根据表单中的用户输入更改PHP文档中的SQL查询很困难

我有一个PHP文档，它根据搜索表单的输入值查询MYSQL数据库。一切都很完美，除了我正在努力增加一个功能，我相信这将使整个应用程序更加用户友好

<form>
    <label for="alcohol_check">Search clubs where alcohol permitted?</label>
       <input type="radio" id="alcohol_check1" value="1" /><label for="radio1">No</label>
       <input type="radio" id="alcohol_check2" value="2" /><label for="radio2">Yes</label>
       <input type="radio" id="alcohol_check3" value="???" /><label for="radio3">Show both</label>

    <label for="club_name">Enter social club name:</label>
       <input type="text" id="club_name" name="club_name"/>
</form>

$user_alcohol_permitted_selection = $_POST['alcohol_check']; //Value sent using jquery .load()
$user_social_club_name_input = $_POST['name']; //Value sent using jquery .load()

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input
        AND
        WHERE alcohol_permitted = $user_alcohol_permitted_selection";

我数据库中的表基本上存储了各种社交俱乐部的信息。它有关于他们的位置、姓名、订票价格和允许饮酒的数据。我需要帮助的部分与酒精有关。在数据库中，its存储1或2，2表示允许饮酒，1表示不允许饮酒

<form>
    <label for="alcohol_check">Search clubs where alcohol permitted?</label>
       <input type="radio" id="alcohol_check1" value="1" /><label for="radio1">No</label>
       <input type="radio" id="alcohol_check2" value="2" /><label for="radio2">Yes</label>
       <input type="radio" id="alcohol_check3" value="???" /><label for="radio3">Show both</label>

    <label for="club_name">Enter social club name:</label>
       <input type="text" id="club_name" name="club_name"/>
</form>

$user_alcohol_permitted_selection = $_POST['alcohol_check']; //Value sent using jquery .load()
$user_social_club_name_input = $_POST['name']; //Value sent using jquery .load()

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input
        AND
        WHERE alcohol_permitted = $user_alcohol_permitted_selection";

当前我的表单如下所示：

<form>
    <label for="alcohol_check">Search clubs where alcohol permitted?</label>
       <input type="radio" id="alcohol_check1" value="1" /><label for="radio1">No</label>
       <input type="radio" id="alcohol_check2" value="2" /><label for="radio2">Yes</label>
       <input type="radio" id="alcohol_check3" value="???" /><label for="radio3">Show both</label>

    <label for="club_name">Enter social club name:</label>
       <input type="text" id="club_name" name="club_name"/>
</form>

$user_alcohol_permitted_selection = $_POST['alcohol_check']; //Value sent using jquery .load()
$user_social_club_name_input = $_POST['name']; //Value sent using jquery .load()

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input
        AND
        WHERE alcohol_permitted = $user_alcohol_permitted_selection";

目前的问题是，我的查询将只检索提供酒精或不提供酒精的社交俱乐部。因为此时从允许形式传递的值为1或2。 我可以对我的查询做些什么，以便如果用户选择“两者”选项，它检索提供酒精的社交俱乐部和不提供酒精的俱乐部？我想不出来！查询中是否需要if语句？

<form>
    <label for="alcohol_check">Search clubs where alcohol permitted?</label>
       <input type="radio" id="alcohol_check1" value="1" /><label for="radio1">No</label>
       <input type="radio" id="alcohol_check2" value="2" /><label for="radio2">Yes</label>
       <input type="radio" id="alcohol_check3" value="???" /><label for="radio3">Show both</label>

    <label for="club_name">Enter social club name:</label>
       <input type="text" id="club_name" name="club_name"/>
</form>

$user_alcohol_permitted_selection = $_POST['alcohol_check']; //Value sent using jquery .load()
$user_social_club_name_input = $_POST['name']; //Value sent using jquery .load()

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input
        AND
        WHERE alcohol_permitted = $user_alcohol_permitted_selection";

---

谢谢你检查这个，如果我没有解释清楚的话，很抱歉

好的，首先，请将alchohol_允许列的数据类型更改为bool（MySQL中的位（1））。只是读书让我觉得很冷

现在，谈谈你的问题。尝试以下where语句：

WHERE $user_alcohol_permitted_selection IS NULL 
OR alcohol_permitted = $user_alcohol_permitted_selection

这应该行得通

$user_alcohol_permitted_selection = $_POST['alcohol_check']; //Value sent using jquery .load()
$user_social_club_name_input = $_POST['name']; //Value sent using jquery .load()

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input";

if ($user_alcohol_permitted_selection == "???")
{
     $query.= "AND WHERE alcohol_permitted = 1 OR alcohol_permitted = 2";
} else {
     $query.= "AND WHERE alcohol_permitted = $user_alcohol_permitted_selection";
}

此外，如果列

仅包含值1或2，则允许使用该列。我建议您将列类型更改为bool，就像TheAllmitchConnors在回答中建议的那样

然后，可以将条件语句缩减为：

$query="SELECT * FROM social_clubs 
        WHERE name = $user_social_club_name_input";    

if ($user_alcohol_permitted_selection != "???")
{
     $query.= "AND WHERE alcohol_permitted = $user_alcohol_permitted_selection";
}

请注意
您确实需要阅读并开始使用准备好的语句来避免SQL注入。
其中酒精允许=1和酒精允许=2
应该是
或
x=1和x=2总是假的：）
if（！$user\u alcohold\u allowed\u selection==“？”）
应该是
if（$user\u alcohold\u allowed\u selection！=“？”）
，而我们在这里。@FrankFarmer甚至想不出任何借口。我能怪JavaScript还是C？