使用LEFT-JOIN-pdo-php-mysql更新

我想同时更新多个表，所以我使用LEFT JOIN为我的更新创建了一个查询。我尝试了两种方法使左连接，但都失败了。我不知道我在哪里犯了错误，所以我希望有人能够正确地剖析查询并指出错误

我已经对查询应用了一些格式设置，以便看起来比以前更可读：

首先是：

"UPDATE "
.
"table1 AS t1 SET t1.Bid = :id "
.
"LEFT JOIN table2 AS t2 SET t2.id = :id ON t1.Bid = t2.id AND t1.status = t2.status "
.
"LEFT JOIN table3 AS t3 SET t3.Bid = :id ON t1.Bid = t3.Bid AND t1.status = t3.status "
.
"LEFT JOIN table4 AS t4 SET t4.id = :id ON t1.Bid = t4.id AND t1.status = t4.status "
.
"LEFT JOIN table5 AS t5 SET t5.Bid = :id ON t1.Bid = t5.Bid AND t1.status = t5.status "
.
"LEFT JOIN table6 AS t6 SET t6.id = :id ON t1.Bid = t6.id AND t1.status = t6.status "
.
"LEFT JOIN table7 AS t7 SET t7.Bid = :id ON t1.Bid = t7.Bid AND t1.status = t7.status "
.
"LEFT JOIN table8 AS t8 SET t8.id = :id ON t1.Bid = t8.id AND t1.status = t8.status "

.
"WHERE t1.Bid = :oldid AND t1.status = :status "

第二个是：

$stmt = $dbh - > prepare("UPDATE "
    .
    "table1 AS t1 "
    .
    "LEFT JOIN table2 AS t2 ON t1.Bid = t2.id AND t1.status = t2.status "
    .
    "LEFT JOIN table3 AS t3 ON t1.Bid = t3.Bid AND t1.status = t3.status "
    .
    "LEFT JOIN table4 AS t4 ON t1.Bid = t4.id AND t1.status = t4.status "
    .
    "LEFT JOIN table5 AS t5 ON t1.Bid = t5.Bid AND t1.status = t5.status "
    .
    "LEFT JOIN table6 AS t6 ON t1.Bid = t6.id AND t1.status = t6.status "
    .
    "LEFT JOIN table7 AS t7 ON t1.Bid = t7.Bid AND t1.status = t7.status "
    .
    "LEFT JOIN table8 AS t8 ON t1.Bid = t8.id AND t1.status = t8.status "
    .
    " SET t1.Bid = :id, "
    .
    " SET t2.id = :id, "
    .
    " SET t3.Bid = :id, "
    .
    " SET t4.id = :id, "
    .
    " SET t5.Bid = :id, "
    .
    " SET t6.id = :id, "
    .
    " SET t7.Bid = :id, "
    .
    " SET t8.id = :id "

    .
    "WHERE t1.Bid = :oldid AND t1.status = :status ");

更新

我正在使用第一个选项，我得到：

致命错误：未捕获异常“PDOException”，消息为“SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；查看与您的MySQL服务器版本对应的手册，了解使用“LEFT JOIN table2 AS t2 SET t2.id=”315-512-613-12“在第1行”的正确语法

第二个是：

致命错误：未捕获异常“PDOException”，消息为“SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；检查与MySQL服务器版本对应的手册，以了解在“SET t2.id=”315-512-613-123V'，SET t3.Bid=”315-512-613-123V'，SE“第1行”附近使用的正确语法

---

我用这个查询让它工作。

"UPDATE 
    table1 AS t1 LEFT JOIN
    table2 AS t2 ON t1.Bid = t2.id AND t1.status = t2.status LEFT JOIN
    table3 AS t3 ON t1.Bid = t3.Bid AND t1.status = t3.status LEFT JOIN
    table4 AS t4 ON t1.Bid = t4.id AND t1.status = t4.status LEFT JOIN
    table5 AS t5 ON t1.Bid = t5.Bid AND t1.status = t5.status LEFT JOIN
    table6 AS t6 ON t1.Bid = t6.id AND t1.status = t6.status LEFT JOIN
    table7 AS t7 ON t1.Bid = t7.Bid AND t1.status = t7.status LEFT JOIN
    table7 AS t8 ON t1.Bid = t8.id AND t1.status = t8.status
    SET t1.Bid = :id,
        t2.id = :id,
        t3.Bid = :id,
        t4.id = :id,
        t5.Bid = :id,
        t6.id = :id,
        t7.Bid = :id,
        t8.id = :id
    WHERE t1.Bid = :oldid
    AND t1.status = :status "

---

那有什么问题？请包括两个请求的预期和实际结果。我相信更新多个表的Mysql语法是

updatet1，t2，t3 SET t1.field1=value1，t2.field2=value2。。。其中t1.pk=t2.fk…

检查MySQL在执行查询时返回的错误。@AlexBlex我希望它会更新，但我得到了错误..我将检查错误并在更新时粘贴它..我不需要使用左连接是什么意思？但是如果表中不存在特定ID，这将意味着它不会更新。这就是一个我使用左连接的原因