Question

我有一个php代码，它汇总了来自同一数据库的两个不同表的同一列的值，并将其存储在变量中。代码如下：

$sql = 'SELECT
        (SELECT SUM( time_spent )
         FROM '.TICKET_RESPONSE_TABLE.'
         WHERE ticket_id='.db_input($id).')
        +(SELECT SUM( time_spent )
          FROM '.TICKET_NOTE_TABLE.'
          WHERE ticket_id='.db_input($id).')
    AS total_time';
$result = db_query($sql);
$cursor = mysql_fetch_row($result);
$total_time = $cursor[0];

现在，我想要更新同一数据库的另一个表中的列，其值存储在变量$ total_time中。请帮助我。

Answer 1

您可以直接更新，而不是使用变量。

$sql = 'UPDATE *tablename*
        SET *columname* = 
        (SELECT SUM( time_spent )
         FROM '.TICKET_RESPONSE_TABLE.'
         WHERE ticket_id='.db_input($id).')
        +(SELECT SUM( time_spent )
          FROM '.TICKET_NOTE_TABLE.'
          WHERE ticket_id='.db_input($id).')';
$result = db_query($sql);

或像这样的后记：

$sql = 'SELECT
        (SELECT SUM( time_spent )
         FROM '.TICKET_RESPONSE_TABLE.'
         WHERE ticket_id='.db_input($id).')
        +(SELECT SUM( time_spent )
          FROM '.TICKET_NOTE_TABLE.'
          WHERE ticket_id='.db_input($id).')
    AS total_time';
$result = db_query($sql);
$cursor = mysql_fetch_row($result);
$total_time = $cursor[0];
$sql = 'SUPDATE *tablename*
            SET *columname* = ' . $total_time
$result = db_query($sql);

Answer 2

您可以进行子选择：

UPDATE table1 t1
SET t1.val1 =
(SELECT val FROM table2 t2 WHERE t2.id = t1.t2_id )
WHERE t1.val1 = '';

Answer 3

为什么不呢 $ sql =“INSERT INTO其他表SET field_name = $ total_time”; $ result = db_query（$ sql）;

如何使用具有动态值的变量更新sql表

3 个答案: