SQL问题：一对多关系和EAV模型

Question

Good evening guys, I'm a newbie to web programming and I need your help to solve a problem inherent to SQL query. The database engine I'm using is MySQL and I access it via PHP, here I'll explain a simplified version of my database, just to fix ideas. Let's suppose to work with a database containing three tables: teams , teams_information , attributes . More precisely:

1) teams is a table containing some basic information about italian football teams (soccer, not american football :D), it is formed by three fields: 'id' ( int, primary key) , 'name' (varchar, team name), nickname ( Varchar , team nickname);

2) attributes is a table containing a list of possible information about a football team, such as city (the city where team plays its home match), captain (team captain's fullname), f_number (number of fans) and so on. This table is formed by three fields: id (int, primary key) , attribute_name ( varchar , an identifier for the attribute), attribute_desc (text, an explanation of the meaning of attribute). Each record of this table represents a single possible attribute of a football team;

3) teams_information is a table where some information, about teams listed in team table, are available. This table contains three fields: id ( int , primary key), team_id ( int , a foreign key which identifies a team), attribute_id ( int , a foreign key which identifies one of the attributes listed in attributes table), attribute_value ( varchar , the value of the attribute). Each record represents a single attribute of a single team. In general, different teams will have a different number of information, so for some teams a large number of attributes will be available while for other teams only a small number of attributes will be available.

Note that relation between teams and teams_information is one to many and the same relation exists between attributes and teams_information

Well, given this model my purpose is to realize a grid (maybe with ExtJS 4.1) to show user the list of italian football team, each record of this grid will represent a single football team and will contain all possible attributes: some fields may be empty (because, for considered team, the correspondent attribute is unknown), while the others will contain the values stored in teams_information table (for the considered team). According to the above grid's field are: id, team_name and a number of fields to represent all the different attributes listed in 'attributes' table.

My question is: can I realize such a grid by using a SINGLE SQL query (maybe a proper SELECT query, to fetch all data I need from database tables) ? Can anyone suggest me how to write a similar query (if it exists) ?

Thanks in advance for helping me.

Regards.

Enrico.

Answer 1

The short answer to your question is no, there is no simple construct in MySQL to achieve the result set you are looking for.

But it is possible to carefully (painstakingly) craft such a query. Here is an example, I trust you will be able to decipher it. Basically, I'm using correlated subqueries in the select list, for each attribute I want returned.

SELECT t.id
     , t.name
     , t.nickname

     , ( SELECT v1.attribute_value 
           FROM team_information v1 
           JOIN attributes a1
             ON a1.id = v1.attribute_id AND a1.attribute_name = 'city'
          WHERE v1.team_id = t.id ORDER BY 1 LIMIT 1
       ) AS city

     , ( SELECT v2.attribute_value
           FROM team_information v2 JOIN attributes a2
             ON a2.id = v2.attribute_id AND a2.attribute_name = 'captain'
          WHERE v2.team_id = t.id ORDER BY 1 LIMIT 1
       ) AS captain

     , ( SELECT v3.attribute_value
           FROM team_information v3 JOIN attributes a3
             ON a3.id = v3.attribute_id AND a3.attribute_name = 'f_number'
          WHERE v3.team_id = t.id ORDER BY 1 LIMIT 1
       ) AS f_number

  FROM teams t
 ORDER BY t.id

For 'multi-valued' attributes, you'd have to pull each instance of the attribute separately. (Use the LIMIT to specify whether you are retrieving the first one, the second one, etc.)

     , ( SELECT v4.attribute_value
           FROM team_information v4 JOIN attributes a4
             ON a4.id = v4.attribute_id AND a4.attribute_name = 'nickname'
          WHERE v4.team_id = t.id ORDER BY 1 LIMIT 0,1
       ) AS nickname_1st

     , ( SELECT v5.attribute_value
           FROM team_information v5 JOIN attributes a5
             ON a5.id = v5.attribute_id AND a5.attribute_name = 'nickname'
          WHERE v5.team_id = t.id ORDER BY 1 LIMIT 1,1
       ) AS nickname_2nd

     , ( SELECT v6.attribute_value
           FROM team_information v6 JOIN attributes a6
             ON a6.id = v6.attribute_id AND a6.attribute_name = 'nickname'
          WHERE v6.team_id = t.id ORDER BY 1 LIMIT 2,1
       ) AS nickname_3rd

---

I use nickname as an example here, because American soccer clubs frequently have more than one nickname, eg Chicago Fire Soccer Club has nicknames: 'The Fire', 'La Máquina Roja', 'Men in Red', 'CF97', et al.)

