用于创建多个表的MySQL语法

Question

I am having a some problem that i don't understand and trying to create tables from a php file , so i created a variable and i added all the create statements in it so i can run a mysql_query on it , i keep getting an error that there is a syntax problem , so i echo the content of the variable and past it in the sql section at phpmyadmin and it word fine but if i try it from php it gives me an error as :

Error creating table:

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 'CREATE TABLE IF NOT EXISTS Materials ( Id int(11) NOT NULL AUTO_INCREMENT,' at line 9

here is the string in the variable that i run the query on :

   // Create Tables Variable
$Tables = "";

// Create Students Table
$Tables = $Tables . "CREATE TABLE IF NOT EXISTS `Students` (
  `Id` int(11) NOT NULL AUTO_INCREMENT,
  `FirstName` varchar(50) DEFAULT NULL,
  `LastName` varchar(50) DEFAULT NULL,
  `Age` int(3) DEFAULT NULL,
  `Major` varchar(50) DEFAULT NULL,
  `Image` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`Id`)
);";

// Create Materials Table
$Tables = $Tables . "CREATE TABLE IF NOT EXISTS `Materials` (
  `Id` int(11) NOT NULL AUTO_INCREMENT,
  `Name` varchar(50) DEFAULT NULL,
  `LastName` varchar(50) DEFAULT NULL,
  PRIMARY KEY (`Id`)
);";

// Create Materials Students intersectons table for many to many relation
$Tables = $Tables . "CREATE TABLE `Material_Student`(
    `Student_Id` INT NOT NULL,  
    `Material_id` INT NOT NULL,    
    FOREIGN KEY (`Student_Id`) REFERENCES Students(`id`) ON UPDATE CASCADE,  
    FOREIGN KEY (`Material_id`) REFERENCES Materials(`id`) ON UPDATE CASCADE
);";

// Create Pages Table
$Tables = $Tables . "CREATE TABLE `Pages`(
    `Id` INT NOT NULL AUTO_INCREMENT,
    PRIMARY KEY(`Id`),
    `Name` varchar(50),
    `Conent` varchar(5000)

);";

// Create News Table
$Tables = $Tables . "CREATE TABLE `News`(
    `Id` INT NOT NULL AUTO_INCREMENT,
    PRIMARY KEY(`Id`),
    `Title` varchar(50),
    `Conent` varchar(5000),
    `Image` varchar(100)
);";

// Create Notifications Table
$Tables = $Tables . "CREATE TABLE `Notifications`(
    `Id` INT NOT NULL AUTO_INCREMENT,
    PRIMARY KEY(`Id`),
    `Title` varchar(50),
    `Conent` varchar(5000),
    `Student_Id` INT NOT NULL,  
    FOREIGN KEY (`Student_Id`) REFERENCES Students(`id`) ON UPDATE CASCADE
);";

so what am I missing here ??

Answer 1

通常，您不能执行许多查询，我建议像执行不同的查询一样执行所有查询，但是MySQLi可以选择一次执行许多查询，您可以在此处找到文档和示例。

Answer 2

Your issue seems to be passing a multiple-query into Php's single query function.

See this link for explanation. http://php.net/manual/en/mysqli.quickstart.multiple-statement.php

I suggest you either break down your statements and execute one at a time, or use multi_query function instead.

http://www.php.net/manual/en/mysqli.multi-query.php

$mysqli = new mysqli("localhost", "yourusername", "yourpassword", "test");
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

if (!$mysqli->multi_query($Tables)) {
    echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

...

HTH