Group function doesnt works

I m trying to do a report with a count  of items that a user técnico has .




SELECT
     COUNT(*)Quantidade,
 item.`codigo` AS item_codigo,
 empresa.`codigo` AS empresa_codigo,
     tecnico_item.`codigo_item` AS tecnico_item_codigo_item,
     tecnico_item.`codigo_tecnico` AS tecnico_item_codigo_tecnico,
     usuario.`codigo` AS usuario_codigo,
     empresa.`nome_fantasia` AS empresa_nome_fantasia,
     item.`codigo_item` AS item_codigo_item,
     item.`criticidade` AS item_criticidade,
     item.`data_item` AS item_data_item,
     item.`descricao` AS item_descricao,
     usuario.`nome` AS usuario_nome,
     empresa.`data_fim` AS empresa_data_fim,
     empresa.`data_ini` AS empresa_data_ini


FROM
     `empresa` empresa INNER JOIN `item` item ON empresa.`codigo` = item.`codigo_empresa`
     INNER JOIN `tecnico_item` tecnico_item ON item.`codigo` = tecnico_item.`codigo_item`
     INNER JOIN `usuario` usuario ON tecnico_item.`codigo_tecnico` = usuario.`codigo`
     AND empresa.`codigo` = usuario.`codigo_empresa`

group by

 item.`codigo`,
 empresa.`codigo`,
     tecnico_item.`codigo_item` ,
     tecnico_item.`codigo_tecnico` ,
     usuario.`codigo` ,
     empresa.`nome_fantasia`,
     item.`codigo_item`,
     item.`criticidade` ,
     item.`data_item`,
     item.`descricao` ,
     usuario.`nome` ,
     empresa.`data_fim`,
     empresa.`data_ini`

Open in new window



The count isn't correct --> see in queryresult


When i pick only the manutomany table the count () its ok.

see bellow:

select 
   count(coditen) 
from tabela
where codtecnico = codtecnico
group by codtecnico

Open in new window

model.jpg
queruresult.jpg
Rafael_MoreiraAsked:
Who is Participating?
 
Kevin CrossConnect With a Mentor Chief Technology OfficerCommented:
I suspect you need to aggregate one table in a derived query, joining to the outer query.  Not sure I can make out exactly what you are after; however, here is a generic example to count records in tableB for a single record in tableA.

SELECT b.cnt, ...
FROM tableA a
-- use LEFT JOIN if you may not have a match to get zero counts
JOIN (
    SELECT col_matching_tableA, COUNT(*) cnt
    FROM tableB
    GROUP BY col_matching_tableA
) b ON b.col_matching_tableA = a.col_matching_tableB

;

I hope that helps!
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Rafael_MoreiraAuthor Commented:
I m trying this .

Take a look at image it seems that something between item and usuário lack.
query.png
tableItem.jpg
tableNxN.jpg
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Kevin CrossChief Technology OfficerCommented:
What are you expecting?  One row with 27?  Just making sure I understand.
0
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Rafael_MoreiraAuthor Commented:
each group row per user very similar with the last result , but the items descriptions  are not complete .

The result is grouping some descriptions .


T0001      X      1969-12-31      VÁLVULA PARA MOTOR
V000011      X      2014-01-02      1FUSIVEL NH 00 80 A 500 V SIEMENS/3NA10202
V000015      X      2014-01-02      5FUSIVEL NH 00 80 A 500 V SIEMENS/3NA10206
V000019      X      2014-01-02      9FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102010
V0000113      X      2014-01-02      13FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102014
V0000117      X      2014-01-02      17FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102018
V0000121      X      2014-01-02      21FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102022
V0000125      X      2014-01-02      25FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102026
V0000129      X      2014-01-02      29FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102030
V0000133      X      2014-01-02      33FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102034
V0000137      X      2014-01-02      37FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102038
V0000141      X      2014-01-02      41FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102042
V0000145      X      2014-01-02      45FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102046
V0000149      X      2014-01-02      49FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102050
V0000153      X      2014-01-02      53FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102054
V0000157      X      2014-01-02      57FUSIVEL NH 00 80 A 500 V SIEMENS/3NA102058



i think that a join is lacking.
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Rafael_MoreiraAuthor Commented:
I had to make many subqueryes to get what i want.

But your solution was very helpful
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Kevin CrossChief Technology OfficerCommented:
I am glad you found what you needed and that I could help.

Best regards and happy coding,

Kevin
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Rafael_MoreiraAuthor Commented:
you where wellcome.
0
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