查询 return 空结果但数据存在

Question

我正在使用 Slim 和 PDO 以及 MySql 用于 return 我的数据库中可用的 matches 的特定列表。我的查询是这样的：

SELECT m.*, 
       t.name AS home_team_name, 
       t2.name AS away_team_name 
FROM `match` m 
LEFT JOIN team t ON m.home_team_id = t.id 
LEFT JOIN team t2 ON m.away_team_id = t2.id
WHERE (home_team_id = 117 OR away_team_id = 117) AND round_id = 488

如果我执行这个 query 我会得到一个列表 matches:

但是在使用 Slim 开发的 API 中，我得到一个空数组。这是方法结构：

$app->get('/match/get_matches_by_team/{round_id}/{team_id}/{type}', function (Request $request, Response $response, array $args)
{
  $query = "SELECT m.*,
  t.name AS home_team_name,
  t2.name AS away_team_name
  FROM `match` m
  LEFT JOIN team t ON m.home_team_id = t.id
  LEFT JOIN team t2 ON m.away_team_id = t2.id
  WHERE ";

  switch($args["type"])
  {
    case "home":
          $query .= "home_team_id = :team_id AND ";
    break;
    case "away":
        $query .= "away_team_id = :team_id AND ";
    break;
    default:
        $query .= "(home_team_id = :team_id OR away_team_id = :team_id) AND ";
    break;
  }

  $query .= "round_id = :round_id";

  $sql = $this->db->prepare($query);
  $sql->bindParam("team_id", $args["team_id"]);
  $sql->bindParam("round_id", $args["round_id"]);
  $sql->execute();

  $result = $sql->fetchAll();
  return $response->withJson($result);
});

我做错了什么？

在此先感谢您的帮助。

更新

如果我这样做 echo $query; return; 我会得到：

SELECT m.*,
t.name AS home_team_name,
  t2.name AS away_team_name
  FROM `match` m
  LEFT JOIN team t ON m.home_team_id = t.id
  LEFT JOIN team t2 ON m.away_team_id = t2.id
  WHERE away_team_id = :team_id AND round_id = :round_id

假设通过 away，如果我通过 all，我将得到：

SELECT m.*,
t.name AS home_team_name,
  t2.name AS away_team_name
  FROM `match` m
  LEFT JOIN team t ON m.home_team_id = t.id
  LEFT JOIN team t2 ON m.away_team_id = t2.id
  WHERE (home_team_id = :team_id OR away_team_id = :team_id) AND round_id = :round_id

更新 2

使用建议的提示更新了方法

$app->get('/match/get_matches_by_team
/{round_id}/{team_id}/{type}', function (Request $request, Response $response, array $args)
{
  $query = "SELECT m.*,
  t.name AS home_team_name,
  t2.name AS away_team_name
  FROM `match` m
  LEFT JOIN team t ON m.home_team_id = t.id
  LEFT JOIN team t2 ON m.away_team_id = t2.id
  WHERE ";

switch($args["type"])
   {
    case "home":
          $query .= "home_team_id = :home_team_id 
    AND ";
        break;
        case "away":
            $query .= "away_team_id = :away_team_id AND ";
        break;
        default:
            $query .= "(home_team_id = :home_team_id OR away_team_id = :away_team_id) AND ";
        break;
      }

      $query .= "round_id = :round_id";

      $sql = $this->db->prepare($query);
      $sql->bindParam("home_team_id", $args["team_id"]);
      $sql->bindParam("away_team_id", $args["team_id"]);
      $sql->bindParam("round_id", $args["round_id"]);
      $sql->execute();

      $result = $sql->fetchAll();
      return $response->withJson($result);
    });

Answer 1

你的：

 $sql->bindParam("team_id", $args["team_id"]);
 $sql->bindParam("round_id", $args["round_id"]);

试试这个，参数可能需要不同的格式

$sql->bindParam(":team_id", $args["team_id"], PDO::PARAM_INT);
$sql->bindParam(":round_id", $args["round_id"], PDO::PARAM_INT);

或

$sql->bindParam(":team_id", $args["team_id"]);
$sql->bindParam(":round_id", $args["round_id"]);

Answer 2

选择默认开关时，您正在尝试使用相同的参数标记（在您的情况下：team_id）两次绑定值。为了使其工作，您必须在 PDO 中打开仿真模式。

http://www.php.net/manual/en/pdo.prepare.php

You must include a unique parameter marker for each value you wish to pass in to the statement when you call PDOStatement::execute(). You cannot use a named parameter marker of the same name more than once in a prepared statement, unless emulation mode is on.

查询 return 空结果但数据存在

Query return empty result but data exist

mysql

sql

pdo

slim