如何在 PHP 中读取表格并将其呈现为 JSON？

Question

我有类别和子类别结构的数据库：

category:

subcategory:

我希望通过 PHP 获得所需格式的 JSON。我尝试了大部分时间，但没有成功。

{
"success":"true",
"category_1":{
"cat_id":"1",
"cat_name":"menu1",
"cat_img":"cat1.jpg",
"sub_category":[
{
"sub_id":"1",
"sub_name":"sub name"
},
{
"sub_id1":"2",
"sub_name1":"sub name"
},
{
"sub_id2":"3",
"sub_name2":"sub name"
},
{
"sub_id3":"4",
"sub_name3":"sub name"
}
]
},
"category_2":{
"cat_id":"2",
"cat_name":"menu2",
"cat_img":"cat2.jpg",
"sub_category":[
{
"sub_id":"1",
"sub_name":"sub name"
},
{
"sub_id1":"2",
"sub_name1":"sub name"
}
]
},
"category_3":{
"cat_id":"3",
"cat_name":"menu3",
"cat_img":"cat3.jpg",
"sub_category":[
{
"sub_id":"1",
"sub_name":"sub name"
},
{
"sub_id1":"2",
"sub_name1":"sub name"
},
{
"sub_id3":"",
"sub_name1":"sub name"
}
]
}
}

我试过这段代码，但是没有用。我更喜欢 PDO 解决方案。

$servername = "localhost";
$username = "res_user";
$password = "Res@123";
$dbname = "res_db";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

 $sql = "SELECT *, category.id AS catId FROM category INNER JOIN sub_category ON category.id = sub_category.category_id ORDER BY category.id, sub_category.category_id";
$res = mysqli_query($conn, $sql);
$categoryArray = array();
$oldCatId = 0;
while ($row = mysqli_fetch_assoc($res)) {
    if ($row['catId'] != $oldCatId) {
        $categoryArray[$row['catId']] = array(
            'success' => true,
            'category_' . $row['catId'] = array(
        'cat_id' => $row["catId"],
        'cat_name' => $row['cat_name'],
        'cat_img' => $row['cat_img'],
        'sub_category' => array(
            'sub_id' => $row['category_id'],
            'sub_name' => $row['sub_name']
        )
            )
        );
     $oldCatId = $row['catId'];
    } else {
        $categoryArray[$row['catId']]['category_' . $row['catId']]['sub_category'][] = array(
            'sub_id' => $row['category_id'],
            'sub_name' => $row['sub_name']
        );
    }
}


  echo  json_encode($categoryArray);

Answer 1

你需要这样的东西：从按类别 id 排序的类别中取出所有行，并与子类别一起加入。

然后你遍历你的结果。您检查一下，当前类别 id 是否等于先前的猫 id。如果没有，则构建主数据数组，并添加第一个子类别，如果没有，则添加下一个子类别。

$sql = "
    SELECT *, category.id AS catId
    FROM category
    INNER JOIN subcategory ON category.id = subcategory.category_id
    ORDER BY category.id, subcategory.category_id
";
$res = mysqli_query($conn, $sql);
$categoryArray = array();
$oldCatId = 0;
while ($row = mysqli_fetch_assoc($res)) {
    if ($row['catId'] != $oldCatId) {
        $categoryArray[$row['catId']] = array(
            'success' => true,
            'category_' . $row['catId'] = array(
        'cat_id' => $row["catId"],
        'cat_name' => $row['cat_name'],
        'cat_img' => $row['cat_img'],
        'sub_category' => array(
            'sub_id' => $row['category_id'],
            'sub_name' => $row['sub_name']
        )
            )
        );
        $oldCatId = $row['catId'];
    } else {
        $categoryArray[$row['catId']]['category_' . $row['catId']]['sub_category'][] = array(
            'sub_id' => $row['category_id'],
            'sub_name' => $row['sub_name']
        );
    }
}