Abrufen Sie Daten aus mehreren Tabellen mit mehreren bis zu vielen Beziehungen ab

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Expand view Topic review: Abrufen Sie Daten aus mehreren Tabellen mit mehreren bis zu vielen Beziehungen ab

by Anonymous » 28 Feb 2025, 06:06

Ich habe 4 Tabellen. < /p>

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project

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--------------------------------------
| project_id | name      | client_id |
--------------------------------------
| 1          | Make Safe | 12        |
| 2          | Quote     | 55        |
| 3          | Assist    | 2         |
--------------------------------------
< /code>
project_job

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---------------------------------------------------------
| project_job_id | project_id | project_number| manager |
---------------------------------------------------------
| 1              | 1          | 1             | Steve   |
| 2              | 1          | 2             | Harry   |
| 3              | 2          | 1             | Paul    |
| 4              | 3          | 1             | Sally   |
---------------------------------------------------------
< /code>
project_job_trade

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---------------------------------------
| project_job_id| project_job_trade_id|
---------------------------------------
| 1             | 1                   |
| 1             | 2                   |
| 1             | 3                   |
| 2             | 4                   |
| 3             | 3                   |
| 4             | 1                   |
| 4             | 2                   |
---------------------------------------
< /code>
master_project_job_trade

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--------------------------------------
| project_job_trade_id | name        |
--------------------------------------
| 1                    | Electrician |
| 2                    | Plumber     |
| 3                    | Capenter    |
| 4                    | Arborist    |
--------------------------------------
< /code>
from top to bottom:
project
.

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project_id
= project_job .

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project_id< /code> = 1: viele
project_job
.

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project_job_id
=

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project_job_trade
.

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project_job_id< /code> = 1: viele
project_job_trade
.

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project_job_trade_id
=

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master_project_job_trade
.

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project_job_trade_id

Zuvor hatte ich das Projekt project_job_trade .

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project_job_trade_id
in einem CSV -Format als Spalte in Project_Job Tabelle, dh:

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project_job

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--------------------------------------------------------------------------------
| project_job_id | project_id | project_number| manager | project_job_trade_id |
--------------------------------------------------------------------------------
| 1              | 1          | 1             | Steve   | 1,2,3                |
| 2              | 1          | 2             | Harry   | 4                    |
| 3              | 2          | 1             | Paul    | 3                    |
| 4              | 3          | 1             | Sally   | 1,4                  |
--------------------------------------------------------------------------------
< /code>
I was able to retrieve the trade name associated with the project>project_number, but only the 1st one in the list with a LEFT JOIN, but any others in the list were not displayed, so I was researching a solution. After a lot of reading, it was clear that storing data in a delimited list is not recommended, so i split the data into project_job_trade
Tabelle.
Das Problem, das ich habe, ist, dass ich nicht herausfinden kann

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-------------------------------------------------------------------------------------------------
| project_id | name      | client_id | project_number| manager | name                           |
-------------------------------------------------------------------------------------------------
| 1          | Make Safe | 12        | 1             | Steve   | Electrician, Plumber, Capenter |
| 1          | Make Safe | 12        | 2             | Harry   | Arborist                       |
| 2          | Quote     | 55        | 1             | Paul    | Capenter                       |
| 3          | Assist    | 2         | 1             | Sally   | Electrician, Arborist          |
-------------------------------------------------------------------------------------------------
< /code>
I did stumble upon GROUP_CONCAT
als Möglichkeit, mehrere Handelsnamen in 1 Spalte anzuzeigen, die das Problem theoretisch beheben würde, hatte ich zuvor nur das erste Element, das zurückgegeben wurde. Mit dieser zusätzlichen Tabelle kann ich meinen Kopf jedoch nicht einwickeln, wie die Daten extrahiert werden und alles zusammenfügen.
Bitte helfen Sie.>

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