为什么查询数据库,返回来的是空null???
来源:1-3 用户注册数据入库
天道酬勤168
2017-12-29 21:55:02
<?php //表单进行了提交处理 if (!empty($_POST['username'])) { include_once ('./lib/fun.php'); $username = trim($_POST['username']); $password = trim($_POST['password']); $repassword = trim($_POST['repassword']); if (!$username) { echo '用户名不能为空'; exit; } if (!$password) { echo '密码不能为空'; exit; } if (!$repassword) { echo '确认密码不能为空'; exit; } if ($password !== $repassword) { echo '两次输入密码不一致,请重新输入。'; exit; } //数据库连接 $con = mysqlInit('127.0.0.1', 'root', 'root', 'hz_mall', '8889'); if (!$con) { echo mysql_errno(); exit; } // echo '这里来了'; // 判断用户表用户是否存在 $sql = "SELECT COUNT(`id`) FROM `hz_user` WHERE `username`='{$username}'"; // var_dump(phpinfo()); // mysqli_query($con, "SELECT * FROM websites"); $obj = mysqli_query($con, $sql); var_dump($obj); echo '这里来了'; $result = mysqli_fetch_assoc($obj); var_dump($result); die; } ?> <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge"> <title>M-GALLARY|用户注册</title> <link type="text/css" rel="stylesheet" href="./static/css/common1.css"> </head> <body> <header> <div class="logo fl"> <img src="./static/image/logo.png" alt="LOGO"> </div> <div class="auth fr"> <ul> <li><a href="login.php">登录</a></li> <li><a href="register.php">注册</li> </ul> </div> </header> <section> <div class="content"> <div class="center"> <div class="center-login"> <div class="login-banner"> <a href="#"><img src="./static/image/login_banner.png" alt="登录界面广告图"></a> </div> <div class="user-login"> <div class="user-box"> <div class="user-title">用户注册</div> <form action="register1.php" method="post" id="register-form"> <div class="login-left"> <label class="username">用户名</label> <input type="username" name="username" id="username" placeholder="请输入用户名"> </div> <div class="login-right"> <label class="password">密码</label> <input type="password" name=password id="password" placeholder="请输入密码"> </div> <div class="login-right"> <label class="password">确认密码</label> <input type="password" name="repassword" id="repassword" placeholder="再次输入密码"> </div> <div class="login-btn"> <button type="submit">注册</button> </div> </form> </div> </div> </div> </div> </div> </section> <footer> <div class="footer"> <p><span>M-GALLARY </span>©2017 POWERED BY HZ.INC</p> </div> </footer> </body> <script src="./static/js/jquery-1.10.2.min.js" type="text/javascript"></script> <script src="./static/js/layer/layer.js"></script> <script> $(function(){ $('#register-form').submit(function(){ var username=$('#username').val(), password=$('#password').val(), repassword=$('#repassword').val(); if(username==''||username.length<=0){ layer.tips("用户名不能为空",'#username',{time:2000,tips:2}); $('#username').focus(); return false; } if(password==''||password.length<=0){ layer.tips('密码不能为空','#password',{time:2000,tips:2}); $('#password').focus(); return false; } if(repassword==''||repassword.length<=0){ layer.tips('请再次输入密码','#repassword',{time:2000,tips:2}); $('#repassword').focus(); return false; } if(password!=repassword){ layer.tips('两次输入的密码不一致','#repassword',{time:2000,tips:2}); $('#repassword').focus(); return false; } return true; }) }) </script> </html> <?php /** * 数据库连接初始化 * @param [type] $host 主机地址 * @param [type] $username 数据库用户名 * @param [type] $password 数据库密码 * @param [type] $dbName 数据库名称 * @return [type] utf8 */ function mysqlInit($host, $username, $password, $dbName, $port) { // 数据库操作 // echo $host, $username, $password; $link = mysqli_init(); // var_dump($link); $con = mysqli_real_connect($link, $host, $username, $password, $dbName, $port); //判断是否连接成功,连接失败返回错误,然后结束。 var_dump($con); if (!$con) { echo '连接成功'; return false; } //选择数据库名称 // echo '这里来了'; // mysql_select_db($dbName); //设置字符集 // mysql_set_charset('utf8'); // var_dump($con); return $con; }
1回答
imooc_澈
2017-12-30
您好,查询语句的语法是没有问题的,小慕本地测试自己的是可以查到数据的,检查一下是不是您的条件字段有问题。如果解决了您的问题,请采纳,祝学习愉快~
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