[英]PHP and SQL INSERT JOIN 2 tables with criteria error
$rentingsquery = "SELECT * FROM TBL_rentings WHERE personId='$personId'
JOIN TBL_rentings ON TBL_properties.propertyId = TBL_rentings.propertyId
JOIN TBL_people ON TBL_rentings.personId = TBL_people.personId";
$rentingsResult=mysql_query($rentingsquery) or die ("Query to get data from TBL_properties failed: ".mysql_error());
This is the error I get when I execute the code in my web page: 这是我在网页中执行代码时遇到的错误:
What I'm trying to do is select only the 'rentings' where the personId = $personId
, (as the current page is a page for each individual person), and display only those 'rentings'. 我想做的是仅选择
personId = $personId
的“租金”(因为当前页面是每个人的页面),并且仅显示那些“租金”。 Also in the code which I haven't posted I'm displaying data about the property which is related to that rent, hence why I'm trying to join the renting and properties table with propertyId so that I call the correct property's details off the database. 同样在我未发布的代码中,我正在显示与该租金相关的财产的数据,因此为什么我要尝试使用propertyId将租金和属性表联接在一起,以便我从数据库。
$rentingsquery = "SELECT * FROM TBL_rentings
JOIN TBL_rentings ON TBL_properties.propertyId = TBL_rentings.propertyId
JOIN TBL_people ON TBL_rentings.personId = TBL_people.personId
WHERE personId='$personId'";
$rentingsResult=mysql_query($rentingsquery) or die ("Query to get data from TBL_properties failed: ".mysql_error());
First of all try this, the where condition has to be after the joins. 首先尝试此操作,连接后必须将where条件设为。 If there are other problems I will edit the answer
如果还有其他问题,我将编辑答案
The next problem is that you are joining the same table, but you are joining it in an external field. 下一个问题是您要联接同一张表,但是要在外部字段中联接它。 A quick fix, if I am supposing correct, is that you wrongly joined with an unwanted table:
如果我认为正确的话,一个快速的解决方法是您错误地加入了不需要的表:
SELECT * FROM TBL_rentings
JOIN TBL_properties ON TBL_properties.propertyId = TBL_rentings.propertyId
JOIN TBL_people ON TBL_rentings.personId = TBL_people.personId
WHERE TBL_rentings.personId='$personId'
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