Question

上下文
该数据库是关于手机的。我在TABLE sav（＆＃34;售后服务＆＃34;的记录）中有一个搜索引擎，其中包含 IMEI的字段phone （独特的id）手机在某些情况下，手机会返回我们的库存（从TABLE sales删除并插入TABLE stock），另一部手机将退回给客户。

问题：
在＆＃34;售后服务＆＃34;的搜索引擎中，我使用LEFT JOIN sales，但如果手机已经回到stock，则无法找到 IMEI 。

总结：

如果找到记录，我需要加入一个表，否则，加入另一个表。

我的实际查询：
我缩短了它（[...]）以使其可读：

SELECT sav.id, sav.phone, [...] FROM sav 
LEFT JOIN sales ON (sav.phone = sales.imei)
LEFT JOIN stock_model ON (sales.model = stock_model.id) [...]
WHERE (sales.imei LIKE :search_1 OR [...])
OR (sales.imei LIKE :search_2 OR [...])
OR [...]

我需要改变第二行，但我不明白怎么做。我之前阅读的问题似乎回答了具体案例。

编辑：表格结构＆amp;虚拟数据
我不会发布不相关的字段（太多）
TABLE sav

| id | phone     | issue   | replacement_imei | ...
----------------------------------------------------
| 35 | 123456789 | battery | NULL             | ...     <-- phone n°1
| 54 | 987654321 | screen  | 147258369        | ...     <-- phone n°2 replaced by phone n°3

TABLE stock

| id  | imei      | model | color | memory | ...
-------------------------------------------------
| 68  | 369258147 | 2     | 4     | 1      | ...
| 92  | 987654321 | 5     | 8     | 3      | ...     <-- phone n°2 replaced by phone n°3

TABLE sales

| id  | imei      | model | color | memory | ...
-------------------------------------------------
| 68  | 123456789 | 2     | 4     | 1      | ...     <-- phone n°1
| 103 | 147258369 | 5     | 8     | 3      | ...     <-- phone n°3 replaced phone n°2

编辑：完整的搜索引擎[PHP] ：
如果整个功能有帮助：

$keywords = explode(" ", $keywords);
$query = "SELECT sav.id, sav.phone, sav.date, sav.nom, sav.issue, sav.replacement_imei, sav.commentaire, sav.attente, sav.pieces, sav.frais, sav.date_envoi, sav.tracking, sav.termine FROM sav LEFT JOIN sales ON (sav.phone = sales.imei) LEFT JOIN stock_memory ON (sales.memory = stock_memory.id) LEFT JOIN stock_color ON (sales.color = stock_color.id) LEFT JOIN stock_model ON (sales.model = stock_model.id) LEFT JOIN stock_grade ON (sales.grade = stock_grade.id) WHERE ";

// Makes a long query returning each row where at least 1 keyword matches 1 column
$query_array = array();
    for ($i = 0; $i < count($keywords); $i += 1) {
        $query .= " ( sales.imei LIKE :search_" . $i;
        $query .= " OR ( sales.snowden = 1 AND 'snowden'= :search_" . $i . " )";
        $query .= " OR stock_model.name LIKE :search_" . $i;
        $query .= " OR stock_color.name LIKE :search_" . $i;
        $query .= " OR stock_memory.amount LIKE :search_" . $i;
        $query .= " OR stock_grade.name LIKE :search_" . $i;
        $query .= " OR sav.id LIKE :search_" . $i;
        $query .= " OR DATE_FORMAT(sav.date, '%d/%m/%Y') LIKE :search_" . $i;
        $query .= " OR sav.nom LIKE :search_" . $i;
        $query .= " OR sav.issue LIKE :search_" . $i;
        $query .= " OR sav.replacement_imei LIKE :search_" . $i;
        $query .= " OR sav.commentaire LIKE :search_" . $i;
        $query .= " OR sav.attente LIKE :search_" . $i;
        $query .= " OR sav.pieces LIKE :search_" . $i;
        $query .= " OR sav.frais LIKE :search_" . $i;
        $query .= " OR DATE_FORMAT(sav.date_envoi, '%d/%m/%Y') LIKE :search_" . $i;
        $query .= " OR sav.tracking LIKE :search_" . $i;
        $query .= " OR ( sav.termine = 1 AND 'termine'= :search_" . $i . " )";
        // $query .= "LIKE :search_" . $i;
        if ($i != (count($keywords) - 1)) {
            $query .= " ) AND ";
        } else {
            $query .= " ) ";
        }
        if (strtolower($keywords[$i]) == 'snowden' || strtolower($keywords[$i]) == 'termine') {
            $query_array['search_' . $i] = $keywords[$i];
        } else {
            $query_array['search_' . $i] = "%" . $keywords[$i] . "%";
        }
    }
    $query .= "ORDER BY sav.id DESC LIMIT $offset, $limit";
    $results = $this->bdd->prepare($query);
    $results->execute($query_array);
    return $results;

Answer 1

首先，我获得全外连接的所有数据，然后是Sales.Model =＆gt;从Stock_Model中选择所有ID。

SELECT sav.id, sav.phone, [...] FROM sav 
FULL OUTER JOIN sales ON (sav.phone = sales.imei)
WHERE sav.phone = sales.imei or sales.model in (select Id from stock_model)

MySQL加入一个表或另一个表

1 个答案: