从一个数据库中选择一个数组，并将其与另一个数据库的数组进行比较

我知道标题不是最好的，但它以某种方式总结了问题：我运行了一个页面，客户可以在该页面上请求某个特定主题的导师。现在，我有一个数据库供我的导师使用，其中包含他们的特定主题，还有一个数据库用于接收客户的请求。我现在想做的是，编写一个SQL命令，将传入请求的主题与我导师的主题进行比较，这样教授该主题的老师在登录时就会看到请求。举个例子：母亲安妮塔请一位数学老师->发送请求。代码检查哪些老师教数学，并在这些老师登录时向他们显示请求。请记住，我是一个该死的初学者。

我目前的代码如下：

// Connect to the teacher databank and select the subjects of the teacher that has logged in
$connect = mysql_connect('xxx', 'xxx', 'xxx');
mysql_set_charset("utf8", $connect);
mysql_select_db('xxx');
$sql = "SELECT subjects FROM teachers WHERE teacher LIKE '%". $_SESSION['user'] ."%'" ;
$erg = mysql_query(@$sql);
if (false === $erg) {
die (mysql_error());
}
$speicher = mysql_fetch_array($erg, MYSQL_ASSOC);
print_r ($speicher);  
// Until here it functions and the array ($speicher) contains the subjects of the teacher
// Now connect to the other Databank (with the requests)
$connect = mysql_connect('xxx', 'xxx', 'xxx');
mysql_set_charset("utf8", $connect);
mysql_select_db('xxx');
$sql = "SELECT street, class, subject, school FROM requests WHERE subject IN ('%s')";
$inElems = array_map(function($elm) {
return "'" . mysql_real_escape_string($elm) . "'";
}, $speicher);
$sql1 = sprintf($sql, join(",", $inElems));
echo $sql1;
// It correctly spills out the SELECT Query but it is wrong because of the '' around the subjects are missing -> see output
$erg = mysql_query($sql1);
if (false === $erg) {
die (mysql_error());
}
$content = mysql_fetch_array($erg, MYSQL_ASSOC);
print_r ($content);

输出：

SELECT street, class, subject, school FROM requests WHERE subject IN ("English, German, Mathematics")
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 'English, German, Mathematics")' at line 1

所以我不知道如何设置它是正确的，目前它不起作用，因为主题之间缺少"，但我不知道其他方法。我是个初学者。我知道，我真的需要把它改成MySQLi，一旦这个问题解决了，我就会这么做。