用于过滤 php 中结果的下拉菜单 从 sql 中获取所有结果

Dropdown menu to filter results in php get all results from sql

我有这个下拉列表:

<select class="w-select filterdropdown" id="Tipo-de-cocina" name="tipofiltro" data-name="Tipo de cocina">
<?php
    if (isset($_GET['tipofiltro'])) {
        echo '<option value="' .$filtrocuisine.  '"> '  .$filtrocuisine. "</option>";

    } else {
        echo '<option value="Todos los tipos">Todos los tipos</option>';
    }

?>


              <option value="Japonesa">Japonesa</option>
              <option value="Mexicana">Mexicana</option>
              <option value="India">India</option>
              <option value="Mediterranea">Mediterranea</option>
              <option value="Italiana">Italiana</option>
              <option value="Americana">Americana</option>
              <option value="Asiatica">Asiatica</option>
              <option value="Thai">Thai</option>
              <option value="China">China</option>
              <option value="Francesa">Francesa</option>
              <option value="Turca">Turca</option>
              <option value="Latina">Latina</option>
              <option value="Africana">Africana</option>
              <option value="Griega">Griega</option>
              <option value="Arabe">Arabe</option>


            </select>

当用户选择字段 "Todos los tipos" 我的 sql 查询 returns 所有类型时,我该如何做到这一点?这是后面的sql:

if (isset($_GET['preciofiltro']) OR isset($_GET['preciofiltro'])) {
    $filtroprecio = $_GET['preciofiltro'];
    $filtrocuisine = $_GET['tipofiltro'];

    $sql = "SELECT Meals.Meal_ID, Meals.Name, Price, Cooks.Cook_ID 
            FROM Meals 
               INNER JOIN Cooks ON Cooks.Cook_ID = Meals.Cook_ID 
            WHERE Cooks.Area = '$area' 
              AND Meals.Capacity > 0 
              AND Meals.Price < '$filtroprecio' 
              AND Meals.Type = '$filtrocuisine'";

    $result = mysqli_query($conn, $sql);

基本上我需要诸如"AND Meals.Type = any"

之类的东西

干杯!

好吧,有更简单的方法,但无需重写所有内容,简单的答案是,如果用户从下拉列表中选择 Todos los tipos,您真正想要做的是从查询中删除此选择条件 AND Meals.Type = '$filtrocuisine'完全即您不再使用该条件限制查询。

所以像这样更改你的脚本:-

我当然假设您已经从 $_GET 数组中获取数据,验证它,并在我们得到这段代码之前清理它。

if (isset($_GET['preciofiltro']) OR isset($_GET['preciofiltro'])) {
    $filtroprecio = $_GET['preciofiltro'];
    $filtrocuisine = $_GET['tipofiltro'];

$sql = "SELECT Meals.Meal_ID, Meals.Name, Price, Cooks.Cook_ID 
        FROM Meals 
           INNER JOIN Cooks ON Cooks.Cook_ID = Meals.Cook_ID 
        WHERE Cooks.Area = '$area' 
          AND Meals.Capacity > 0 
          AND Meals.Price < '$filtroprecio'";

if ( isset($filtrocuisine) && $filtrocuisine == 'Todos los tipos' ) {
    $sql .= " AND Meals.Type = '$filtrocuisine'";
}

$result = mysqli_query($conn, $sql);