查询以从给定范围的下拉菜单中选择值

Question

我正在制作临时网站，我想根据下拉菜单中具有不同年龄组的条件进行搜索查询

<select class="form-input" id="age" name="age"  >
    <option value="Age"> <div class="dot"></div> Age</option>
    <option value="1"> <div class="dot"></div> 0-20/option>
    <option value="2"> <div class="dot"> </div>21-30</option>
    <option value="3"> <div class="dot"></div> 31-40</option>
    <option value="4"> <div class="dot"></div> >40 </option>
</select>

我想显示满足上述任何年龄条件的所有用户我尝试在sql中使用上面的开关大小写，但没有得到任何结果。

$age = mysqli_real_escape_string($db, $_POST['age']);
echo $age;
switch ($age) {
    case 1: $age = ' AND age BETWEEN 100 AND 130 ';  break;
    case 2: $age = ' AND age BETWEEN 131 AND 150 ';  break;
    case 3: $age = ' AND age BETWEEN 151 AND 200 ';  break;
}
$result = mysqli_query($db,
    "SELECT * 
    FROM approved_user 
    WHERE 
        first_name= '$fn'
        OR age=$age "
) or die(mysqli_error($db));
while ($row3 = mysqli_fetch_array($result)) {
    $id     = $row3['id'];
    $fname  = $row3['first_name'];
    $lname  = $row3['surname'];
    $dob    = $row3['dob'];
    $gender = $row3['gender'];
    $age    = $row3['age'];
}

所以请帮忙。

Answer 1

您生成的查询没有意义。

例如，给定以下值：

$age = "  AND age BETWEEN 100 AND 130 ";

这个表达式：

"SELECT * FROM approved_user WHERE (first_name= '$fn') OR  (age=$age)"

实际将生成：

"SELECT * FROM approved_user WHERE (first_name= '$fn') OR  (age= AND age BETWEEN 100 AND 130)"

这不是有效的 SQL。您可能想要：

switch ($age) {
    case 1: $filter = "100 AND 130";  break; 
    case 2: $filter = "131 AND 150";  break;  
    case 3: $filter = "151 AND 200";  break;   
 }
 $result = mysqli_query($db,
    "SELECT *
    FROM approved_user 
    WHERE 
        first_name= '$fn' 
        OR age BETWEEN $filter"
) or die(mysqli_error($db));