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PHP代码中不反映操作的错误_Php_Database - Fatal编程技术网

PHP代码中不反映操作的错误

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

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["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?