即时通讯仍在开发android我的android项目我真的需要帮助,我的问题是我无法获得在我的系统登录的用户的用户ID,所以当他们把一个用户ID将附加到它的数据记录..我想要在我的系统中输出他们自己的数据。希望有人能帮助。它唯一的PHP代码谢谢有人帮助。在如何在php中获取用户的用户名?
<?php
// Connection Details altered to hide actual values.
$con = mysqli_connect("localhost", "db_user", "db_password", "db_name");
$username = $_POST['username'];
$password = $_POST['password'];
$sql = "SELECT * FROM tbl_userinfo WHERE username = '$username' AND password='$password' LIMIT 1";
$res = mysqli_query($con,$sql);
$response = array();
$response["success"] = false;
$row = mysqli_fetch_array($res);
if(mysqli_num_rows($res)> 0){
$response["success"] = true;
session_start();
$_SESSION['user_id'] =$userID;
}
echo json_encode($response);
?>
这就是日志,这里有保存数据..
<?php
session_start();
$userID ="";
// Connection Details altered to hide actual values.
$con = mysqli_connect("localhost", "db_user", "db_password", "db_name");
if(!isset($_SESSION['user_id'])){
$userID = $_SESSION['user_id'];
$checkdate = $_POST["checkdate"];
$checkno = $_POST["checkno"];
$datepaid = $_POST["datepaid"];
$clientname = $_POST["clientname"];
$bank = $_POST["bank"];
$amount = $_POST["amount"];
$status = "UNFINISHED";
$statement = mysqli_prepare($con, "INSERT INTO tbl_checkinfo (user_id,checkno, checkdate, datepaid, clientname, bank, amount, status) VALUES (?,?, ?, ?, ?,?,?,?)");
mysqli_stmt_bind_param($statement, "iissssis", $userID, $checkno, $checkdate, $datepaid, $clientname, $bank, $amount, $status);
mysqli_stmt_execute($statement);
$response = array();
$response["success"] = false;
if($statement){
$response["success"] = true;
}
echo json_encode($response);
}
?>
和用于显示用户数据。
<?php
// Connection Details altered to hide actual values.
$con = mysqli_connect("localhost", "db_user", "db_password", "db_name");
$checkdate = $_POST["checkdate"];
$checkno = $_POST["checkno"];
$datepaid = $_POST["datepaid"];
$clientname = $_POST["clientname"];
$bank = $_POST["bank"];
$amount = $_POST["amount"];
$status = "UNFINISHED";
$sql = "Select * from tbl_checkinfo";
$result = mysqli_query($con, $sql);
// $statement = mysqli_prepare($con, "Select * from tbl_checkinfo");
// mysqli_stmt_execute($statement);
// mysqli_stmt_store_result($statement);
// mysqli_stmt_bind_result($statement, $user_id, $checkdate, $checkno, $datepaid, $clientname, $bank, $amount, $status);
$response = array();
$info=array();
$flag = array();
$response["success"] = false;
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result))
{
$flag[checkdate]=$row[checkdate];
$flag[checkno]=$row[checkno];
$flag[datepaid]=$row[datepaid];
$flag[clientname]=$row[clientname];
$flag[bank]=$row[bank];
$flag[amount]=$row[amount];
$flag[status]=$row[status];
array_push($info, $flag);
}
$response["success"] = true;
$response["message"] = $info;
echo json_encode($response);
}
else
{
$response["success"] = 0;
$response["message"] = "No entries yet";
echo json_encode($response);
}
?>
我怎样才能使它动态? –
使其动态替换USER_ID_INDEX与数据库表的列名称。 –
我在保存数据php中发现一个错误,它不再保存了?帮帮我? –