如何检查MySQLi中是否存在表？

Question

多数民众赞成我正在做的事情：

$mysqli = new mysqli($db_host,$db_user,$db_pass,$db_name);
$result = $mysqli->query("SELECT COUNT(*) FROM information_schema.tables WHERE table_schema = ".$db_name." AND table_name = users");
$nr = $result->num_rows; // 77 line
var_dump($nr);
$mysqli->close();

但错误说：

Notice: Trying to get property of non-object in test.php on line 77
NULL

Answer 1

你也可以使用这个语句来检查SHOW TABLES LIKE 'tablename';

Answer 2

要使您的方法起作用，您需要围绕字符串常量引用：

SELECT COUNT(*)
FROM information_schema.tables
WHERE table_schema = '".$db_name."' AND table_name = 'users';

但是，要使您的方法真正起作用，请执行以下操作：

SELECT 1
FROM information_schema.tables
WHERE table_schema = '".$db_name."' AND table_name = 'users';

您正在查看返回的行数，而不是count(*)的值。

Answer 3

SELECT COUNT(*) FROM `table_name` WHERE FALSE;

:-)

成功意味着表存在，失败......意味着它没有！