Question

我正在学习MySQL / PHP，我只是想熟悉它，但我收到了这个错误：

“表'Daniel.food'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";

?>

我期待看到任何回复。

Answer 1

首先，你的创建声明：

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

包含您的错误检查的查询报价。尝试在'Position VARCHAR（8）'之后关闭你的引用而不是'die（mysql_error（）'。

PHP / MySQL：关于表创建的简单问题

1 个答案: