Question

我正在尝试执行以下查询：

SELECT g1.Name_Gene as "Name_ref"
     , g1.Type_Gene
     , gr.Class_Name
     , g2.Name_Gene as "Name_avr"
     , path.Name_SpeciePa
     , dis.Name_Disease
     , g1.Id_Genes as "Id_ref"
     , g2.Id_Genes as "Id_avr"
     , dis.Id_Diseases
     , path.Id_Pathogens
FROM plants_genes as pg
   , genes_reference as gr
   , genes as g1
LEFT JOIN avirulence_reference as ar ON (g1.Id_Genes=ar.Id_Genes_ref) 
LEFT JOIN genes as g2 ON (ar.Id_Genes_avr=g2.Id_Genes) 
LEFT JOIN reference_diseases as rd ON (gr.Id_Genes=rd.Id_Genes) 
LEFT JOIN pathogens_diseases as pd ON (rd.Id_Diseases=pd.Id_Diseases) 
LEFT JOIN diseases as dis ON (pd.Id_Diseases=dis.Id_Diseases) 
LEFT JOIN pathogens as path ON (pd.Id_Pathogens=path.Id_Pathogens) 
WHERE pg.Id_Plants='196' AND pg.Id_Genes=g1.Id_Genes AND g1.Id_Genes=gr.Id_Genes

我收到了这个错误：

ERROR 1054（42S22）：未知栏＆＃39; gr.Id_Genes＆＃39;在＆＃39; on条款＆＃39;

我不明白为什么会收到此错误，因为genes_reference定义为gr且Id_Genes是现有列。

Answer 1

这是因为您混合了隐式和显式连接符号：

FROM genes as g1 
JOIN genes_reference as gr on g1.Id_Genes=gr.Id_Genes 
JOIN plants_genes as pg on pg.Id_Genes=g1.Id_Genes
<...>
WHERE pg.Id_Plants='196'

Answer 2

你应该改变你的内部联接： FROM plants_genes as pg 加入genes_reference作为gr on（{common keys}）将基因作为g1加入（{common keys}）

为什么在我的代码中有左连接的未知列的错误？

2 个答案: