[英]How to add a variable value to an existing table value in SQL using PHP
[英]How to update sql table by using a variable which has dynamic value
我有一個PHP代碼,它從同一個數據庫的兩個不同表中匯總同一列的值,並將其存儲在一個變量中。 代碼如下:
$sql = 'SELECT
(SELECT SUM( time_spent )
FROM '.TICKET_RESPONSE_TABLE.'
WHERE ticket_id='.db_input($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 = $cursor[0];
現在,我想要更新同一數據庫的另一個表中的列,其值存儲在變量$ total_time中。 請幫助我。
您可以直接更新,而不是使用變量。
$sql = 'UPDATE *tablename*
SET *columname* =
(SELECT SUM( time_spent )
FROM '.TICKET_RESPONSE_TABLE.'
WHERE ticket_id='.db_input($id).')
+(SELECT SUM( time_spent )
FROM '.TICKET_NOTE_TABLE.'
WHERE ticket_id='.db_input($id).')';
$result = db_query($sql);
或像這樣的后記:
$sql = 'SELECT
(SELECT SUM( time_spent )
FROM '.TICKET_RESPONSE_TABLE.'
WHERE ticket_id='.db_input($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);
您可以進行子選擇:
UPDATE table1 t1
SET t1.val1 =
(SELECT val FROM table2 t2 WHERE t2.id = t1.t2_id )
WHERE t1.val1 = '';
為什么不$ sql =“INSERT INTO其他表SET field_name = $ total_time”; $ result = db_query($ sql);
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