PHP代码中不反映操作的错误
当我在我的浏览器中运行下面的代码时,我得到了以下两个错误,代码本身似乎可以工作,因为我可以登录、注册并添加到数据库中,但错误仍然存在。 代码是:PHP代码中不反映操作的错误,php,database,Php,Database,当我在我的浏览器中运行下面的代码时,我得到了以下两个错误,代码本身似乎可以工作,因为我可以登录、注册并添加到数据库中,但错误仍然存在。 代码是: <?php require "conn.php"; $doctor_number = $_POST["doctor_number"]; $doctor_name = $_POST["doctor_name"]; $email = $_POST["email"]; $password = $_POST["passw
<?php
require "conn.php";
$doctor_number = $_POST["doctor_number"];
$doctor_name = $_POST["doctor_name"];
$email = $_POST["email"];
$password = $_POST["password"];
$mysql_qry = "insert into doctors(doctor_number, doctor_name, email, password) values ('$doctor_number', '$doctor_name', '$email', '$password')";
$result = mysqli_query($conn ,$mysql_qry);
if($conn->query($mysql_qry) === TRUE){
echo "Insert Successful";
}
else{
echo "Error: " . $mysql_qry . "<br>" . $conn->error;
}
$conn->close();
将下面的代码置于设置条件下
<?php
require "conn.php";
if(isset($_POST)){
$doctor_number = $_POST["doctor_number"];
$doctor_name = $_POST["doctor_name"];
$email = $_POST["email"];
$password = $_POST["password"];
$mysql_qry = "insert into doctors(doctor_number, doctor_name, email, password) values ('$doctor_number', '$doctor_name', '$email', '$password')";
$result = mysqli_query($conn ,$mysql_qry);
if($conn->query($mysql_qry) === TRUE){
echo "Insert Successful";
}
else{
echo "Error: " . $mysql_qry . "<br>" . $conn->error;
}
$conn->close();
}?>
将下面的代码置于设置条件下
<?php
require "conn.php";
if(isset($_POST)){
$doctor_number = $_POST["doctor_number"];
$doctor_name = $_POST["doctor_name"];
$email = $_POST["email"];
$password = $_POST["password"];
$mysql_qry = "insert into doctors(doctor_number, doctor_name, email, password) values ('$doctor_number', '$doctor_name', '$email', '$password')";
$result = mysqli_query($conn ,$mysql_qry);
if($conn->query($mysql_qry) === TRUE){
echo "Insert Successful";
}
else{
echo "Error: " . $mysql_qry . "<br>" . $conn->error;
}
$conn->close();
}?>
提交表单(method=POST)后,$\u POST的内容只存在一次。事实上,这不是一个数据库错误…我不擅长PHP,你能建议我如何修复它吗?这与PHP无关。这是所有语言的正常行为。只需将您的登录凭据放入$\u会话并保持登录即可。但是,请避免在每次请求时向数据库发送登录凭据好的,谢谢,我将结束这个问题,因为它似乎是重复的。提交表单(method=POST)后,$\u POST的内容只存在一次。事实上,这不是一个数据库错误…我不擅长PHP,你能建议我如何修复它吗?这与PHP无关。这是所有语言的正常行为。只需将您的登录凭据放入$\u会话并保持登录即可。但是避免在每次请求时向数据库发送登录凭据好的,谢谢,我将结束这个问题,因为它似乎是重复的。我尝试过,但错误持续存在。请添加html代码好吗?我尝试过,但错误持续存在。请添加html代码好吗?