PHP/MySQL：关于表创建的简单问题

我正在学习MySQL/PHP，我只是想熟悉它，但我遇到了以下错误：

“Daniel.food表不存在”

当我运行此代码时

<?php

mysql_connect("localhost", "USER", "PASSWORD") or die(mysql_error());

mysql_query("CREATE DATABASE Daniel") or die(mysql_error());

echo "Database created<br/><br/>";

mysql_select_db("Daniel") or die(mysql_error());

mysql_query("CREATE TABLE food(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
Meal VARCHAR(15),
Position VARCHAR(8)) or die(mysql_error()");

echo "Table: \"food\" created successfully<br/><br/>";

mysql_query("CREATE TABLE family(
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
Position VARCHAR(15),
Age INT) or die(mysql_error()");

echo "Table: \"family\" created successfully<br/><br/>";

mysql_query("INSERT INTO food
(Meal, Position) VALUES ('Steak', 'Dad')") or die(mysql_error());

mysql_query("INSERT INTO food
(Meal, Position) VALUES ('Salad', 'Mom')") or die(mysql_error());

mysql_query("INSERT INTO food
(Meal, Position) VALUES ('Spinach Soup', '')") or die(mysql_error());

mysql_query("INSERT INTO food
(Meal, Position) VALUES ('Tacos', 'Dad')") or die(mysql_error());

mysql_query("INSERT INTO family
(Position, Age) VALUES ('Dad', '41')") or die(mysql_error());

mysql_query("INSERT INTO family
(Position, Age) VALUES ('Mom', '45')") or die(mysql_error());

mysql_query("INSERT INTO family
(Position, Age) VALUES ('Daughter', '17')") or die(mysql_error());

mysql_query("INSERT INTO family
(Position, Age) VALUES ('Dog', '')") or die(mysql_error());

echo "Values entered succussfully";

?>

---

首先，您的create语句：

mysql_查询（“创建表食物（id INT NOT NULL AUTO_INCREMENT、主键（id）、餐变量字符（15）、位置变量字符（8））或骰子（mysql_error（）”

包含错误检查的查询引号。请尝试在“Position VARCHAR（8））”而不是“die（mysql_error（）”之后关闭引号