用PHP变量创建mysql表 [英] Creating a mysql table with a PHP variable

问题描述

我无法使用PHP变量创建SQL表.

I am having trouble creating a SQL table with a PHP variable.

比方说$ tbl_date =数据";

Let's say $tbl_date = "data";

mysql_query('CREATE TABLE '.$tbl_date.'
(
Test varchar(15),
Yes varchar(15),
Very int
)');

它不会创建表格.

推荐答案

似乎不起作用，因为您的表名不在引号中.试试这个:

It looks like it's not working because your table name is not in the quotes. Try this:

mysql_query('CREATE TABLE `'.$tbl_date.'`
(
`Test` varchar(15),
`Yes` varchar(15),
`Very` int
)');

字段名称也是如此.
也可以尝试一下，这样更容易被人阅读:

And the same applies for field names.
Also try this, it's easier to read by humans:

mysql_query("CREATE TABLE `{$tbl_date}`
(
`Test` varchar(15),
`Yes` varchar(15),
`Very` int
)");

请注意不同类型的引号:
-单引号和双引号用于在PHP中构建某些字符串(例如SQL查询)
-反引号特别用作表名和列名的SQL查询的一部分

Please note different types of quotes:
- single quotes and double quotes for building some string (eg.SQL query) in PHP
- back quotes used specially as a part of SQL query for table names and column names

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