php - 如何正确地将 "$data"从一个 mySQL 查询输入到另一个查询中？

Question

我正在尝试从数据表中提取一些数据并将该数据输入到另一个查询中。我相信我的语法不正确，因为我没有收到任何数据传输。当我的 $data2->destination 被回显时，结果是正确的。

$query2 = 'SELECT * FROM booking WHERE status=1 AND userid= '.$user_ID.'';
$result2 = mysql_query($query2);
$data2 = mysql_fetch_object($result2);
echo $data2->destination;


$query3 = 'UPDATE booking SET source=$data2->destination WHERE status=0 AND 
userid='.$user_ID.'';
$result3 = mysql_query($query3);

Answer 1

$query3 = 'UPDATE booking SET source=$data2->destination WHERE status=0 AND userid='.$user_ID.'';

应该(可以)

$query3 = sprintf(
    "UPDATE booking SET source = %s WHERE status = 0 AND userid = %s",
    $data2->destination, 
    $user_ID
);

或

$query3 = "UPDATE booking SET source = $data2->destination WHERE status = 0 AND userd = $user_ID";

或

$query3 = 'UPDATE booking SET source = '.$data2->destination.' WHERE status = 0 AND userid = '.$user_ID;