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~
相似问题