尝试加入MySQL数据库中的多个表并在Web表中显示

Question

Ok, so I'm very new to this database stuff, and I'm trying to figure out how to query multiple tables at the same time. Apparently you can query as many different tables as you'd like, although you might experience poor performance if you do too many joins. However I'm having issues getting 2 tables to join, let alone the 7 or so I'd need to join later.

I've done some reading on the various ways you can join tables, and most people seem to favor the union option, since MySQL doesn't support the Full Join . The examples shown for this method not only confuse me, but look like it would get massively complicated when trying to join 7 tables at once. Then I saw an article HERE that says MySQL can use a comma operator that simulates a full join. This not only looked much easier to understand, but much easier to use when joining very many tables. But I can't seem to get it to work for me, so hopefully someone can help me get this figured out.

EDIT - Here is some more information, hopefully it will help.

I have two tables

test_dogs which has breed names and basic breed information

lifestyle which as breed characteristics such as how much exercise a dog needs, average health and such.

Both tables have a column named Breed_Name , which is the foreign key in the lifestyle table.

I want to create a query where in this case I can join the two tables and select breeds that fit the following conditions:

Breed_Size = Large

Exercise < 7 (traits in the lifestyle table are on a scale of 1 - 10)

I am able to connect to my database and perform queries on individual tables.

SQL

    $large = $db->query('
SELECT * 
from test_dogs 
where breed_size = "Large" 
order by breed_name '); // this query works

    $exercise = $db->query('
SELECT * 
from lifestyle 
where exercise < 7 
order by exercise DESC'); // this query works

    $join = $db->query(' 
SELECT test_dogs.*, lifestyle.* 
from test_dogs, lifestyle 
ON test_dogs.breed_name = lifestyle.breed_name 
where test_dogs.breed_size = "Large" 
and lifestyle.exercise < 7 
order by exercise DESC'); // THIS QUERY DOES NOT WORK

PHP

<h1> Large Breeds </h1> <!--This table works-->

    <table>
        <tr>
            <th>Breed Name</th>
            <th>Size</th>
        </tr>

        <tr>
        <?php
        while ($rows = $large->fetch()){
            echo "<tr><td>" . $rows['Breed_Name'] . "</td><td>" . $rows['Breed_Size'] . "</td></tr>";
        };
        ?>
    </table>

    <h1> Not High Exercise </h1> <!--This table works-->

    <table>
        <tr>
            <th>Breed Name</th>
            <th>Exercise Needs</th>
        </tr>

        <tr>
        <?php
        while ($rows = $exercise->fetch()){
            echo "<tr><td>" . $rows['Breed_Name'] . "</td><td>" . $rows['Exercise'] . "</td></tr>";
        };
        ?>
    </table>

    <h1> Large AND Not High Exercise </h1> <!--This table DOES NOT work-->

    <table>
        <tr>
            <th>Breed Name</th>
            <th>Size</th>
            <th>Exercise Needs</th>
        </tr>

        <tr>
        <?php
        while ($rows = $join->fetch()){
            echo "<tr><td>" . $rows['test_dogs.Breed_name'] . "</td><td>" . $rows['test_dogs.Breed_Size'] ."</td><td>" . $rows['lifestyle.Exercise'] . "</td></tr>";
        };
        ?>
    </table>

I've seen how some people can include information about the database they're trying to access, but I don't know how to do that. If there's any more information I could offer to help make this question clearer, please let me know.

Answer 1

just a wee pointer as to the various types of joins you can have between tableshttp://blog.codinghorror.com/a-visual-explanation-of-sql-joins/

With reference to your sql above that you say does not work - personally I find that syntax is unclear which doesn't help you at this point when you are trying to learn. To my mind, and others will likely disagree, a clearer syntax would be something akin to:-

select * from `test_dogs` td
    left outer join `lifestyle` l on l.`breed_name`=td.`breed_name`
    where 
        td.`breed_size`='Large'
        and
        l.`exercise` < '7'
    order by l.`exercise` desc;

Where td and l after the table names are known as aliases and allow you reference the table by that unique identifier throughout. Rather than selecting all records ( indicated with * ) once you have the alias you can reference specific fields directly using the alias and the field name. ie: td. dog_type etc so the query could be more specific:-

select 
    td.`dog_type` as 'type',
    td.`dog_weight` as 'weight',
    l.`walks_per_day` as 'walkies'
    from `test_dogs` td
        left outer join `lifestyle` l on l.`breed_name`=td.`breed_name`
        where 
            td.`breed_size`='Large'
            and
            l.`exercise` < '7'
        order by l.`exercise` desc;

etc.... the use of " as 'type'" allows you to supply a short moniker to the referenced field if you wish.

Without seeing your database we cannot categorically state why your query does not work - make sure the columns you are trying to link tables with contain common information

Answer 2

Ok, so I got this figured out. RamRaider was very helpful in pointing out the issues with my syntax, and I will be accepting his answer. I'm posting this here for those who might understand what I was getting at and wanted help with the same thing.

So I was trying to figure out how to write my $db->query so that I could query 2 tables at once. My test_dogs table and lifestyle table. It turns out, as RamRaider pointed out, I was missing quotations. The proper syntax was:

SQL

SELECT td.`breed_name`, td.`breed_size`, l.`Exercise` 
FROM test_Dogs as td 
LEFT OUTER JOIN Lifestyle as l 
ON td.`Breed_Name` = l.`Breed_Name` 
WHERE td.`breed_size` like "%Large%" 
AND l.`exercise` < 7 
ORDER BY l.`Exercise` ASC'

I changed the ORDER BY from Descending ( DESC ) to Ascending ( ASC ), as I decided that made more sense.

Then, I wanted to populate a table with the results of my query. Which looked as follows:

PHP

<table>
        <tr>
            <th>Breed Name</th>
            <th>Size</th>
            <th>Exercise Needs</th>
        </tr>

        <tr>
        <?php
        while ($rows = $join->fetch()){
            echo "<tr><td>" . $rows['breed_name'] . "</td><td>" . $rows['breed_size'] ."</td><td>" . $rows['Exercise'] . "</td></tr>";
        };
        ?>
    </table>

What I have discovered after getting the SQL to work is that whatever capitalization I use in the query, I need to use the same capitalization in the php, or it will not work. Notice in my SQL in the SELECT section, I wrote the first 2 columns in lower-case, but oddly decided to capitalize the last column. Then, when I used all lower-case in the PHP, it didn't work, and it took some time to figure out why. Also, I realized I did not need to specify which table each column came from. Or at least not in this instance.