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我在使用下面的脚本时遇到了这个错误，我在这个网站和谷歌上搜索了很多线程，但我找不到我的问题的解决方案。

    select ct.tons as total_tons,
       sum(IF(sum(wr.tons) is not null,sum(wr.tons),0)) as total_delivered,            
       sum(ct.tons - IF(sum(wr.tons) is not null,sum(wr.tons),0)) as total_balance,
             sum(IF(sum(f.tons_fixed) is not null, sum(f.tons_fixed), 0)) as total_fixed,
             sum(IF(sum(f.tons_fixed) is not null, ct.tons - sum(f.tons_fixed), 0)) as total_unfixed,
             avg(sum(IF(f.tons_fixed * f.hedge_price is not null,f.tons_fixed * f.hedge_price, 0))/IF(sum(f.tons_fixed) is not null, sum(f.tons_fixed), 1) + ct.differential_fob) as avg_contract_price,
           avg(db.current_basis) as avg_market_price,            
             avg(ct.contract_price_foreign - db.current_basis) as avg_outright_fob,
             avg(ct.differential_fob) as avg_contract_fob_diff,
             avg(IF(mf.diff is not null, mf.diff, 0)) as avg_market_fob,
             avg(ct.differential_fob - IF(mf.diff is not null, mf.diff, 0)) as avg_diff_fob 
from contract ct 
left join grade_master gm on ct.grade_id = gm.id
left join movement m on m.contract_id = ct.id
left join warehouse_receipt wr on wr.movement_id = m.id
left join daily_basis db on ct.terminal_month = db.id 
left join contract_price_fixation f on f.contract_id = ct.id
left join market_fob_diff mf on mf.grade_id = ct.grade_id
left join grade_master gmt on mf.grade_id = gmt.id
left join company_master cm on cm.id = ct.supplier_buyer_id
GROUP BY ct.id

我认为问题可能是总和（平均）内的总和，因为在删除最左边的总和时，脚本有效。请帮我纠正它！
谢谢

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1 回答 1

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像这样换行

sum(IF(sum(wr.tons) is not null,sum(wr.tons),0)) as total_delivered

至

COALESCE(SUM(wr.tons), 0) AS total_delivered

返回它的COALESCE()第一个不为空的参数。

于 2013-03-27T11:49:48.917 回答