我是编程新手,非常感谢您的帮助。:)所以,我有一张USER
桌子和一张SALES
桌子。在SALES
桌子上,我只有用户的名字和姓氏。在USER
我有表name
,last name
,USER_ID
,email
等..
当和是匹配项时,我需要USER_ID
从USER
表复制到SALES
表。NAME
LAST NAME
结构如下:
USER_TABLE_A
USER_ID_A
NAME_LASTNAME_A
SALES_TABLE_B
ROW_ID_B
NAME_B
LASTNAME_B
USER_ID_B (empty)
到目前为止,我让两个表在它们相交时都显示数据,但不知道从这里到哪里。谁能帮忙吗?
$sql1 = mysql_query("SELECT name_B, lastname_B, user_id_B, row_id_B FROM sales_table_B WHERE name_B IS NOT NULL AND lastname_B IS NOT NULL", $db);
$sql2 = mysql_query("SELECT name_lastname_A, user_id_A FROM user_table_A WHERE name_lastname_A IS NOT NULL", $db);
$a1 = array();
while ($row = mysql_fetch_array($sql1)) {
$id = $row['row_id_B'];
$name1.$id = $row['name_B']." ".$row['lastname_B'];
array_push($a1, $name1.$id);
}
$a2 = array();
while ($row2 = mysql_fetch_array($sql2)) {
$id2 = $row2['user_id_A'];
$name2.$id2 = $row2['name_lastname_A'];
array_push($a2, $name2.$id2);
}
$result = array_intersect($a1,$a2);
print_r($result);
提前致谢!
非常感谢Darshan!在限制1之后,您的答案缺少),但是经过调整后,效果很好!这是起作用的代码:
UPDATE sales_table_b
SET user_id_b = (SELECT user_table_a.user_id_a
FROM user_table_a
WHERE user_table_a.name_lastname_a = CONCAT(sales_table_b.name_b, ' ' , sales_table_b.lastname_b) LIMIT 1)
WHERE EXISTS (SELECT *
FROM user_table_a
WHERE user_table_a.name_lastname_a = CONCAT(sales_table_b.name_b, ' ' , sales_table_b.lastname_b))
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我来说两句