Question

<html>
<body>
<?php include './includes/connect_db.php' // connecting to mysql server  ?>   
<?php include './includes/select_db.php' //  selecting database ?>
<?php
    $username=$_POST["username"];
    $password=$_POST["pass"];
    $email=$_POST["email"];             
$sql="insert into table 'users' values ('$username','$password','$email')";
if(mysql_query($sql,$connect))
{
    echo "query executed";
}
else
{
    echo "query did not run ".mysql_error();
} ?>

<?php include './includes/close_db.php' ; //closing connection ?>

该表由4个字段组成（id，用户名，密码，电子邮件）。 “id”是主键并自动递增。一切都工作正常，除了查询和PHP不是diplaying查询中的错误尽管display_error是ON是.ini文件。

Answer 1

那里有几个问题..首先，你不需要多次启动和关闭php解释器（使用php打开和关闭标签为你的包括）..只需打开一次。

您尚未定义连接 - 我建议将其保留在if语句之外。从我在那里看到的，它无论如何都无法连接......但我看不出你是如何定义它的。

首先设置数据库连接，然后将$ sql更改为read;

$sql="INSERT INTO 'users' 
values ('{$username}','{$password}','{$email}')";

那应该对你有用。

祝你好运！

锐衡

Answer 2

试试这个。

$sql="INSERT INTO users (username, password, email) VALUES ('$username','$password','$email')";
$link = mysql_query($sql) or die(mysql_error());

将(username, password, email)更改为您自己的列值。

Answer 3

您可以回显$ sql行，然后将其复制并粘贴到PHPmyadmin的SQL选项卡中。这将告诉您SQL是否正确。并将字段名称放在查询中。

Answer 4

请使用以下sql命令

$sql="insert into users values ('".$username."','".$password."','".$email."')";

由于

Answer 5

尝试

mysql_query($sql,$connect) or die mysql_error()

获取实际错误

Answer 6

尝试以下方法：

$sql = "INSERT INTO users (username, password, email) VALUES('$username','$password','$email')";

如果您的错误仍然存在，请复制以下代码：

<?php
  $servername = "localhost";
  $username = "root";
  $password = "pass";
  $dbname = "myDB";

  // Create connection
  $conn = mysqli_connect($servername, $username, $password, $dbname);
  // Check connection
  if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
  }
  $myusername = mysqli_real_escape_string($conn,$_POST['name']);
  $myusername1 = mysqli_real_escape_string($conn,$_POST['name1']);
  $mylastname = mysqli_real_escape_string($conn,$_POST['lastname']);
  $myemail = mysqli_real_escape_string($conn,$_POST['email']);
  #the if statement is just an example of when you to control the flow of your code
  if($myusername == $myusername1){
    $sql = "INSERT INTO MyGuests (name, lastname, email)
    VALUES ('$myusername', '$mylastname', '$myemail')";
    if (mysqli_query($conn, $sql)) {
      echo "New record created successfully";
    } else {
      echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
  }else{
    echo "Error: Check your username!!!";
  }

  mysqli_close($conn);
?>

Php注册错误

6 个答案: