简单的 PHP/MySQL 创建表脚本
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/17688387/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Simple PHP/MySQL Create Table Script
提问by user2581841
So I have this script just to create a table in my database. I've copied it over from an old script I did that is working right now. How come this one is not working? Anyone?
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。这个怎么不行啊？任何人？
The error I am getting is "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"
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";
?>
回答by Amal Murali
Remove all =as already suggested:
=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
)";
Each and every table should have a primary key and you must specify AUTO_INCREMENTcolumn as PRIMARY KEY. In this case, the AUTO_INCREMENTcolumn is column_twoand I've set that as the PRIMARY KEY.
每个表都应该有一个主键，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.
回答by Ravi Chauhan
MySQLi Procedural
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)
(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;
?>
回答by CRISTIAN STEVEN MONTERROSO PRE
The solution is to assign all the privileges to the BD user, like this:
解决方法是将所有权限分配给BD用户，如下所示：
GRANT ALL PRIVILEGES ON *. * TO 'user_name' @ 'localhost';
