PHP，在一个数组中获取所有特定类型的项。
PHP，在一个数组中获取所有特定类型的项。
提问于 2017-09-02 10:31:13
好的，我有两张桌子，分类和盘子。每个类别都有一个id。每一道菜都属于特定的类别。现在，我想要的是，所有的菜肴属于一个特定的类别分组。所以所有的汤都在汤阵列{甘蓝，价格，描述，}，{番茄，价格，描述，}。所有的鸡都在鸡的排列中等等。我现在要做的是：
$query = "Select id from Category";
$Id = $mysqli->query($query);
while($row = $Id->fetch_assoc()) {
$idlist[] = $row["id"];
}
foreach ($idlist as $id){
$query = "SELECT * from Dishes where Dishes.cat_id = '$id'";
$Listings = $mysqli->query($query);
while($row = $Listings->fetch_assoc()) {
$AllDishes[] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' => $row['Description'],
'cat_id' => $row['cat_id'],
'Price' => $row['Price'],
'ImagePath' => $row ['ImagePath']
);
}
}但这会导致所有的菜肴被组合在一起。我如何根据菜肴的类别来区分它们？这也是构造JSON数据的正确方式吗？
回答 3
Stack Overflow用户
回答已采纳
发布于 2017-09-02 10:41:28
1.使用JOIN进行单次查询(最佳解决方案)：-
$query = "Select Dishes.*,Category.id as cat_id from Category LEFT JOIN Dishes on Dishes.cat_id = Category.id";
$Listings = $mysqli->query($query);
$AllDishes = array();
while($row = $Listings->fetch_assoc()) {
$AllDishes[$row['cat_id']][] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' => $row['Description'],
'cat_id' => $row['cat_id'],
'Price' => $row['Price'],
'ImagePath' => $row ['ImagePath']
);
}
print_r($AllDishes);如果您想以您的编码方式执行，那么：-
2.根据类别的id区分它们：-
$query = "Select id from Category";//change column name
$Id = $mysqli->query($query);
while($row = $Id->fetch_assoc()) {
$idlist[] = $row['id'];
}
foreach ($idlist as $id){
$query = "SELECT * from Dishes where Dishes.cat_id = '$id'";
$Listings = $mysqli->query($query);
while($row = $Listings->fetch_assoc()) {
$AllDishes[$id][] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' => $row['Description'],
'cat_id' => $row['cat_id'],
'Price' => $row['Price'],
'ImagePath' => $row ['ImagePath']
);
}
}3.根据类别的name区分它们：-
$query = "Select id,category_name from Category";//change column name
$Id = $mysqli->query($query);
while($row = $Id->fetch_assoc()) {
$idlist[] = $row;
}
foreach ($idlist as $id){
$id = $id['id'];
$category_name = $id['category_name']; //change column-name
$query = "SELECT * from Dishes where Dishes.cat_id = '$id'";
$Listings = $mysqli->query($query);
while($row = $Listings->fetch_assoc()) {
$AllDishes[$category_name][] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' => $row['Description'],
'cat_id' => $row['cat_id'],
'Price' => $row['Price'],
'ImagePath' => $row ['ImagePath']
);
}
}注意：-请在第三个解决方案中更改列名(因为我不知道要存储类别名称的列名是什么)。
Stack Overflow用户
发布于 2017-09-02 10:37:21
我不确定，但我认为这里将是您期望的结果循环。
$AllDishes = array();
while($row = $Listings->fetch_assoc()) {
$AllDishes[$row['Name']][] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' => $row['Description'],
'cat_id' => $row['cat_id'],
'Price' => $row['Price'],
'ImagePath' => $row ['ImagePath']
);
}Stack Overflow用户
发布于 2017-09-02 11:09:56
您可以使用SQL联接而不是两个单独的SQL查询。
I hope the following may help you.
$query = "SELECT * FROM Dishes INNER JOIN Category
ON Dishes.id = Category.id ORDER BY Dishes.id";
$res = $mysqli->query($query);
while($row = $res->fetch_assoc())
{
$AllDishes[$$row['category_name']][] = array(
'id' => $row['id'],
'Name' => $row['Name'],
'Description' =>
$row['Description'],
'cat_id' =>
$row['cat_id'],
'Price' => $row['Price'],
'ImagePath' =>
$row['ImagePath']
);//Array Close
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/46012840
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