Question

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Php form for MySQL</title>
</head>
<body>
<form action="home.php" method="post">
<p>
    <label for="table">Name your Table</label>
    <input type="text" name="table" id="table">
</p>
<p>
    <label for="firstName">First Name:</label>
    <input type="text" name="firstname" id="firstName">
</p>
<p>
    <label for="lastName">Last Name:</label>
    <input type="text" name="lastname" id="lastName">
</p>
<p>
    <label for="emailAddress">Email Address:</label>
    <input type="text" name="email" id="emailAddress">
</p>
<button name="submit" type="submit" value="submit">Create Table</button>
</form>

  if (isset($_POST['submit'])){

  //enter database details
  $servername = "localhost";
  $username = "username";
  $password = "password";
  $dbname = "database";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
$table = $_POST['table'];

// Create a MySQL table in the selected database
mysqli_query("CREATE TABLE $table (
firstname VARCHAR(30), 
lastname VARCHAR(30),
email VARCHAR(50))") or die(mysql_error());  

 if ($conn->query($sql) === TRUE) {
    echo "Table account created successfully";
  } else {
    echo "Error creating table: " . $conn->error;
 }

  // Check connection
  if (!$conn) {
      die("Connection failed: " . mysqli_connect_error());
    }
     echo "Connected successfully";

  $firstname = $_POST['firstname'];
  $lastname = $_POST['lastname'];
  $email = $_POST['email'];

  if(isset($_POST['firstname']) || ($_POST['lastname']) || ($_POST['email']))
{   
  $order="INSERT INTO $table (firstname,lastname,email) VALUES ('$firstname','$lastname','$email')";
$result = mysqli_query($order) or die (mysql_error());
}


  mysqli_close($conn);
  }

 ?>
 </body>
 </html>

我正在尝试允许用户为数据库选择一个名称并被卡住了。我收到错误

用户''@ localhost'拒绝访问（使用密码：否）

我想要的图片：

Here is a picture of what I want

我不知道该怎么办。希望你能帮忙。感谢

Answer 1

您获得的错误意味着您没有提供有效的凭据来连接数据库。请勿在您的问题中更新它们，因为这些详细信息应该是私有的，但您需要更改代码：

#include <stdlib.h>
#include <stdio.h>
// self - referential structure
struct listNode {
char data; // each listNode contains a character
struct listNode *nextPtr; //pointer to next node
};

typedef struct listNode ListNode;  // synonym for struct listNode
typedef ListNode *ListNodePtr; // synonym for struct listnode*



void insert(ListNodePtr *sPtr, char value)
{
ListNodePtr newPtr = malloc(sizeof(ListNode)); // create node

if (newPtr != NULL) // is space availale?
{
    newPtr->data = value; // place value in node
    newPtr->nextPtr = NULL; // node does not link to another node

     ListNodePtr previousPtr = NULL;
    ListNodePtr currentPtr = *sPtr;

    // loop to find the correct location in the list
     while (currentPtr != NULL && value > currentPtr->data)
     {
        previousPtr = currentPtr; // walk to...
         currentPtr = currentPtr->nextPtr; // ... next node
     }

    if (previousPtr == NULL) // insert new node at beginning of list
     {
        newPtr->nextPtr = *sPtr;
        *sPtr = newPtr;
    }
    else
    {
        previousPtr->nextPtr = newPtr;
        newPtr->nextPtr = currentPtr;
    }
}

else 
{
    printf("%c not inserted. No memory available. \n", value);
}
}

是实际值。例如，如果您的mysql用户名是$servername = "localhost"; $username = "username"; $password = "password"; $dbname = "database";，则用户名行将如下所示： 'tomi'

你真的没有名为'database'的数据库吗？如果您的实际密码是“密码”，则需要将其更改为更安全的密码。

Answer 2

您打开mysqli连接，然后使用mysql_query进行查询，该时间应为mysqli_query。 mysql_*和mysqli_*是完全不同的东西，你永远不应该使用mysql_*。更改后，请检查您的身份验证详细信息。只要您使用私人计算机，也可以使用var_dump()查看它们并确保它们符合您的预期。看起来它们可能会作为空字符串传递到mysqli。如果您的身份验证详细信息绝对正确，请仔细检查数据库中的给定用户是否确实具有所需的权限。

Answer 3

使用我认为适合您的完整php代码编辑我的最佳答案：

<?php

if (isset($_POST['submit'])){

  //enter database details
  $servername = "localhost";
  $username = "username";
  $password = "password";
  $dbname = "database";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    echo "Failed to connect: (" . $conn->connect_error . ") ";
}
$table = $_POST['table'];

// SQL to Create a table in the selected database
$sql = "CREATE TABLE $table (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30), 
lastname VARCHAR(30),
email VARCHAR(50))";  


 if ($conn->query($sql) === TRUE) {
    echo "Table account created successfully";
  } else {
    echo "Error creating table: " . $conn->error;
 }

  // Check connection
  if (!$conn) {
      die("Connection failed: " . $conn->connect_error);
    }
     echo "Connected successfully";

  $firstname = $_POST['firstname'];
  $lastname = $_POST['lastname'];
  $email = $_POST['email'];

if(isset($_POST['firstname']) || isset($_POST['lastname']) || isset($_POST['email'])) {   
    $order="INSERT INTO $table (firstname,lastname,email) VALUES ('$firstname','$lastname','$email')";

    if ($conn->query($order) === TRUE) {
    echo "New record created successfully";
    } else {
    echo "Error: " . $order . "<br>" . $conn->error;
    }

}


  $conn->close();
  }

 ?>

重要提示：

1）确保在php代码的顶部输入正确的数据库详细信息。例如，对于我在Windows上使用xampp，它是：

  $servername = "localhost";
  $username = "root";
  $password = "";
  $dbname = "database";

我之前已经手动创建了数据库“database”。

2）如果您想在代码的后半部分确保所有字段都不为空，您可能更愿意使用＆amp;＆amp; （和）而不是|| （或）你在if的条件。目前，只要填写了任何字段，它就会运行，即使其余字段是EMPTY也是如此。

示例：

if(isset($_POST['firstname']) && isset($_POST['lastname']) && isset($_POST['email'])) {

而不是：

if(isset($_POST['firstname']) || isset($_POST['lastname']) || isset($_POST['email'])) {

我还强烈建议您查看使用php实现MySQLi的不同方法;例如创建表格：http://www.w3schools.com/php/php_mysql_create_table.asp

允许用户在SQL中输入我的表的名称

3 个答案: