从MySQL查询结果创建和使用关联数组-PHP

我有一个sql查询，它通过数字键选择用餐选项。我想将每个数字键与餐名相关联。下面的查询只显示餐号（这是正确的），我无法匹配ID，因为我会得到不正确的结果

Array
(
    [meal_1] => 1
    [meal_2] => 2
    [meal_3] => 1
)
Array
(
    [meal_1] => 2
    [meal_2] => 1
    [meal_3] => 2
)
Array
(
    [meal_1] => 0
    [meal_2] => 3
    [meal_3] => 3
)
Array
(
    [meal_1] => 2
    [meal_2] => 4
    [meal_3] => 0
)

但我想这样输出

Array
(
    [meal_1] => Soup
    [meal_2] => Salmon
    [meal_3] => Torte
)
Array
(
    [meal_1] => Salad
    [meal_2] => Chicken
    [meal_3] => Tart
)
Array
(
    [meal_1] => No
    [meal_2] => Pasta
    [meal_3] => Brulee
)
Array
(
    [meal_1] => Salad
    [meal_2] => Burger
    [meal_3] => Pate
)

$result = mysql_query("SELECT Meal_table_1.meal_1, Meal_table_1.meal_2, Meal_table_1.meal_3 FROM Meal_table_1 ");
echo "<pre>";
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
}
echo "</pre>";
mysql_free_result($result);

//Meal_table_1

+----+--------+-------+-------+
| ID | meal_1 | meal_2| meal_3|
+----+--------+-------+-------+
|  1 | 1      | 2     | 1     |
|  2 | 2      | 1     | 2     |
|  3 | 0      | 3     | 3     |
|  4 | 2      | 4     | 0     |
+----+--------+-------+-------+

$meal_1 = array('0' => 'No','1' => 'Soup','2' => 'Salad','3' => 'Pate');

$meal_2 = array('0' => 'No','1' => 'Chicken','2' => 'Salmon','3' => 'Pasta','4' => 'Burger');

$meal_3 = array('0' => 'No','1' => 'Torte','2' => 'Tart','3' => 'Brulee',);

// Meal_table_2 (should this be in three separate tables?)

+----+--------+--------+--------+
| ID | starter| main   | dessert|
+----+--------+--------+--------+
|  1 | Soup   | Chicken| Torte  |
|  2 | Salad  | Salmon | Tart   |
|  3 | Pate   | Pasta  | Brulee |
|  4 |        | Burger |        |
+----+--------+--------+--------+

数组
(
[一餐]=>汤
[膳食2]=>鲑鱼
[膳食3]=>玉米饼
)
排列
(
[一餐]=>沙拉
[套餐2]=>鸡肉
[套餐3]=>蛋挞
)
排列
(
[膳食1]=>否
[膳食2]=>意大利面
[膳食3]=>布鲁利
)
排列
(
[一餐]=>沙拉
[套餐2]=>汉堡
[膳食3]=>肉馅
)
$result=mysql_query（“从餐表_1中选择餐表_1.1、餐表_1.2、餐表_1.3”）；
回声“；
while（$row=mysql\u fetch\u assoc（$result））{
打印（行）；
}
回声“；
mysql_free_result（$result）；
//饭桌1
+----+--------+-------+-------+
|ID |膳食1 |膳食2 |膳食3|
+----+--------+-------+-------+
|  1 | 1      | 2     | 1     |
|  2 | 2      | 1     | 2     |
|  3 | 0      | 3     | 3     |
|  4 | 2      | 4     | 0     |
+----+--------+-------+-------+
$MEIN_1=数组（'0'=>'No'、'1'=>'Soup'、'2'=>'Salad'、'3'=>'Pate'）；
$MEINE_2=数组（'0'=>'否'，'1'=>'鸡肉'，'2'=>'鲑鱼'，'3'=>'意大利面'，'4'=>'汉堡'）；
$MEINE_3=数组（'0'=>'No'、'1'=>'Torte'、'2'=>'Tart'、'3'=>'Brulee'，）；
//餐表2（是否应在三张单独的桌子中？）
+----+--------+--------+--------+
|ID |开胃菜|主菜|甜点|
+----+--------+--------+--------+
|1 |汤|鸡|玉米饼|
|2 |沙拉|鲑鱼|馅饼|
|3 |肉酱|意大利面|布鲁利|
|4 | |汉堡||
+----+--------+--------+--------+

您现在的查询是：

SELECT tstarter.starter, tmain.main, tdessert.dessert
FROM Meal_table_1 meal
LEFT OUTER JOIN Meal_table_2  tstarter
ON meal.ID = tstarter.id
LEFT OUTER JOIN Meal_table_2  tmain
ON meal.ID = tmain.id
LEFT OUTER JOIN Meal_table_2  tdessert
ON meal.ID = tdessert.id

您的查询可能是：

+----+---------+---------+
| ID | type    | main    |
+----+---------+---------+
|  1 | starter | Soup    |
|  1 | main    | Chicken | 
|  1 | dessert | Torte   |
|  2 | starter | Salad   |
|  2 | main    | Salmon  |
|  2 | dessert | Tart    |
|  3 | starter | Pate    |
|  3 | main    | Pasta   |
|  3 | dessert | Brulee  |
|  4 | main    | Burger  |
+----+---------+---------+

但是最好改变

餐表2的结构（并且给你的桌子起一个更好的名字！）

除非ID必须相同，否则最好为每一行指定自己的ID。也许您甚至不再需要该类型了？
餐食名称存储在哪里？如果您有另一个表将餐食ID映射到名称，请编写一个联接查询。一如既往：请停止使用
mysql\ucode>扩展名。它已弃用、不安全、未积极维护和过时。切换到一个现代、更安全、维护和全面更好的替代方案：
PDO
，或
mysqli.*
（
i
用于改进…）
+----+---------+---------+
| ID | type    | main    |
+----+---------+---------+
|  1 | starter | Soup    |
|  1 | main    | Chicken | 
|  1 | dessert | Torte   |
|  2 | starter | Salad   |
|  2 | main    | Salmon  |
|  2 | dessert | Tart    |
|  3 | starter | Pate    |
|  3 | main    | Pasta   |
|  3 | dessert | Brulee  |
|  4 | main    | Burger  |
+----+---------+---------+