如果变量等于值 php [英] If variable equals value php

问题描述

我正在尝试在数据插入 MySQL 查询之前进行检查.这是代码；

I am trying to do a check before the data inserts into the MySQL query. Here is the code;

$userid = ($vbulletin->userinfo['userid']);
$sql3 = mysql_query("SELECT * FROM table WHERE ID='$_POST[hiddenID]'");

while ($row = mysql_fetch_array($sql3)){

$toon = $row['toonname'];
$laff = $row['tlaff'];
$type = $row['ttype'];

if ($type == 1){
$type == "Bear";
} elseif ($type == 2){
$type == "Cat";
} elseif ($type == 3){
$type == "Dog";
}            

}

但是，这不起作用.基本上，每种类型的表"中都有不同的值.1 表示熊，2 表示猫，3 表示狗.

However, this isn't working. Basically, there are different values in the 'table' for each type. 1 means Bear, 2 means Cat, and 3 means Dog.

感谢能帮我发现我脚本中的问题的人！

Thanks to whomever can help see a problem in my script!

推荐答案

你是在比较，而不是在分配:

You are comparing, not assigning:

if ($type == 1){
  $type = "Bear"; 
}

您将值与 == 或 === 进行比较.

You compare values with == or ===.

您使用 = 分配值.

您也可以编写更少的代码来获得相同的结果，使用 switch 语句，或者只是一堆没有 elseifif>s.

You could write less code to achieve the same result too, with a switch statement, or just a bunch of ifs without the elseifs.

if ($type == 1) $type = "Bear";
if ($type == 2) $type = "Cat";
if ($type == 3) $type = "Dog";

我会为它创建一个函数，如下所示:

I would make a function for it, like this:

function get_species($type) {
    switch ($type):
        case 1: return 'Bear';
        case 2: return 'Cat';
        case 3: return 'Dog';
       default: return 'Jeff Atwood';
    endswitch;
}

$type = get_species($row['ttype']);

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