Php 如何从多维数组创建json数据（每年每月）

我需要为highchart导出一组json数据。但我有一个循环的问题。 我有两套阵列：

年份数组（20152014201312012）-表单MySQL

月数组（01,02,03,04,05,06,07,08,09,10,11,12）

所以我把它做成数组-多维的。用这个代码

$sql_arrYr=mysqli_query($con,"select year(bk_date1) as arrYr from booking_db group by year(bk_date1) order by arrYr asc");
while($rec_arrYr=mysqli_fetch_array($sql_arrYr)){
    $arrYr[$rec_arrYr['arrYr']]=array("01","02","03","04","05","06","07","08","09","10","11","12");
}

foreach($arrYr as $key=>$val){
    $rows=array();
    $rows[type]='spline';
    $rows[name]=$key;

    foreach($val as $mon){
        $sql_nBk=mysqli_query($con,"select count(*) as nBkr from booking_db where year(bk_date1)='$key' and month(bk_date1)='$mon'");
        $rec_nBk=mysqli_fetch_array($sql_nBk);
            $rows[data][]=$rec_nBk['nBkr']; 
    }
}

echo json_encode($rows);

然后我有：

Array ( [2015] => Array ( [0] => 01 [1] => 02 [2] => 03 [3] => 04 [4] => 05 [5] => 06 [6] => 07 [7] => 08 [8] => 09 [9] => 10 [10] => 11 [11] => 12 )...);

这看起来不错，因为我在一组数组中有我需要的所有数据-年值和每月值

因此，我继续使用这段代码从mySQL获取数据

$sql_arrYr=mysqli_query($con,"select year(bk_date1) as arrYr from booking_db group by year(bk_date1) order by arrYr asc");
while($rec_arrYr=mysqli_fetch_array($sql_arrYr)){
    $arrYr[$rec_arrYr['arrYr']]=array("01","02","03","04","05","06","07","08","09","10","11","12");
}

foreach($arrYr as $key=>$val){
    $rows=array();
    $rows[type]='spline';
    $rows[name]=$key;

    foreach($val as $mon){
        $sql_nBk=mysqli_query($con,"select count(*) as nBkr from booking_db where year(bk_date1)='$key' and month(bk_date1)='$mon'");
        $rec_nBk=mysqli_fetch_array($sql_nBk);
            $rows[data][]=$rec_nBk['nBkr']; 
    }
}

echo json_encode($rows);

问题发生在这里。这不是循环。我只有一句话

{"type":"spline","name":2015,"data":["9","8","0","0","0","0","0","0","0","0","0","0"]}

我希望：

{"type":"spline","name":2015,"data":["9","8","0","0","0","0","0","0","0","0","0","0"]}
{"type":"spline","name":2014,"data":["9","8","0","0","0","0","0","0","0","0","0","0"]}
{"type":"spline","name":2013,"data":["9","8","0","0","0","0","0","0","0","0","0","0"]}
{"type":"spline","name":2012,"data":["9","8","0","0","0","0","0","0","0","0","0","0"]}

---

请给我一个不是循环的原因。甚至我都把它们放在了一个右循环中。

每次通过外循环，您都将

$rows

设置为空数组。此外，您不能有重复的密钥。第一个

类型

会被下一个类型（也就是

名称

和

数据

）覆盖，因此必须使其多维化。您可以使用

$key

变量来简化此操作，或者实现计数器

$i

并使用

$i++

或其他方法

$i = 0;
foreach($arrYr as $key=>$val){
    $rows[$i]['type']='spline';
    $rows[$i]['name']=$key;

    foreach($val as $mon){
        $sql_nBk=mysqli_query($con,"select count(*) as nBkr from booking_db where year(bk_date1)='$key' and month(bk_date1)='$mon'");
        $rec_nBk=mysqli_fetch_array($sql_nBk);
            $rows[$i]['data'][]=$rec_nBk['nBkr']; 
    }
    $i++;
}
echo json_encode($rows);

---

在循环的每次迭代中，您都要将

$rows

重置为空数组…我不明白。请具体一点。试着按照你自己的代码逻辑，特别是循环的最后一次迭代。请给我一个线索。我应该在第二个循环中使用

foreach

，还是其他什么。代码生成了错误的格式：

“2015”：{“类型”：“样条线”、“名称”：2015，“数据”：[“9”、“8”、“0”、“0”、“0”、“0”、“0”、“0”、“0”、“0”、“0”]}

已编辑。尝试一个基于0的计数器。完美！尸体。格式正确，代码现在非常出色。非常感谢你。