3

所以我有这个脚本只是为了在我的数据库中创建一个表。我已经从我现在正在工作的旧脚本中复制了它。这个怎么不行？任何人？

我得到的错误是"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near "'= varchar (20) NOT NULL, column_two = int NOT NULL auto_increment, column_thre' at line 2"

<?php

include("server_connect.php");

mysql_select_db("assignment5");

$create = "CREATE TABLE tbltable (
column_one = varchar (20) NOT NULL,
column_two = int NOT NULL auto_increment,
column_three = int NOT NULL,
column_four = varchar (15) NOT NULL,
column_five = year,
PRIMARY KEY = (column_one)
)";

$results = mysql_query($create) or die (mysql_error());

echo "The tables have been created";

?>
4

3 回答 3

8

=按照已经建议的方式删除所有内容：

$create = "CREATE TABLE tbltable (
column_one varchar (20) NOT NULL,
column_two int NOT NULL auto_increment PRIMARY KEY,
column_three int NOT NULL,
column_four varchar (15) NOT NULL,
column_five year
)";

每个表都应该有一个主键，并且您必须将AUTO_INCREMENT列指定为PRIMARY KEY. 在这种情况下，该AUTO_INCREMENT列是column_two，我已将其设置为PRIMARY KEY.

于 2013-07-16T23:03:33.413 回答
1

MySQLi 程序

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

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

// sql to create table
$sql = "CREATE TABLE MyGuests (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";

if (mysqli_query($conn, $sql)) {
    echo "Table MyGuests created successfully";
} else {
    echo "Error creating table: " . mysqli_error($conn);
}

mysqli_close($conn);
?>

(PDO)

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDBPDO";

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    // sql to create table
    $sql = "CREATE TABLE MyGuests (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
    firstname VARCHAR(30) NOT NULL,
    lastname VARCHAR(30) NOT NULL,
    email VARCHAR(50),
    reg_date TIMESTAMP
    )";

    // use exec() because no results are returned
    $conn->exec($sql);
    echo "Table MyGuests created successfully";
    }
catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
?>
于 2015-07-07T13:22:12.933 回答
-1

解决方案是将所有权限分配给 BD 用户，如下所示：

GRANT ALL PRIVILEGES ON *. * TO 'user_name' @ 'localhost';
于 2017-12-07T03:23:10.837 回答