Question

Am试图通过利用三个数据库表的信息来创建json数组。

利用此处的信息和解决方案source
我意识到我必须使用外键进行引用，还必须使用联接查询。

三个表的创建过程如下，每个表都有其参考键

Create table categorys(id int primary key auto_increment, category_name varchar(30), buyer_userid int(30));

Create table product(id int primary key auto_increment, product_name varchar(30), buyer_userid int(30),category_id int,
foreign key (category_id) references categorys(id));


Create table sales(items_id int primary key auto_increment, item_name varchar(30),quantity varchar(30), buyer_userid int(30),
product_id int, foreign key (product_id) references product(id));

这是我期望的Json格式...

[
{"id":"1","cat_name":"Provision","catbuy_userid":"100",
"products_info":[{"productid":"2","product_name":"Malt","buyer_userid":"100",
"sales_Info":[{"items_id":"1","item_name":"malt","buyer_userid":"100","quantity":"72"}]
}]},
{"id":"2","cat_name":"Cosmetics","catbuy_userid":"200",
"products_info":[{"productid":"3","product_name":"soapy","buyer_userid":"200",
"sales_Info":[{"items_id":"2","item_name":"cream","buyer_userid":"200","quantity":"83"}]
}]}
]

这是到目前为止我的编码

<?php

$host = "localhost"; /* Host name */
$user = "root"; /* User */
$password = ""; /* Password */
$dbname = "mydb"; /* Database name */

$con = mysqli_connect($host, $user, $password,$dbname);
// Check connection
if (!$con) {
    die("Connection failed: " . mysqli_connect_error());
}

    $res_arr = array();
$query = '
    SELECT 
c.id,
c.category_name,
c.buyer_userid,
p.id,
p.product_name,
p.buyer_userid,
p.category_id,
s.items_id,
s.item_name,
s.quantity
    FROM `categorys` c
        INNER JOIN `product` p ON c.id=p.category_id
        INNER JOIN `sales` s ON p.id=s.product_id
    ORDER BY c.id';

    $result = mysqli_query($con,$query);
    while($row = mysqli_fetch_array($result)){    

        $res_arr[] = $row;
}
    echo json_encode($res_arr);
    exit;

?>

这是我的json输出

[{"0":"1","id":"1","1":"Provisions","category_name":"Provisions","2":"100","buyer_userid":"100","3":"4","4":"Malt","product_name":"Malt","5":"100","6":"1","category_id":"1","7":"4","items_id":"4","8":"malt","item_name":"malt","9":"93","quantity":"93"},
{"0":"2","id":"2","1":"Cosmetics","category_name":"Cosmetics","2":"200","buyer_userid":"200","3":"3","4":"soapy","product_name":"soapy","5":"200","6":"2","category_id":"2","7":"3","items_id":"3","8":"cream","item_name":"cream","9":"83","quantity":"83"}
]

请按照上面的json格式循环播放结果。

更新部分

错误警告示例，当我运行代码时，它们是如此之多。下面只是其中的几个

警告：在第51行的C：\ xampp \ htdocs \ api \ test.php中使用未定义的常量products_info-假定为'products_info'（这会在PHP的未来版本中引发错误）

警告：在第51行的C：\ xampp \ htdocs \ api \ test.php中使用未定义的常量product_name-假定为“ product_name”（这会在PHP的未来版本中引发错误）

Answer 1

当您说获得结果时，您是说要访问这些值吗？如果是这样，请将您的json数据分配给变量data。

然后： $json = json_decode($data);

foreach($json[0] as $values1) {
    var_dump($values1);
}

foreach($json[1] as $values2) {
    var_dump($values2);
}

例如，要从第一个对象获取产品名称，则需要执行$ values1-> product_name。对于其他属性，依此类推。

Answer 2

这可以通过将(array) $row转换为(object) $row来实现：

$res_arr = array();
while($row = $result->fetch_assoc()){    
    array_push($res_arr, (object) $row);
}
header("Content-type: application/json; charset=utf-8");
die(json_encode($res_arr));

我会使用fetch_assoc()来防止这些重复编号的字段，这些字段只会使有效负载膨胀……整个嵌套基本上是无用的，并且只会产生垃圾代码质量，而没有什么地方比嵌套的嵌套多了项（因此，它比在客户端进行访问所必需的更为复杂）。例如，嵌套会有意义。首先获取所有categories，然后为该products嵌套所有category，然后继续下一个category。

Answer 3

我对您的php代码进行了验证，因此您可以获得所需的输出。

根据您的要求，而不是从mysql查询返回的assignind行创建关联数组，然后转换为json


<?php

$host = "localhost"; /* Host name */
$user = "root"; /* User */
$password = ""; /* Password */
$dbname = "mydb"; /* Database name */

$con = mysqli_connect($host, $user, $password,$dbname);
// Check connection
if (!$con) {
    die("Connection failed: " . mysqli_connect_error());
}

    $res_arr = array();
$query = '
    SELECT 
c.id,
c.category_name,
c.buyer_userid,
p.id,
p.product_name,
p.buyer_userid,
p.category_id,
s.items_id,
s.item_name,
s.quantity
    FROM `categorys` c
        INNER JOIN `product` p ON c.id=p.category_id
        INNER JOIN `sales` s ON p.id=s.product_id
    ORDER BY c.id';

    $result = mysqli_query($con,$query);
    while($row = mysqli_fetch_array($result)){    

        $res['id'] = $row[0];
        $res['cat_name'] = $row[1];
        $res['catbuy_userid'] = $row[2];
        $res['products_info']['id'] = $row[3];

        $res['products_info']['productid'] = $row[4];
        $res['products_info']['product_name'] = $row[5];
        $res['products_info']['buyer_userid'] = $row[6];

        $res['products_info']['sales_info']['items_id'] = $row[7];
        $res['products_info']['sales_info']['items_name'] = $row[8];
        $res['products_info']['sales_info']['buyer_userid'] = $row[6];
        $res['products_info']['sales_info']['quantity'] = $row[9];

        $res_arr[] = $res;
  }

}
  echo json_encode($res_arr);

?>

如何使用php

3 个答案: