我有一个问题，我只是一个编程新手，逻辑问题有点低。我想从MySQL查询结果生成一个数组值，它必须嵌套。
我正在处理的实际代码：
<?php
$mostSold_arr = array();
$subData_array = array();
$day_arr = array();
$data = array();
// Connect to MySQL
$link = new mysqli( 'localhost', 'root', '', 'db' );
if ( $link->connect_errno ) {
die( "Failed to connect to MySQL: (" . $link->connect_errno . ") " . $link->connect_error );
}
$fetch_data = mysqli_query($link, "SELECT products.id as prodID, title, SUM(total_price) as value, orders.order_date as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id GROUP BY order_details.product_id ORDER BY value DESC LIMIT 2 ") or die(mysqli_error($link));
while ($row_mainData = mysqli_fetch_assoc($fetch_data)) {
$mostSold_arr['category'] = $row_mainData['title'];
$mostSold_arr['value'] = $row_mainData['value'];
$prod_id['prodID'] = $row_mainData['prodID'];
$mostSold_arr['url'] = "#";
$mostSold_arr['description'] = "click to drill-down";
$mostSold_arr['data'] = array();
$fetch_subData = mysqli_query($link, "SELECT title, SUM(total_price) as value, MONTHNAME(orders.order_date) as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id WHERE order_details.product_id=" . $prod_id['prodID'] ." GROUP BY title, date ORDER BY order_date LIMIT 3 ") or die(mysqli_error($link));
while ($row_subData = mysqli_fetch_assoc($fetch_subData)) {
$subData_arr['category']=$row_subData['date'];
$subData_arr['value']=$row_subData['value'];
$day_arr['data'] = array();
array_push($mostSold_arr['data'],$subData_arr);
$fetch_subDataDay = mysqli_query($link, "SELECT title, SUM(total_price) as value, DATE_FORMAT(orders.order_date, '%M-%d') as date FROM products LEFT JOIN order_details ON order_details.product_id=products.id LEFT JOIN orders ON orders.id=order_details.order_id WHERE order_details.product_id=" . $row_mainData['prodID'] ." AND DATE_FORMAT(orders.order_date, '%M-%d') LIKE '%March%' GROUP BY date ORDER BY order_date LIMIT 3 ") or die(mysqli_error($link));
while ($row_subDataDay = mysqli_fetch_assoc($fetch_subDataDay)) {
$subDataDay_arr['category']=$row_subDataDay['date'];
$subDataDay_arr['value']=$row_subDataDay['value'];
//array_push($day_arr['data'],$subDataDay_arr);
if($subData_arr['category'] == 'March'){
array_push($mostSold_arr['data'],$subDataDay_arr);
}
//array_push($day_arr['data'],$subDataDay_arr);
//var_dump($subData_arr);
}
}
array_push($data, $mostSold_arr);
}
$jsonData = json_encode($data, JSON_PRETTY_PRINT);
$fp = fopen('dataColMultiNested.json', 'w');
fwrite($fp, $jsonData);
fclose($fp);
die('<pre>' . print_r($jsonData, 1));
mysqli_close($link);
?>
以上代码的实际输出：
[
{
"category": "ACER E5-473-53C0",
"value": "2455200.00",
"url": "#",
"description": "click to drill-down",
"data": [
{
"category": "March",
"value": "809100.00"
},
{
"category": "March-02",
"value": "27900.00"
},
{
"category": "March-03",
"value": "27900.00"
},
{
"category": "March-04",
"value": "55800.00"
},
{
"category": "April",
"value": "781200.00"
},
{
"category": "May",
"value": "418500.00"
}
]
},
{
"category": "ACER ASPIRE E5-551G-F57K LAPTOP",
"value": "2339220.00",
"url": "#",
"description": "click to drill-down",
"data": [
{
"category": "March",
"value": "449850.00"
},
{
"category": "March-02",
"value": "29990.00"
},
{
"category": "March-03",
"value": "59980.00"
},
{
"category": "March-07",
"value": "59980.00"
},
{
"category": "April",
"value": "689770.00"
},
{
"category": "May",
"value": "779740.00"
}
]
}
预期输出（应该是这样）
[
{
"category": "ACER E5-473-53C0",
"value": "2455200.00",
"url": "#",
"description": "click to drill-down",
"data": [
{
"category": "March",
"value": "809100.00",
"data": [
{
"category": "March-02",
"value": "27900.00"
},
{
"category": "March-03",
"value": "27900.00"
},
{
"category": "March-04",
"value": "55800.00"
}
]
},
{
"category": "April",
"value": "781200.00"
},
{
"category": "May",
"value": "418500.00"
}
]
},
{
"category": "ACER ASPIRE E5-551G-F57K LAPTOP",
"value": "2339220.00",
"url": "#",
"description": "click to drill-down",
"data": [
{
"category": "March",
"value": "449850.00",
"data": [
{
"category": "March-02",
"value": "29990.00"
},
{
"category": "March-03",
"value": "59980.00"
},
{
"category": "March-07",
"value": "59980.00"
}
]
},
{
"category": "April",
"value": "689770.00"
},
{
"category": "May",
"value": "779740.00"
}
]
}
]
我正在为此工作约2晚或更长时间，但我可以找到解决问题的正确方法。 我希望你们能帮助我。感谢。
答案 0 :(得分：0)
这是一个完全退款的答案，基于您的问题及其标签中缺少的信息：您想要的结果是JSON格式（但我必须承认您的代表符合它）。
在PHP中，您必须首先创建一个如下所示的等效数组：
[
0 => [
"year" => "2016",
"data" => [
0 => [
"month" => "Dec",
"data" => [
0 => [
"day": "Dec 04",
],
],
],
],
],
]
它可以使用离初始代码不远的东西构建：
$fetch_data1 = "Select * From tbl1";
$year = [];
while($row_year = mysqli_fetch_assoc($fetch_data1)){
$year['year'] = $row_year['year'];
$fetch_data2 = "Select * From tbl2";
/* my query needs to loop inside because
it will contain condition depends on the retrieved data row */
$data = [];
while($row_month = mysqli_fetch_assoc($fetch_data2)){
$data['month'] = $row_month['month'];
$fetch_data3 = "Select * From tbl3";
/* my query needs to loop inside because
it will contain condition depends on the retrieved data row */
$day = [];
while($row_day = mysqli_fetch_assoc($fetch_data3)){
$day['day'] = $row_month['month'] . ' ' . $row_day['day'];
}
$data['data'] = $day;
}
$year[] = $data;
}