Question

此查询工作正常：

SELECT  posts.titulo as value,
                    posts.id as id,
                    posts.img_src as img,
                    posts.id_familia,
                    posts.tiempo,
                    posts.votos,
                    familias.clave,
                    familias.id as fm,
                    textos.clave,
                    textos.texto as familia,
            FROM posts,familias,textos
            WHERE posts.id_familia = familias.id AND familias.clave = textos.clave AND textos.lengua = ".detectarIdioma()." 
            and posts.id_usuario = $term 
            ORDER BY posts.id DESC

但是现在我想添加一篇帖子有多少评论，这是在comentarios表中

SELECT  posts.titulo as value,
                    posts.id as id,
                    posts.img_src as img,
                    posts.id_familia,
                    posts.tiempo,
                    posts.votos,
                    familias.clave,
                    familias.id as fm,
                    textos.clave,
                    textos.texto as familia,
                    count(comentarios.id)
            FROM posts,familias,textos
            JOIN comentarios ON comentarios.id_post = posts.id
            WHERE posts.id_familia = familias.id AND familias.clave = textos.clave AND textos.lengua = ".detectarIdioma()." 
            and posts.id_usuario = $term 
            ORDER BY posts.id DESC

问题是mysql错误是

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '*) FROM posts,familias,textos JOIN comentarios ON ' at line 12

知道我在这里缺少什么吗？

Answer 1

尝试这样的事情：

SELECT posts.titulo AS value,
  posts.id AS id,
  posts.img_src AS img,
  posts.id_familia,
  posts.tiempo,
  posts.votos,
  familias.clave,
  familias.id AS fm,
  textos.clave,
  textos.texto AS familia,
  COALESCE(COM_COUNT.NUM_COMMENTS,0) AS num_comments
FROM posts 
INNER JOIN familias ON posts.id_familia = familias.id
INNER JOIN textos familias.clave = textos.clave
LEFT JOIN 
    ( SELECT id_post, COUNT(*) AS NUM_COMMENTS
      FROM comentarios
      GROUP BY id_post
    ) COM_COUNT ON COM_COUNT.id_post = posts.id   
WHERE AND textos.lengua = ".detectarIdioma()."
  AND posts.id_usuario = $TERM
ORDER BY posts.id DESC

这将使每个帖子的评论数量加入，如果JOIN不匹配则显示为0.

Answer 2

试试这个：

SELECT  posts.titulo as value,
        posts.id as id,
        posts.img_src as img,
        posts.id_familia,
        posts.tiempo,
        posts.votos,
        familias.clave,
        familias.id as fm,
        textos.clave,
        textos.texto as familia,
        count(comentarios.id)
FROM posts INNER JOIN familias ON posts.id_familia = familias.id 
INNER JOIN textos ON familias.clave = textos.clave 
LEFT OUTER JOIN comentarios ON comentarios.id_post = posts.id
WHERE textos.lengua = ".detectarIdioma()." 
AND posts.id_usuario = $term 
GROUP BY posts.titulo,
        posts.id,
        posts.img_src,
        posts.id_familia,
        posts.tiempo,
        posts.votos,
        familias.clave,
        familias.id,
        textos.clave,
        textos.texto
ORDER BY posts.id DESC

您只是混合了两种JOIN语法。 ......除了你正在计算的那一列之外，你可能需要按每列进行分组。

编辑：要不将结果限制为仅包含评论的人，您需要在该表上执行LEFT OUTER JOIN。

Answer 3

您在第一个查询中的FROM之前有一个逗号，但在第二个查询中没有FROM之前的逗号

将另一个表中的计数添加到现有查询

3 个答案: