MySQL结合了几个选择语句

Question

想知道如何/是否可以合并两个选择语句。

        $searchCat = $mysqli->query("SELECT category_id, category.name FROM category WHERE category.name = '$searchTerm'");

    while ($r=$searchCat->fetch_assoc()) {
        $category_id=$r["category_id"];

        $searchBus = $mysqli->query("SELECT business.name, business.street FROM business WHERE business.category_id = '$category_id'");             

        while ($r=$searchBus->fetch_assoc()) {
            $name=$r["name"];
            $street=$r["street"];

            echo "$name, $street<br>";
        }
    } // end of while

对于输出，我只需要Business表中的数据，第一个选择将找到搜索词的相关ID，该ID也包含在Business表中。

那么我可以在第一个选择获得ID并将其传递给第二个选择时让我跳过第一个While的联接或联合吗？

Answer 1

您也可以使其成为单个查询。

$query = "SELECT `business`.`name`, `business`.`street` 
          FROM `business`, `category` 
          WHERE 
             `business`.`category_id` = `category`.`category_id` 
              AND 
             `category`.`name` = '$searchTerm'";

$searchBus = $mysqli->query($query);

Answer 2

SELECT category_id, category.name 
FROM category c inner join business b using (category_id) 
WHERE category.name = {Term}

尽管当前可以进行SQL注入，但是请避免查询

Answer 3

您可以使用exists运算子：

SELECT business.name, business.street 
FROM   business
WHERE  EXISTS (SELECT *
               FROM   category
               WHERE  category.name = '$searchTerm' AND
                      category.category_id = business.category_id)

Answer 4

SELECT business.name, business.street, category.name
FROM business 
JOIN category ON category.category_id = business.category_id
WHERE category.name = '$searchTerm'

Answer 5

SELECT business.name, business.street 
FROM category c, business b
WHERE c.category_id = b.category_id and c.name = '$searchTerm'