Question

我需要获得如下所示的json格式：

<?php
include("db.php");
date_default_timezone_set('Asia/Kolkata');
$thisyear =  date("Y");
$months = array();
for($i=0;$i<=12;$i++)
{
   array_push($months, 0);
    unset($months[0]);
}
$stack = array(
    1 => 'JAN',
    2 => 'FEB',
    3 => 'MAR',
    4 => 'APR',
    5 => 'MAY',
    6 => 'JUN',
    7 => 'JUL',
    8 => 'AUG',
    9 => 'SEP',
    10 => 'OCT',
    11 => 'NOV',
    12 => 'DEC'
);
$year[0] = 0;

$stmt = $conn1->prepare("SELECT * FROM salessummary WHERE dateoforder LIKE :key");
$stmt->execute(array('key' => "%{$thisyear}%"));

foreach ($stmt as $row) {
    $month = date('m', strtotime($row['dateoforder']));
    $months[intval($month)]++;  
}
$high = max($months) + 10;
$datayearly = array(
    'labels' => array_values($stack),
    'series' => array(array(
        'name' => 'series-1',
        'data' => array_values($months)
    ))
);
$temp = json_encode($datayearly);
echo $temp;
?>

..... ... 目前我使用下面给出的代码获得的格式如下。有什么办法可以将这个JSON输出转换为上面的那个。或者我怎么能重写下面的PHP代码。因为我是php的新手，所以对我来说似乎很困惑。感谢您提前提供的宝贵帮助。

{  
   "labels":[  
      "JAN",
      "FEB",
      "MAR",
      "APR",
      "MAY",
      "JUN",
      "JUL",
      "AUG",
      "SEP",
      "OCT",
      "NOV",
      "DEC"
   ],
   "data":[  
            0,
            0,
            0,
            0,
            0,
            0,
            0,
            0,
            1,
            2,
            1,
            0
  ]

当前输出：

{{1}}

}

Answer 1

尝试以下方法：

$datayearly = [];
for($i=1;$i <= count($months);$i++)
{
     $datayearly[] = [
        'month' => $stack[$i],
        'data' => $months[$i]   
    ];
}

这将提供输出作为month和data的对象。这只是使用for循环来创建一个新数组。输出将如下：

[{
    "month": "JAN",
    "data": 0
}, {
    "month": "FEB",
    "data": 0
}, {
    "month": "MAR",
    "data": 0
}, {
    "month": "APR",
    "data": 0
}, 
...........

Answer 2

我会用这样的东西来生成你的阵列：

$datayearly=[];
foreach(array_values($stack) as $i=> $m){
  $datayearly[]['month']=$m;
  $datayearly[]['data']=$array_values($months)[$i];
}

将一种json格式转换为另一种格式

2 个答案: