我目前正在编写Android应用程序以进行登录和注册。我发现错误来自我的php文件,因为当我尝试打开文件时它不会获取任何数据,即使它正确连接到数据库。
这是我的PHP代码:
<?php
$con=mysqli_connect("mysql7.000webhost.com","a3736257_root","********","a3736257_UD");
if ($con->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//echo 'connection successful';
$username = 'jack';
$password = 'jackjack';
$statement = mysqli_prepare($con, "SELECT * FROM Parent WHERE username = ? AND password = ?");
mysqli_stmt_bind_param($statement, "ss", $username, $password);
mysqli_stmt_execute;
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement,$parentID ,$username, $password, $email, $created, $rewardOwed);
$parent = array();
while(mysqli_stmt_fetch($statement)) {
$parent[username] = $username;
$parent[password] = $password;
$parent[email] = $email;
$parent[created] = $created;
}
echo json_encode($parent);
mysqli_stmt_close($statement);
mysqli_close($con); ?>
这是执行代码时的输出
答案 0 :(得分:0)
mysqli_stmt_execute函数不知道您希望它执行什么,除非您提供准备好的语句。函数也应该有括号()
。
所以:
mysqli_stmt_execute;
应该是
mysqli_stmt_execute($statement);