NOT AN ANSWER TO YOUR QUESTION, BUT ...

Have I mentioned numerous times before, how much I dislike working with EAV database implementations? What should IMO be a very simple query turns into an overly complicated beast of a potentially light dimming query.

Wouldn't it be much simpler to create a table where each "attribute" is a separate column? Then queries to return reasonable result sets would look more reasonable...

SELECT id, name, nickname, city, captain, f_number, ... FROM team

But what really makes me shudder is the prospect that some developer is going to decide that the LDQ should be "hidden" in the database as a view, to enable the "simpler" query.

If you go this route, PLEASE PLEASE PLEASE resist any urge you may have to store this query in the database as a view.

Answer 2

I'm going to take a slightly different route. Spencer's answer is fantastic, and it addresses the issue quite well, but there's still a large underlying problem.

The data that you are trying to display on the site is over-normalized in the database. I won't elaborate, since, again, Spencer's answer highlights the issue pretty well.

Rather, I'd like to recommend a solution that denormalizes the data a bit.

Convert all of your Team data into a single table with many columns. (If there is Player data that isn't covered in the question, that would be a second table, but I'll gloss over that for now.)

Sure, you'll have a whole bunch of columns, and a lot of the columns might be NULL for a lot of the rows. It's not normalized, and it's not pretty, but here's the huge advantage that you gain.

Your query becomes:

SELECT * FROM Teams

That's it. That gets displayed right to the website and you are done. You might have to go out of your way to realize this schema, but it would be totally worth the time investment.

Answer 3

I think what you're saying is that you want the rows in the attributes table to appear as columns in the result recordset. If this is correct, then then in SQL you would use PIVOT. A quick search on SO seems to indicate that there is no PIVOT equivalent in MySql.

Answer 4

I wrote a simple PHP script to generalize spencer's idea to solve my issue. Here's the code:

<?php
    require_once('includes/db.config.php'); //this file performs connection to mysql

/*
 * Following function requires a table name ($table)
 * and a number of service fields ($num). Given those parameters
 * it returns the number of table fields (excluding service fields). 
 */

function get_fields_number($table,$num,$conn)
{   
    $query = "SELECT * FROM $table";
    $result = mysql_query($query,$conn);
    return mysql_num_fields($result)-$num; //remember there are $num service fields
}

/*
 * Following function requires a table name ($table) and an array
 * containing a list of service fields names. Given those parameters, 
 * it returns the list of field names. That list is contained within an array and
 * service fields are excluded.
 */

function get_fields_name($table,$service,$conn)
{
    $query = "SELECT * FROM $table";
    $result = mysql_query($query,$conn);
    $name = array(); //Array to be returned
    for ($i=0;$i<mysql_num_fields($result);$i++)
    {
        if(!in_array(mysql_field_name($result,$i),$service))
    {
       //currently selected field is not a service field
       $name[] = mysql_field_name($result,$i); 
    }
    }
    return $name;
}

//Below $conn is db connection created in 'db.config.php'

$query = "SELECT `name` FROM  `detail_arg` WHERE visibility = 0";
$res = mysql_query($query,$conn);
if($res===false)
{
    $err_msg = mysql_real_escape_string(mysql_error($conn));
    echo "{success:false,data:'".$err_msg."'}";
    die();
}
$arg = array(); //list of argument names
while($row = mysql_fetch_assoc($res))
{
    $arg[] = $row['name'];
}

//Following function writes the select subquery which is
    //necessary to build a column containing a single attribute.

function make_subquery($attribute) //$attribute contains attribute name
{
    $query = "";
    $query.="(SELECT incident_detail.arg_value ";
    $query.="FROM incident_detail ";
    $query.="INNER JOIN detail_arg ";
    $query.="ON incident_detail.arg_id = detail_arg.id AND      detail_arg.name='".$attribute."' ";
    $query.="WHERE incident.id = incident_detail.incident_id) ";
    $query.="AS $attribute";
    return $query;
}

/*

echo make_subquery("date"); //debug code

*/

$subquery = array(); //list of subqueries
for($i=0;$i<count($arg);$i++)
{
    $subquery[] = make_subquery($arg[$i]); 
}

$query = "SELECT "; //final query containing subqueries

$fields = get_fields_name("incident",array("id","visibility"),$conn); 
    //list of 'incident' table's fields

for($i=0;$i<count($fields);$i++)
{
    $query.="incident.".$fields[$i].", ";
}

//insert the subqueries

$sub = implode($subquery,", ");
$query .= $sub;

    $query.=" FROM incident ORDER BY incident.id";
echo $query;
?>