第一次写api程序,用的原生php,没有restful要求。如果只实现功能返回json数据的话感觉很容易就能实现大部分功能,那么除了功能实现之外需要注意什么问题呢?新手求指导。。已经贴上代码 一共4个文件
<code><?php class Response { /** *按json方式输出通信数据 * @param integer $code 状态码 * @param string $message 提示信息 * @param array $data 数据 */ public static function show($code,$message = '', $data = array()){ if(!is_numeric($code)){ return ''; } $result = array( 'code' => $code, 'message' => $message, 'data' => $data ); echo json_encode($result); } }</code>
<code><?php /* * 单例模式 数据库连接 */ require_once ('Response.php'); class Db { static private $_instance; static private $_connectSource; private $_DbConfig = array( 'host' => 'localhost', 'user' => 'root', 'password' => 'xxx', 'database' => 'sportsstore' ); private function __construct() { } static public function getInstace(){ if(!(self::$_instance instanceof self)){ self::$_instance = new self(); } return self::$_instance; } public function connect(){ if(!self::$_connectSource) { self::$_connectSource = mysqli_connect($this->_DbConfig['host'], $this->_DbConfig['user'], $this->_DbConfig['password'],$this->_DbConfig['database']); if(!self::$_connectSource){ throw new Exception("mysql connect error" . mysql_errno()); } } return self::$_connectSource; } } </code>
这个是登录的代码
<code>/* *login.phh 登录接口 */ <?php require_once ('Db.php'); require_once ('Response.php'); try{ $link = Db::getInstace()->connect(); } catch (Exception $ex) { return Response::show(402,"数据库连接失败"); } $account = mysqli_real_escape_string($link, trim($_POST['account'])); $password = mysqli_real_escape_string($link, trim($_POST['password'])); $query = "select * from users where account = \"$account\""; $result = mysqli_query($link, $query); if(mysqli_num_rows($result)==1){ $row = mysqli_fetch_array($result); if($row['password'] == $password){ return Response::show(200, '登录成功'); }else{ return Response::show(202,'密码不正确'); } }else{ return Response::show(201,'账户不存在'); }</code>
<code>/* *getPersonalInfo.phh 查询个人信息接口 */ <?php require_once ('Db.php'); require_once ('Response.php'); try{ $link = Db::getInstace()->connect(); } catch (Exception $ex) { return Response::show(402,'数据库连接失败'); } $account = isset($_GET['account']) ? mysqli_real_escape_string($link, trim($_GET['account'])) : null; if($account === null){ return Response::show(401,'未登录'); } $query = "select * from users where account=\"$account\""; $result = mysqli_query($link, $query); $row = mysqli_fetch_assoc($result); return Response::show(200,"操作成功",$row);</code>
回复内容:
第一次写api程序,用的原生php,没有restful要求。如果只实现功能返回json数据的话感觉很容易就能实现大部分功能,那么除了功能实现之外需要注意什么问题呢?新手求指导。。已经贴上代码 一共4个文件
<code><?php class Response { /** *按json方式输出通信数据 * @param integer $code 状态码 * @param string $message 提示信息 * @param array $data 数据 */ public static function show($code,$message = '', $data = array()){ if(!is_numeric($code)){ return ''; } $result = array( 'code' => $code, 'message' => $message, 'data' => $data ); echo json_encode($result); } }</code>
<code><?php /* * 单例模式 数据库连接 */ require_once ('Response.php'); class Db { static private $_instance; static private $_connectSource; private $_DbConfig = array( 'host' => 'localhost', 'user' => 'root', 'password' => 'xxx', 'database' => 'sportsstore' ); private function __construct() { } static public function getInstace(){ if(!(self::$_instance instanceof self)){ self::$_instance = new self(); } return self::$_instance; } public function connect(){ if(!self::$_connectSource) { self::$_connectSource = mysqli_connect($this->_DbConfig['host'], $this->_DbConfig['user'], $this->_DbConfig['password'],$this->_DbConfig['database']); if(!self::$_connectSource){ throw new Exception("mysql connect error" . mysql_errno()); } } return self::$_connectSource; } } </code>
这个是登录的代码
<code>/* *login.phh 登录接口 */ <?php require_once ('Db.php'); require_once ('Response.php'); try{ $link = Db::getInstace()->connect(); } catch (Exception $ex) { return Response::show(402,"数据库连接失败"); } $account = mysqli_real_escape_string($link, trim($_POST['account'])); $password = mysqli_real_escape_string($link, trim($_POST['password'])); $query = "select * from users where account = \"$account\""; $result = mysqli_query($link, $query); if(mysqli_num_rows($result)==1){ $row = mysqli_fetch_array($result); if($row['password'] == $password){ return Response::show(200, '登录成功'); }else{ return Response::show(202,'密码不正确'); } }else{ return Response::show(201,'账户不存在'); }</code>
<code>/* *getPersonalInfo.phh 查询个人信息接口 */ <?php require_once ('Db.php'); require_once ('Response.php'); try{ $link = Db::getInstace()->connect(); } catch (Exception $ex) { return Response::show(402,'数据库连接失败'); } $account = isset($_GET['account']) ? mysqli_real_escape_string($link, trim($_GET['account'])) : null; if($account === null){ return Response::show(401,'未登录'); } $query = "select * from users where account=\"$account\""; $result = mysqli_query($link, $query); $row = mysqli_fetch_assoc($result); return Response::show(200,"操作成功",$row);</code>
Update
数据完整
语义简洁
版本可控
响应灵活
安全可靠(https可以考虑,token机制,ip白名单等)
响应速度
风格统一(不要隔三差五就变,返回的格式统一)
记住一点,不要写json_ecode($array);exit;这样的代码就好。
用自己轻量级的类似apiRespnse($data) 分装,虽然你只返回json.
楼上说的不错,我再加一条。
格式统一
比如返回的数据一定带一个统一名称的“请求状态结果”,比如返回的数据一定统一放一个变量里。
避免出现前台不同的地方要判断不同的名字。
注册的接口 还有抽奖的接口 必须安全性得到保障,防止恶意刷!!!
你先贴上你的代码。我也学习一下。

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