PHP development API interface code sharing

小云云
Release: 2023-03-20 13:34:01
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This article mainly shares with youphpThe code that implements the APIinterface, realizes the functions of user registration, login, and query user information, is to learn PHP APIA good example of development, I hope it can help everyone.

PHP APIServer part of the interface

Code example:

mysql_query()($sql, $conn); $count = mysql_num_rows($query); if ($count > 0) { exit(json_encode(1)); //返回1表示注册失败 } else { $addsql = "insert into `member` (username,password,email) values ('$username','$password','$email')"; mysql_query($addsql); exit(json_encode(0)); //返回0表示注册成功 } break; //查询用户信息 case"selectuserinfo"; $username = lib_replace_end_tag($_GET['username']); $sql = "select id,username,nickname,mobile from `member` where username='$username'"; $query = mysql_query($sql, $conn); $row = mysql_fetch_array($query); foreach ($row as $key => $v) { $res[$key] = urlencode($v); } exit(json_encode($res)); break; //会员登录 case"userlogin"; $username = lib_replace_end_tag($_GET['username']); $password2 = lib_replace_end_tag(trim($_GET['userpassword'])); $password = md5("$password2" . ALL_PS); $sqluser = "select id,username,password from `member` where username='" . $username . "' and password='" . $password . "'"; $queryuser = mysql_query($sqluser); $rowuser = mysql_fetch_array($queryuser); if ($rowuser && is_array($rowuser) && !empty($rowuser)) { if ($rowuser['username'] == $username && $rowuser['password'] == $password) { if ($rowuser['password'] == $password) { $res = urlencode("登录成功"); exit(json_encode($res)); } else { $res = urlencode("密码错误"); exit(json_encode($res)); } } else { $res = urlencode("用户名不存在"); exit(json_encode($res)); } } else { $res = urlencode("用户名密码错误"); exit(json_encode($res)); } /* * 0:表示登录成功,1:表示密码错误,2:用户名不存在,3:用户名密码错误 */ break; default: exit(json_encode(error)); } }
Copy after login

PHP APIClient part

Code example:


        
Copy after login

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