在尝试将JOIN语句添加到下面的查询语句中并删除“ JOIN table_name2”后,我一直在遇到iisnode时遇到错误,然后该查询有效。如何修正该语句以使其生效?有什么建议吗?
const getUserById = (request, response) => {
const id = parseInt(request.params.id)
pool.query('SELECT * FROM table_1 JOIN table_2 WHERE "GTIN" = $1', [id], (error, results) => {
if (error) {
throw error
}
response.status(200).json(results.rows)
})
}
表1
GTIN | gtinName
--------+----------------+---------
26001087006846 | Product Name
表2
GTIN | nutrientTypeCode
--------+-------------------+------------
26001087006846 | Protein
--------+-------------------+------------
26001087006846 | Energy
所需结果
GTIN | gtinName | nutrientTypeCode
--------+---------------+---------------------+------------
26001087006846 | Product Name | FASAT
答案 0 :(得分:0)
您应该使用JOIN和相关的on子句来将两个表都加入
const getUserById = (request, response) => {
const id = parseInt(request.params.id)
pool.query('SELECT table_1.GTIN, table_1.gtinNAme, table_2.nutrientTypeCode
FROM table_1
JOIN table_2 ON table_1.GTIN = table_2.GTIN
NWHERE table_1.GTIN = $1', [id], (error, results) => {
if (error) {
throw error
}
response.status(200).json(results.rows)
})
}