Question

你能帮我解决一下我的疑问。我检查我之前的查询表是否有效，但这不起作用。

                SELECT cj.customer_jd,
                customertable.name,
                customertable.ordersize,
                cj.job_no,
                cj.id,
                cj.ordered_quantity, 
                cj.ordered_quantity,st.id AS sandingID,
                st.*
                FROM sandingtable AS st
                LEFT JOIN customer_job  AS cj  ON 
                customertable.id=cj.customer_jd
                RIGHT JOIN  st ON cj.id = st.`job_id`
               WHERE st.date= '$date' AND st.shift = '$shift' AND 
                st.sandingno=".$row['sandingno']."");

Answer 1

您尚未在var arr = [2017, 4, 30]; function dateAsArrayYYYYMMDD(array) { return new Date(array.toString().replace(/(\d+)/g, "'$1'")) .toJSON().slice(0, 10).split("-"); } // get last date in April as array of string values var date = dateAsArrayYYYYMMDD(arr); console.log(date); // set date reflecting last date in March, // assign to `date` variable identifier arr[1] -= 1; arr[2] += 1; date = dateAsArrayYYYYMMDD(arr); console.log(date);子句中定义customer_table：

from

我不确定如何修复它，但也许是这样的：

FROM sandingtable st LEFT JOIN
     customer_job cj 
     ON customertable.id = cj.customer_jd RIGHT JOIN
     st
     ON cj.id = st.`job_id`
WHERE st.date = '$date' AND st.shift = '$shift' AND 
      st.sandingno = ".$row['sandingno']."");

注意：

我会为from st left join customer_job cj on cj.id = st.job_id left join customertable on customertable.id = cj.customer_jd . . .使用别名ct，但查询的其余部分使用customertable。
您应该在查询中始终使用customertable s。混合左右连接只会令人困惑。

查询无效：不唯一的表/别名：'st'

1 个答案: