JqueryAjax+php实现简单的注册登录,jqueryajaxphp
HTML结构
<div class="container"> <form> <label>用户名</label> <input id="username" type="text" name="username" class="form-control" /> <label>密码</label> <input id="password" type="password" name="password" class="form-control" /> <button type="button" id="login" class="btn btn-primary">登录</button> <button type="button" id="sign" class="btn btn-danger">注册</button> </form> <div id="div"></div> </div>前台JS
$("#login").click(function(){ var sendData = {"username":$("#username").val(),"password":$("#password").val()} $.ajax({ url:"action/login.php", type:"POST", data:sendData, success:function(data){ if(data==1){ $("#div").html("密码正确") }else if(data==2){ $("#div").html("密码不正确") }else if(data==3){ $("#div").html("账号不存在") } } }) }) $("#sign").click(function(){ var sendData = {"username":$("#username").val(),"password":$("#password").val()} $.ajax({ url:"action/addUser.php", type:"POST", data:sendData, success:function(data){ if(data==1){ $("#div").html("用户存在不能注册") }else if(data==2){ $("#div").html("注册成功") } } }) })后台login.php
$username = $_POST['username']; $password = $_POST['password']; $conn = mysqli_connect("localhost","root","","login") or die("连接失败"); mysqli_query($conn,"set names utf8"); $result = mysqli_query($conn,"select * from user where username='$username'"); if($row=mysqli_fetch_array($result)){ if($row["password"]==$password){ echo 1;//密码正确 }else{ echo 2;//密码不正确 } }else{ echo 3;//账号不正确 }后台addUser.php
$username = $_POST['username'];
$password = $_POST['password'];
$conn = mysqli_connect("localhost","root","","login") or die("连接失败");
mysqli_query($conn,"set names utf8");
$result = mysqli_query($conn,"select * from user where username='$username'");
if($row=mysqli_fetch_array($result)){
echo 1;//"用户存在不能注册"
}else{
mysqli_query($conn,"insert into `user` (`username`,`password`) values ('$username','$password')");
echo 2;//注册成功
}
希望能对新手玩家有点帮助。
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