错误：“和”处或附近的语法错误答案

【问题标题】：ERROR: syntax error at or near "and"错误：“和”处或附近的语法错误
【发布时间】：2014-09-19 09:24:24
【问题描述】：

我有一个这样的查询。

select 
  ad.escore,
  ad.mscore,
  round(sum(ps.cnt) / sum(n.cnt) * 100,1) as percent
from 
(
  select 
    account_no,
    -- 602 becomes '595-604'
    to_char(trunc(empirica_score - 5, -1) + 5, '9999') || '-' || to_char(trunc(empirica_score - 5, -1) + 14, '9999') as escore,
    -- 97 becomes '76-100'. Change the expression to group differently.
    cast(((mfin_score - 1) / 25) * 25 + 1 as text) || '-' || cast(((mfin_score - 1) / 25) * 25 + 25 as text) as mscore
  from account_details
) ad
join 
(
  select custno, count(*) as cnt
  from paysoft_results 
  where result = 'Successful' 
  and resultdate >= '13/08/2014'     <------- HERE
  and resultdate <= '12/19/2014'     <------- HERE
  group by custno
) ps on ps.custno = ad.account_no
join 
(
  select customer_code, count(distinct start_date) as cnt
  from naedo 
  and start_date >= '13/08/2014'      <------- HERE
  and start_date <= '12/19/2014'      <------- HERE
  group by customer_code
) n on n.customer_code = ad.account_no
group by ad.escore, ad.mscore;

如果我没有像上面那样安装日期，它就完美了。

如果我输入日期，我会收到错误ERROR: syntax error at or near "and"

有什么想法吗？

更新

好的，我想我可以问一个现在的问题，所以如果我可以追加这个问题。

ERROR: date/time field value out of range: "13/08/2014"

我的查询中的日期比较。正确的做法是什么？

【问题讨论】：

早上，更新可能应该是一个新的问题，但我已经在我原来的答案的旁边提到了它。
不要不依赖隐式数据类型转换。始终使用正确的日期文字，而不是字符串常量。 '13/08/2014' 是一个字符串，而不是日期。您应该使用 Oracle 的 to_date() 函数：to_date('13/08/2014', 'DD/MM/YYYY') 或（稍微短一点）ANSI 日期文字：date '2014-08-13'。

标签： sql postgresql

【解决方案1】：

好吧，这个位行不通：

select customer_code, count(distinct start_date) as cnt
from naedo 
and start_date >= '13/08/2014'      <------- HERE
and start_date <= '12/19/2014'      <------- HERE
group by ...

因为where 子句必须以where 开头，而不是and。否则，我们都将其称为and 子句:-)

必须是：

select customer_code, count(distinct start_date) as cnt
from naedo 
where start_date >= '13/08/2014'
  and start_date <= '12/19/2014'
group by ...

您用HERE 标记的另一位（第二段，第一个join 子句）看起来很好，它应该可以正常工作。

顺便说一句，您的日期中至少有一个格式不正确。片段：

and start_date >= '13/08/2014'
and start_date <= '12/19/2014'

有一个日期是 12 月 8 日或 12^th，好吧，我什至不知道是什么知道十九的拉丁语前缀（或基于已经不同步的真实月份的十七）是。

您需要确定您的数据库支持mm/dd/yyyy 或dd/mm/yyyy 中的哪一个，然后只使用那个。

鉴于您质疑更新状态它抱怨13/08/2014，您可能会发现它应该写为08/13/2014，采用mm/dd/yyyy 格式。

【讨论】：

facePalm ...是的，刚刚注意到。

【解决方案2】：

    select customer_code, count(distinct start_date) as cnt
      from naedo 
      Where start_date >= '13/08/2014'      <------- HERE
      and start_date <= '12/19/2014'      <------- HERE
      group by customer_code

【讨论】：

【解决方案3】：

查询中缺少“Where”：

"    select customer_code, count(distinct start_date) as cnt
      from naedo where
      start_date >= '13/08/2014'      <------- HERE"
"
============================
select 
      ad.escore,
      ad.mscore,
      round(sum(ps.cnt) / sum(n.cnt) * 100,1) as percent
    from 
    (
      select 
        account_no,
        -- 602 becomes '595-604'
        to_char(trunc(empirica_score - 5, -1) + 5, '9999') || '-' || to_char(trunc(empirica_score - 5, -1) + 14, '9999') as escore,
        -- 97 becomes '76-100'. Change the expression to group differently.
        cast(((mfin_score - 1) / 25) * 25 + 1 as text) || '-' || cast(((mfin_score - 1) / 25) * 25 + 25 as text) as mscore
      from account_details
    ) ad
    join 
    (
      select custno, count(*) as cnt
      from paysoft_results 
      where result = 'Successful' 
      and resultdate >= '13/08/2014'     <------- HERE
      and resultdate <= '12/19/2014'     <------- HERE
      group by custno
    ) ps on ps.custno = ad.account_no
    join 
    (
      select customer_code, count(distinct start_date) as cnt
      from naedo where
      start_date >= '13/08/2014'      <------- HERE
      and start_date <= '12/19/2014'      <------- HERE
      group by customer_code
    ) n on n.customer_code = ad.account_no
    group by ad.escore, ad.mscore;

【讨论】：