使用php做融云的服务器端的开发

使用xampp做安卓项目的服务器端的开发,但是连接服务器的时候出现问题了,
在浏览器输入:https://127.0.0.1/phpmyadmin/chat/reg.php访问数据库
但是出现下面的问题:Notice: Undefined index: username in D:\xxrj\xampp\phpMyAdmin\chat\reg.php on line 7
不是学php的,但是现在开发需要到这个,想要请教下这个怎么解决?
连接数据的php文件
<?php


$conn = mysql_connect("localhost", "root", "");
mysql_select_db("ditucon", $conn);

?>


登录的判断代码
<?php
error_reporting(0);
require_once('./conn.php');

if($_POST['phone'] != "" && $_POST['password'] != "")
{
$sql = "select * from `user` where phone='$_POST[phone]'";
$query = mysqli_query($connect, $sql);
$list = mysqli_fetch_array($query);
if($_POST['phone'] == $list['phone'] && md5($_POST['password']) == $list['password'])
{
$result = array('status' => "success", "token"=>$list['token']);
echo json_encode($result);
}
else
{
$result = array('status'=>"error");
echo json_encode($result);
}
}
else
{
$result = array('status'=> "error");
echo json_encode($result);
}
?>

注册页面的代码

<?php


require_once('./conn.php');
require_once('./ServerAPI.php');

if($_POST['username'] != "" && $_POST['password'] != ""&& $_POST['phone'] != "")
{
$password = md5($_POST['password']);

$sql = "select * from `user` where username='$_POST[username]'";
$query = mysql_query($connect, $sql);
$row = mysql_num_rows($query);
if($row == 0) {
$p = new ServerAPI("pvxdm17jxad6r","wzGkFpJsV4kjM");
$r = $p->getToken($_POST['phone'], "", "");
$obj = json_decode($r);
if($obj->code != 200) {
$result = array("status" => "error");
echo json_encode($result);
}
else {
$token = $obj->token;
$sql2 = "insert into `user` (username, password, phone,token) values('$_POST[username]', '$password','$phone', '$token')";
$query = mysqli_query($connect, $sql2);
$result = array("status"=>"success", "token"=>$token);
echo json_encode($result);
}
}
else {
$result = array("status"=>"exists");
echo json_encode($result);
}
}
?>

添加朋友的代码
<?php
error_reporting(0);
require_once('./conn.php');

if($_POST['phone'] != "" && $_POST['target'] != "")
{
$sql = "select * from `user` where phone='$_POST[phone]' or phone='$_POST[target]'";
$query = mysqli_query($connect, $sql);

$rows = mysqli_num_rows($query);
if($rows == 2)
{
$list1 = mysqli_fetch_array($query);
$list2 = mysqli_fetch_array($query);
$userid1 = $list1['ID'];
$userid2 = $list2['ID'];
if($userid1 > $userid2)
{
$t = $userid1;
$userid1 = $userid2;
$userid2 = $t;
}

$sql = "select * from friend where userid1 = $userid1 and userid2 = $userid2";
$query = mysqli_query($connect, $sql);
$rows = mysqli_num_rows($query);
if ($rows == 0) {
$sql = "insert into friend (userid1, userid2) values($userid1, $userid2)";
$query = mysqli_query($connect, $sql);
$result = array("status"=>"success");
echo json_encode($result);
}
else {
$result = array("status"=>"error");
echo json_encode($result);
}
}
else
{
$result = array("status"=>"error");
echo json_encode($result);
}
}
?>


取得朋友的
<?php
error_reporting(0);
require_once('./conn.php');

if(isset($_POST['phone']) && $_POST['phone'] != "")
{
$username = $_POST['phone'];
$sql = "select id from user where phone = '$phone'";

$query = mysqli_query($connect, $sql);
$list = mysqli_fetch_array($query);
$userid = $list['id'];

$sql2 = "select * from friend where userid1 = $userid or userid2 = $userid";
$query = mysqli_query($connect, $sql2);
$friendList = array();
while($list = mysqli_fetch_array($query)) {
$targetUserid = ($list['userid1'] == $userid ? $list['userid2'] : $list['userid1']);
$sql = "select username from user where id = $targetUserid";
$query2 = mysqli_query($connect, $sql);
$list2 = mysqli_fetch_array($query2);
$friendList[] = $list2['username'];
}

echo json_encode($friendList);
}

?>




...全文
714 5 打赏 收藏 转发到动态 举报
写回复
用AI写文章
5 条回复
切换为时间正序
请发表友善的回复…
发表回复
  • 打赏
  • 举报
回复
引用 4 楼 qq564480704 的回复:
可以给个联系方式吗……最近也在学习做融云即时通讯这块,我是负责后台这边的,但我不太清楚和安卓那边是怎样进行联系或者对接的,能指导一下我吗,谢谢!
谢谢,但是我也是初学的,我知道的东西不多的。
qq564480704 2016-02-16
  • 打赏
  • 举报
回复
可以给个联系方式吗……最近也在学习做融云即时通讯这块,我是负责后台这边的,但我不太清楚和安卓那边是怎样进行联系或者对接的,能指导一下我吗,谢谢!
  • 打赏
  • 举报
回复
引用 1 楼 Novolee 的回复:
那就加个条件 把出错的那行改成:
if(isset($_POST['username']) && $_POST['username'] != "" && $_POST['password'] != ""&& $_POST['phone'] != "") 
可以了,谢谢!
李睿_Lee 2016-01-14
  • 打赏
  • 举报
回复
那就加个条件 把出错的那行改成:
if(isset($_POST['username']) && $_POST['username'] != "" && $_POST['password'] != ""&& $_POST['phone'] != "") 

21,886

社区成员

发帖
与我相关
我的任务
社区描述
从PHP安装配置,PHP入门,PHP基础到PHP应用
社区管理员
  • 基础编程社区
加入社区
  • 近7日
  • 近30日
  • 至今
社区公告
暂无公告

试试用AI创作助手写篇文章吧