mysql-php 判断一个表是否存在

function table_exists($tablename, $link, $database = false) {

if(!$database) {
  $res = mysql_query("SELECT DATABASE()", $link);
  $database = mysql_result($res, 0);
}

$res = mysql_query("SELECT COUNT(*) AS count FROM information_schema.tables WHERE table_schema = '$database' AND table_name = '$tablename'", $link);

return mysql_result($res, 0) == 1;
}

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