Question

我尝试使用LEFT OUTER JOIN从多个故事中获取数据，但我收到致命错误。

表名，字段名，数据库连接是正确的。

$sql = "SELECT shipping_info.shipping_id, service1.service, package1.package_type, countries1.country AS fromCountry, countries2.country AS toCountry, countries3.country AS resiCountry, customer1.name, 
FROM shipping_info 
LEFT OUTER JOIN service_types AS service1 ON shipping_info.service_type = service_types.serviceType_id 
LEFT OUTER JOIN package_types AS package1 ON shipping_info.package_type = package_types.packageType_id 
LEFT OUTER JOIN customer_info AS customer1 ON shipping_info.customer_id = customer_info.customer_id 
LEFT OUTER JOIN countries AS countries1 ON shipping_info.from_loc = countries1.country_id 
LEFT OUTER JOIN countries AS countries2 ON shipping_info.to_loc= countries2.country_id 
LEFT OUTER JOIN countries AS countries3 ON shipping_info.to_id = countries3.country_id 
ORDER BY shipping_info.order_date DESC";

致命错误：在....中的非对象上调用成员函数fetchAll（）

Answer 1

尝试将您的查询更改为：

SELECT s1.shipping_id, 
       s1.service, 
       p1.package_type, 
       c1.country fromCountry, 
       c2.country toCountry, 
       c3.country resiCountry, 
       c1.name
FROM shipping_info si
LEFT JOIN service_types s1 ON si.service_type = s1.serviceType_id 
LEFT JOIN package_types p1 ON si.package_type = p1.packageType_id 
LEFT JOIN customer_info c1 ON si.customer_id = c1.customer_id 
LEFT JOIN countries c1 ON si.from_loc = c1.country_id 
LEFT JOIN countries c2 ON si.to_loc= c2.country_id 
LEFT JOIN countries c3 ON si.to_id = c3.country_id 
ORDER BY si.order_date DESC;

你在查询中有多个拼写错误，语法不正确..还有LEFT OUTER JOIN和LEFT JOIN完全相同

您也可以发布执行此查询的方式吗？您可能会遇到执行它的实际方法的问题。

Answer 2

我不是MySQL专家，但这看起来不对。考虑你的第一次加入：

LEFT OUTER JOIN service_types AS service1 
         ON shipping_info.service_type = service_types.serviceType_id

您为table service_types提供了service1的别名（相关名称），但之后不在连接的ON部分使用它。我要尝试的第一件事是要么摆脱相关名称：

LEFT OUTER JOIN service_types
         ON shipping_info.service_type = service_types.serviceType_id

......或者使用它：

LEFT OUTER JOIN service_types AS service1 
         ON shipping_info.service_type = service1.serviceType_id

由于您在实际选择的列的名称中使用了它，因此我会在连接的ON部分使用它。无论如何，请使用package_types和customer_info重复，然后尝试。

Answer 3

马上就可以看到你在FROM子句之前有一个额外的逗号。这会导致错误，并且可能导致您在非对象上运行fetchAll时出现的错误，即未正确格式化的查询。

你的表别名很长，所以你可以缩短它们或完全抛弃它们，只使用完整的表名。

SELECT
    si.shipping_id,
    st.service,
    pt.package_type,
    c1.country AS fromCountry,
    c2.country AS toCountry,
    c3.country AS resiCountry,
    ci.name
FROM shipping_info si
LEFT JOIN service_types st
    ON si.service_type = st.serviceType_id 
LEFT JOIN package_types pt
    ON si.package_type = pt.packageType_id 
LEFT JOIN customer_info ci
    ON si.customer_id = ci.customer_id 
LEFT JOIN countries c1
    ON si.from_loc = c1.country_id 
LEFT JOIN countries c2
    ON si.to_loc= c2.country_id 
LEFT JOIN countries c3
    ON si.to_id = c3.country_id 
ORDER BY si.order_date DESC;

另外，我最近转向使用MySQL Workbench。这绝对值得一试。我比PHPMyAdmin更喜欢它。对我来说这是一个更好的工作流程，并且有一些很酷的工具，比如反向工程工具，它会根据你的表格为你建立一个ERD。它非常适合在PHP代码中使用它们之前测试您的查询。

MySQL Workbench http://www.mysql.com/products/workbench/

使用LEFT OUTER JOIN连接多个表

3 个答案: