如何在MySQL中创建一个使用PHP中另一个数据库视图名称的视图？

Question

I am trying to create a view in a MySQL database that has the same name as a view in another database using this line of code in php.

mysql_query('CREATE VIEW `' . $targetDB . '`.`' . $row[0] . '` LIKE `' . $sourceDB . '`.`' . $row[0] . '`') or die(mysql_error());

However when I run my code I get this error

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LIKE copyinrdb.abnormal_limit_view' at line 1

What am I doing wrong my full code is if it helps:

    <?php
set_time_limit(0);
function duplicateTables($sourceDB=NULL, $targetDB=NULL) {
    $link = mysql_connect('my_host', 'my_user', 'my_pass') or die(mysql_error()); // connect to database
                                $results = mysql_query('SHOW FULL TABLES FROM copyinrdb WHERE TABLE_TYPE= "VIEW"') or die(mysql_error());
        while($row = mysql_fetch_row($results)) {
            mysql_query('CREATE VIEW `' . $targetDB . '`.`' . $row[0] . '` LIKE `' . $sourceDB . '`.`' . $row[0] . '`') or die(mysql_error());
            mysql_query('INSERT INTO `' . $targetDB . '`.`' . $row[0] . '` SELECT * FROM `' . $sourceDB . '`.`' . $row[0] . '`') or die(mysql_error());
        }
        mysql_free_result($results);

    mysql_close($link);
} // end duplicateTables()
duplicateTables('copyinrdb', 'copytest1');
?>

Answer 1

CREATE VIEW db1.viewname LIKE db2.viewname simply isn't valid mysql code.

The LIKE keyword can't be used there.

The syntax is this:

CREATE
    [OR REPLACE]
    [ALGORITHM = {UNDEFINED | MERGE | TEMPTABLE}]
    [DEFINER = { user | CURRENT_USER }]
    [SQL SECURITY { DEFINER | INVOKER }]
    VIEW view_name [(column_list)]
    AS select_statement
    [WITH [CASCADED | LOCAL] CHECK OPTION]

[http://dev.mysql.com/doc/refman/5.7/en/create-view.html]