我正在考虑将以下内容作为 PHP 多查询来完成。但我想弄清楚如何将列值从选择查询传递到插入和更新查询。
$query = "SELECT tbl_links.link, link_id
FROM tbl_links
INNER JOIN tbl_items ON tbl_links.item_id = tbl_items.item_id
WHERE tbl_items.item_name like '".$items_name[$counter]."'
AND NOT EXISTS (
select link_id
from tbl_clickedlinks
where tbl_clickedlinks.link_id = tbl_links.link_id
AND tbl_clickedlinks.cust_id = '$items_custID[$counter]'
)
limit 0, 1;" ;
$query .= "INSERT INTO tbl_claimedlinks (cust_id, link_id, claim_time) VALUES ('$items_custID', $row['link_id'], NOW()) ;";
$query .= "UPDATE tbl_links SET click_count = click_count+1 where link_id = '$linkID' ;";*/
问题是,我不确定如何将 link_id 值传递给其他查询。所以我在想我可能不得不将查询重新排列成一个,但同样,我只是不确定如何实现它。
有人有什么建议吗?
最佳答案
您需要先执行选择查询,然后使用其输出执行第二和第三个查询。
$query = "SELECT tbl_links.link, link_id
FROM tbl_links
INNER JOIN tbl_items ON tbl_links.item_id = tbl_items.item_id
WHERE tbl_items.item_name like '".$items_name[$counter]."'
AND NOT EXISTS (
select link_id
from tbl_clickedlinks
where tbl_clickedlinks.link_id = tbl_links.link_id
AND tbl_clickedlinks.cust_id = '$items_custID[$counter]'
)
limit 0, 1;" ;
$result = mysql_query($query);
while($row = mysql_fetch_array($result)) {
$query2 = "INSERT INTO tbl_claimedlinks (cust_id, link_id, claim_time) VALUES ('$items_custID', $row['link_id'], NOW()) ;";
$query3 = "UPDATE tbl_links SET click_count = click_count+1 where link_id = '$linkID' ;";*/
mysql_query($query2);
mysql_query($query3);
}
关于php - 将列值从 Select 查询传递到另一个查询以在 PHP 中插入,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16743742/