Question

我有一个通用的SQL查询构建器，它为我提供了包含以下列的统一表：

Name, Name2, Amount

查询以GROUP BY Name2, Name结尾。

现在我想让它适合使用以下格式塔将其转换为JSON：

{dim2: Name2, data: [{label: Name, y: Amount},{label: Name, y: Amount},{label: Name, y: Amount},...]},
{dim2: Name2, data: [{label: Name, y: Amount},{label: Name, y: Amount},{label: Name, y: Amount},...]},
{dim2: Name2, data: [{label: Name, y: Amount},{label: Name, y: Amount},{label: Name, y: Amount},...]},
{dim2: Name2, data: [{label: Name, y: Amount},{label: Name, y: Amount},{label: Name, y: Amount},...]},
...

到目前为止，我做到了这一点：

while($result = mysqli_fetch_array($results))
    {
    if(!isset($return[$result['Name2']]))
        $return[$result['Name2']] = array();

    $return[$result['Name2']][] =
        array(
            'Label' => $result['Name'],
            'y'  =>  $result['Anzahl']
        );
}
$filestring = json_encode($return, JSON_NUMERIC_CHECK);

哪个不起作用。整天工作真的很烦人，我现在非常感谢一些帮助，提前谢谢!!!

更新

这是一个如何计算到目前为止的例子：

[{"dim2":"Consumer Entertainment","data":{"Label":"Entertainment","y":1}},
{"dim2":"Digital Life","data":{"Label":"Electronic Devices","y":1}},
{"dim2":"Digital Life","data":{"Label":"Entertainment","y":1}},
{"dim2":"Digital Life","data":{"Label":"Financial Services","y":2}},
{"dim2":"Digital Life","data":{"Label":"IT","y":2}},
{"dim2":"Digital Life","data":{"Label":"Media & Publishing","y":1}},
{"dim2":"Digital Life","data":{"Label":"Transportation & Mobility","y":1}},
{"dim2":"Healthy Life","data":{"Label":"Fashion & Accessories","y":1}},
...]

我想：

[{"dim2":"Consumer Entertainment","data":[{"Label":"Entertainment","y":1}]},
{"dim2":"Digital Life","data":[{"Label":"Electronic Devices","y":1},
{"Label":"Entertainment","y":1},
{"Label":"Financial Services","y":2},
{"Label":"IT","y":2},
{"Label":"Media & Publishing","y":1},
{"Label":"Transportation & Mobility","y":1}]},
{"dim2":"Healthy Life","data":[{"Label":"Fashion & Accessories","y":1}]},
...]

等等

Answer 1

现在它更容易理解：）

    $ar = array('Label'=>'Entertainment', 'y'=>1);
    $q[]=array('dim2'=>'Consumer Entertainment', 'data'=>array($ar));
    $ar = array('Label'=>'Electrodivices', 'y'=>1);
    $q[]=array('dim2'=>'Digital Life', 'data'=>array($ar));
    $ar = array('Label'=>'Entertainment', 'y'=>2);
    $q[]=array('dim2'=>'Digital Life', 'data'=>array($ar));
    echo json_encode($q);

它给出了

[{“dim2”：“Consumer Entertainment”，“data”：[{“Label”：“Entertainment”，“y”：1}]}，

{“dim2”：“Digital Life”，“data”：[{“Label”：“Electrodivices”，“y”：1}]}，

{“dim2”：“数字生活”，“数据”：[{“标签”：“娱乐”，“y”：2}]}]

所以，你应该尝试写一些像

这样的东西

$return[$result['Name2']][] =
    array(array(
        'Label' => $result['Name'],
        'y'  =>  $result['Anzahl']
    ));

它看起来多么愚蠢。可能它不明白它正在采取阵列？：P

Multi SQL-Group by与JSON兼容的表单

1 个答案: