[英]Join three table codeigniter 4
Hi everyone i'am new on codeigniter 4 and currently working on a small project in the project i'am trying to join three tables and display there data in single table.大家好,我是 codeigniter 4 的新手,目前正在项目中的一个小项目上工作,我正在尝试加入三个表并在单个表中显示数据。
Table_1
id unique_id Name
1 1111 Sam
2 2222 Charlote
Table_2
id unique_id Name
1 1212 Jhon
2 5151 Alex
Table_3
id author title
1 1111 Book_1
2 5151 Book_2
3 1111 Book_3
Result
------------------------
| No | Author | Title |
------------------------
| 1 | Sam | Book_1 |
| 2 | Alex | Book_2 |
| 3 | Sam | Book_3 |
------------------------
I've tried to join with, but not working.我试过加入,但没有工作。
$this->join('Table_1', 'Table_1.unique_id = Table_3.author ');
$this->join('Table_2', 'Table_2.unique_id = Table_3.author ');
$this->select('Table_1.Name');
$this->select('Table_2.Name');
$this->select('Table_3.*');
$this->orderBy('Table_3.id');
return $this->findAll();
Is there another way to Join them?还有其他方法可以加入他们吗? Thank you
谢谢
What you currently have won't work because your Table_1 and Table_2 are effectively the same table.您当前所拥有的将无法使用,因为您的 Table_1 和 Table_2 实际上是同一张表。
Taking your Attempt and correcting it to use LEFT JOIN尝试并更正它以使用 LEFT JOIN
public function example_1() {
$this->join('Table_1', 'Table_1.unique_id = Table_3.author', 'LEFT');
$this->join('Table_2', 'Table_2.unique_id = Table_3.author', 'LEFT');
$this->select('Table_1.Name');
$this->select('Table_2.Name');
$this->select('Table_3.*');
$this->orderBy('Table_3.id');
$result = $this->findAll();
echo $this->db->getLastQuery();
return $result;
}
You would get...你会得到...
SELECT `Table_1`.`Name`, `Table_2`.`Name`, `Table_3`.*
FROM `Table_3`
LEFT JOIN `Table_1` ON `Table_1`.`unique_id` = `Table_3`.`author`
LEFT JOIN `Table_2` ON `Table_2`.`unique_id` = `Table_3`.`author`
ORDER BY `Table_3`.`id`
array(3) {
[0]=>
array(4) {
["Name"]=>
NULL
["id"]=>
string(1) "1"
["author"]=>
string(4) "1111"
["title"]=>
string(6) "Book_1"
}
[1]=>
array(4) {
["Name"]=>
string(4) "Alex"
["id"]=>
string(1) "2"
["author"]=>
string(4) "5151"
["title"]=>
string(6) "Book_2"
}
[2]=>
array(4) {
["Name"]=>
NULL
["id"]=>
string(1) "3"
["author"]=>
string(4) "1111"
["title"]=>
string(6) "Book_3"
}
}
Note that you have, two occurrences of name in your query.请注意,您的查询中有两次出现的名称。 So which one will win?
那么哪一个会赢呢? It appears that Table_2.name is only used and any reference to Table_1.name is NULL as it's not used.
似乎只使用了 Table_2.name 并且对 Table_1.name 的任何引用都是 NULL 因为它没有被使用。
You could give them different names using aliases but then you would have something like name_1 and name_2 so which one is it?您可以使用别名给它们起不同的名称,但是您会得到类似 name_1 和 name_2 的名称,那么它是哪一个? This is due to the duplication in your Table_1 and Table_2 and you asking for both.
这是由于您的 Table_1 和 Table_2 中的重复,并且您要求两者。
The Better way So in this case you would need to perform an UNION on Table_1 and Table_2.更好的方法所以在这种情况下,您需要对 Table_1 和 Table_2 执行 UNION。
I don't think that there is a UNION command in the CI query builder.我认为 CI 查询生成器中没有 UNION 命令。
Using mysql, it would be...使用 mysql,它将是...
public function get_book_and_author() {
$sql = "SELECT Table_3.id, T12.name as author, Table_3.title
FROM (
SELECT Table_1.* FROM Table_1
UNION
SELECT Table_2.* FROM Table_2
) as T12
LEFT JOIN Table_3 ON T12.unique_id = Table_3.author
WHERE Table_3.author IS NOT NULL
";
$result = $this->db->query($sql);
return $result->getResultArray();
}
So in this example, we have specified the 3 fields you require in the Select.所以在这个例子中,我们在 Select 中指定了您需要的 3 个字段。 Note the T12.name is renamed author.
注意 T12.name 被重命名为作者。 (See the output below)
(见下面的 output)
Then an UNION
has to be performed on Table_1 and Table_2 and the result is named (aliased) as T12 (shorthand for Table_1 and Table_2) as the result requires a new name.然后必须对 Table_1 和 Table_2 执行
UNION
,并将结果命名(别名)为 T12(Table_1 和 Table_2 的简写),因为结果需要一个新名称。
Then a LEFT JOIN
is performed against Table_3, which will give all combinations where there will be NULLS, so the WHERE statement filters those out using "IS NOT NULL" on Table_3.author.然后对 Table_3 执行
LEFT JOIN
,这将给出所有存在 NULL 的组合,因此 WHERE 语句使用 Table_3.author 上的“IS NOT NULL”过滤掉这些组合。
I left out the ORDER BY
as it's not really needed and you can add that back in if you wish to.我省略了
ORDER BY
,因为它并不是真正需要的,如果您愿意,可以将其添加回来。
A var_dump() of the result gives...结果的 var_dump() 给出了...
array(3) {
[0]=>
array(3) {
["id"]=>
string(1) "1"
["author"]=>
string(3) "Sam"
["title"]=>
string(6) "Book_1"
}
[1]=>
array(3) {
["id"]=>
string(1) "2"
["author"]=>
string(4) "Alex"
["title"]=>
string(6) "Book_2"
}
[2]=>
array(3) {
["id"]=>
string(1) "3"
["author"]=>
string(3) "Sam"
["title"]=>
string(6) "Book_3"
}
}
So that will give you the id,author and title for each matching row as you have requested using your example Tables.因此,这将为您提供每个匹配行的 ID、作者和标题,正如您使用示例表所请求的那样。
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