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我有这个下拉：

<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”时，我该如何做到这一点，我的 sql 查询返回所有类型？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!

4

1 回答 1

1

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Todos los tipos从下拉列表中选择您真正想要做的是AND Meals.Type = '$filtrocuisine'从查询中完全删除此选择条件，即您不再使用该条件限制查询。

所以像这样改变你的脚本： -

我当然假设您已经从$_GET数组中获取数据，对其进行验证，并在我们获取此代码之前对其进行清理。

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);
于 2015-09-24T15:28:22.870 回答