Php 获取slim框架上的请求

我正在使用slim框架构建一个restful web服务。 我有两个表，其中包含以下条目：

Table team
id_team    name
1          Brasil
2          England
3          Portugal

Table Person
id_pesorn id_team name     age ...
1          1       Rafinha 23  ...

以及以下代码：

require '../Slim/Slim/Slim.php';
\Slim\Slim::registerAutoloader();
$app = new \Slim\Slim();
$app->response()->header('Content-Type', 'application/json;charset=utf-8');

$app->get('/person/team/:name/:team','getPerson');
$app->run();

function getConn(){
...
}

function getPerson($name, $team){
    $conn = getConn();
    $sql = "SELECT id_team FROM team WHERE name=:team";
    $stmt = $conn->prepare($sql);
    $stmt->execute();
    $pessoa = $stmt->fetchObject();

    $sql1 = "SELECT * FROM person WHERE id_team=".$pessoa->id_team."and name=:name";
    $stmt1 = $conn->prepare($sql1);
    $stmt1->bindParam("team", $team);
    $stmt1->bindParam("name", $name);
    $stmt1->execute();
    $pessoa1 = $stmt1->fetchObject();

    echo "{pessoas:".json_encode($pessoa1)."}";
}

当我通过GET:localhost/services/pesorn/team/Rafinha/Brasil调用它时，给了我一个错误：

Slim Application Error
Details

Type: ErrorException
Code: 2
Message: PDOStatement::execute(): SQLSTATE[HY093]: Invalid parameter number: no parameters were bound
File: /Applications/XAMPP/xamppfiles/htdocs/aw014/webservice/services/index.php
Line: 38

但是应该以JSON返回：

{"id_person":"1","id_team":"1","nome":"Rafinha","idade":"23",...}

---

我不知道我做错了什么。谁能帮帮我吗。谢谢。

只需使用

$stmt1->bindParam（“：name”，“$name”）<代码>$stmt1->bindParam（“团队”，“团队”）未绑定任何内容，因此您可以将其删除。您可以在doc页面上看到这一点和更多内容：而不是绑定param，这在第一个“team”查询中是不做的。使用：$stmt->execute（数组（':team'=>$team））；注意键上的前导“：”。有用链接：