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在一次使用mysql数据库查询的时候出现错误提示 Column "languageid" in ，这个错误in 多半是因为多表查询的时候几个表中同时出现了某个相同的列名，而在查询条件WHERE后面又没有指定是那个表，而引起的
又或者是查询结果里面有两个相同的列名，而没有指定是哪个表
使用的时候可以这样,mysql查询前面加表名可避免出现错误Column "languageid" in
SELECT tablea.id aid table.id bid WHERE tablea.id = tableb.id
SELECT *
FROM tbl_listings
WHERE postcode = "var1" AND catID = "var2" AND stateID = "var3"
ORDER BY listingName ASC
using the following variables
name: var1
default value: -1
run-time value: $_GET["postcode"]
name: var2
default value: -1
run-time value: $_GET["catID"]
name: var3
default value: -1
run-time value: $_GET["stateID"]
I, however, want to get the category name from the category table using a join so I tried this:
Code:
SELECT *
FROM tbl_listings LEFT JOIN tbl_category ON tbl_listings.catID = tbl_category.catID
WHERE postcode = "var1" AND catID = "var2" AND stateID = "var3"
ORDER BY listingName ASCusing the following variables
name: var1
default value: -1
run-time value: $_GET["postcode"]
name: var2
default value: -1
run-time value: $_GET["catID"]
name: var3
default value: -1
run-time value: $_GET["stateID"]
This gives me an error saying: Column: "catID" in where clause in ambiguous.
WHERE postcode = "var1" AND tbl_listings.catID = "var2"
OR
PHP Code:
WHERE postcode = "var1" AND tbl_category.catID = "var2"