistant": 
        $usr_rights  = 4;
    break;
    case "External":
        $usr_rights  = 5;
    break;
}

或使用数组。

$valid_types = array(
                "Architect" => 0,
                "Director" => 1,
                "Manager" => 2,
                "Associate" => 3,
                "Assistant" => 4,
                "External" => 5
            );

if(isset($valid_types[$usr_type])) {
    $usr_rights = $valid_types[$usr_type];
}

Answer 2

我从你当前的解决方案中看到了两种不同的解决方案。

1。像这样制作一个开关语句：

switch($usr_type) {

    case "Architect":
        $usr_rights = 0;
    break;

    case "Director":
        $usr_rights = 1;
    break;

    //...

}

有关switch()的详情，请参阅手册：http://php.net/manual/en/control-structures.switch.php

2。或者制作一个数组并在其中搜索：

$arr = array(0 => "Architect", 1 => "Director", 2 => "Manager", 3 => "Associate", 4 => "Assistant", 5 => "External");           

if(array_search($usr_type, $arr) !== false)
    $usr_rights = array_search($usr_type, $arr);

Answer 3

您需要使用switch statement

switch($usr_type) {

    case "Architect":
        $usr_rights = 0;
        break;

    case "Director":
        $usr_rights = 1;
        break;
    case "Manager":
        $usr_rights = 2;

    etc..

}

Answer 4

您可以使用数组函数array_map。如果我们考虑＆＃34; Architect＆＃34;，＆＃34; Director＆＃34;等是关键。然后我们可以使用地图结构来组合键及其值。

<?php
$usr_rights = array(0, 1, 2, 3, 4, 5);
$usr_types = array("Architect", "Director", "Manager", "Associate", "Assistant","External");
$c = array_combine($usr_types,$usr_rights);
print_r($d);

// Test
echo $c["Architect"]; // output: 0
?>

请参阅文档：http://php.net/manual/en/function.array-map.php

简化多个if语句

4 个答案: