从 foreach 创建动态 JSON

Question

我使用jquery flot charts 来表示我的数据。这是我制作的示例 JSFiddle，它显示了图表所需的 JSONS 的外观。

数据源来自 MySql 存储过程，其输出示例如下：

我需要在图表中表示 count 值堆叠在 y 轴上的不同 innumber 上，name 值在 x 轴上，在另一个图表中，outnumber 的值. （在堆积条形图中）。

-数据系列应匹配，因此特定标签应针对客户显示。

这是我目前拥有的 PHP：

$query = $this->db->query("call GetAllCustomersV2($id, $year, $month, $day)");
$customers = $query->result_array();

foreach ($customers as $customer) {

  if($customer['innumber'] != null){

      $chartInbound['name'] = $customer['name'];
      $chartInbound['label'] = $customer['innumber'];
      $chartInbound['count'] = $customer['count'];
      $chartInbound['customerid'] = $customer['id'];

      array_push($out['chartInbound'], $chartInbound);
   }

   if($customer['outnumber'] != null){

      $chartOutbound['name'] = $customer['name'];
      $chartOutbound['label'] = $customer['outnumber'];
      $chartOutbound['count'] = $customer['count'];
      $chartOutbound['customerid'] = $customer['id'];

      array_push($out['chartOutbound'], $chartOutbound);
   }
}

print_r($out['chartInbound']); 的输出为：

Array
(
[0] => Array
    (
        [name] => 1st Online Solutions
        [label] => 01-02
        [count] => 577
        [customerid] => 129
    )

[1] => Array
    (
        [name] => Bookngo
        [label] => 01-02
        [count] => 2
        [customerid] => 95
    )

[2] => Array
    (
        [name] => Boutixury
        [label] => 07
        [count] => 1
        [customerid] => 14
    )

[3] => Array
    (
        [name] => Cruise Village
        [label] => 01-02
        [count] => 16
        [customerid] => 25
    )

[4] => Array
    (
        [name] => Cruise Village
        [label] => 00
        [count] => 1
        [customerid] => 25
    )

[5] => Array
    (
        [customer] => Cruise Village
        [label] => 07
        [countInbound] => 16
        [minsInbound] => 125
        [customerid] => 25
    )
  ...................
)

print_r(json_encode($out['chartInbound'])); 的输出是：

[
{
    "name": "1st Online Soultions"
    "label": "01-02",
    "count": "577",
    "customerid": "129"
},
{
    "name": "Bookngo"
    "label": "01-020",
    "count": "2",
    "customerid": "129"
},
{
    "name": "Boutixury"
    "label": "07",
    "count": "1",
    "customerid": "14"
},
{
    "name": "Cruise Village"
    "label": "07",
    "count": "16",
    "customerid": "25"
},
 .................
]

但这不是很有帮助。

问：如何根据查询数据创建上述 jsfiddle 中显示的动态 JSON？

Answer 1

使用循环遍历您的数据并构建 newData 和 newTicks 数组以供 flot 使用：

var newData = [];
var newLabels = []; // only used to get index since newData has objects in it
var newTicks = [];

for (var i = 0; i < dataFromServer.length; i++) {
    var datapoint = dataFromServer[i];

    var tick = newTicks.indexOf(datapoint.name);
    if (tick == -1) {
        tick = newTicks.length;
        newTicks.push(datapoint.name);
    }

    var index = newLabels.indexOf(datapoint.label);
    if (index == -1) {
        index = newLabels.length;
        newLabels.push(datapoint.label);

        newDataPoint = {
            label: datapoint.label,
            data: []
        };
        newDataPoint.data[tick] = [tick, datapoint.count];
        newData.push(newDataPoint);
    } else {
        newData[index].data[tick] = [tick, datapoint.count];
    }
}
for (var i = 0; i < newTicks.length; i++) {
    newTicks[i] = [i, newTicks[i]];
}
newLabels = null;

我还必须更改您的工具提示生成，因为您的代码仅在所有数据序列都完成并排序后才有效。现在也更简单了。

complete fiddle

Answer 2

只是一个想法，我想您在存储过程中使用了 group by。如果您可以修改它并添加 WITH ROLLUP，那么数据库将为您计算计数...请参阅https://dev.mysql.com/doc/refman/5.0/en/group-by-modifiers.html 或搜索 SO 以获取建议

Answer 3

您可以在客户端执行此操作（尽管理想情况下应该在服务器端执行），方法是：

var table = [
    {name: 'a', label: 'l1', count: '15', customerid: '1'},
    {name: 'a', label: 'l2', count: '1', customerid: '1'},
    {name: 'a', label: 'l3', count: '7', customerid: '1'},
    {name: 'b', label: 'l1', count: '3', customerid: '2'},
    {name: 'b', label: 'l2', count: '9', customerid: '2'},
    {name: 'b', label: 'l3', count: '2', customerid: '2'},
    {name: 'c', label: 'l1', count: '1', customerid: '3'},
    {name: 'c', label: 'l2', count: '7', customerid: '3'},
    {name: 'a', label: 'l3', count: '5', customerid: '4'},
    {name: 'a', label: 'l2', count: '6', customerid: '4'}
];

var customers = {};
var labels = {};

var i;
for (i = 0; i < table.length; ++i) {
    customers[table[i].customerid] = table[i].name;
    labels[table[i].label] = labels[table[i].label] || [];
    labels[table[i].label].push([+table[i].customerid, +table[i].count]);
}

var chartData = [];
var chartTicks = [];

for (customer in customers) {
    if (customers.hasOwnProperty(customer)) {
        chartTicks.push([+customer, customers[customer]]);
    }
}
for (label in labels) {
    if (labels.hasOwnProperty(label)) {
        chartData.push({label: label, data: labels[label]});
    }
}

它考虑了具有相同名称的不同客户（不同的 customerid）（尽管 Flot 并不能很好地处理这一点），以及缺少某些标签数据的客户。将此逻辑转移到 PHP 中并在服务器端执行应该不会太难。

编辑： 好的，我没有注意到当有 labelID“间隙”时它的行为很奇怪。这是修改后的代码：

var table = [
    {name: 'a', label: 'l1', count: '15', customerid: '1'},
    {name: 'a', label: 'l2', count: '1', customerid: '1'},
    {name: 'a', label: 'l3', count: '7', customerid: '1'},
    {name: 'b', label: 'l1', count: '3', customerid: '2'},
    {name: 'b', label: 'l2', count: '9', customerid: '2'},
    {name: 'b', label: 'l3', count: '2', customerid: '2'},
    {name: 'c', label: 'l1', count: '1', customerid: '3'},
    {name: 'c', label: 'l2', count: '7', customerid: '3'},
    {name: 'a', label: 'l3', count: '5', customerid: '7'},
    {name: 'a', label: 'l2', count: '6', customerid: '7'}
];

var customers = {};
var labels = {};

var chartData = [];
var chartTicks = [];

var i;
var customerNo = 0;
for (i = 0; i < table.length; ++i) {
    if(!customers.hasOwnProperty(table[i].customerid)) {
        customers[table[i].customerid] = table[i].name;
        chartTicks.push([customerNo, table[i].name]);
        customerNo++;
    }
    labels[table[i].label] = labels[table[i].label] || [];
    labels[table[i].label].push([customerNo - 1, +table[i].count]);
}

for (label in labels) {
    if (labels.hasOwnProperty(label)) {
        chartData.push({label: label, data: labels[label]});
    }
}

标签 ID 按照它们在来自服务器的表中出现的顺序给出。 （虽然它仍然区分了两个同名但customerID不同的客户）

Answer 4

您将不得不自己改造这些结构。您可以在服务器端或客户端执行此操作。在任何一种情况下，运行结果并构建您想要的结构。 小心尝试在 json 中编码 php 关联数组，并注意 NUMERIC_CHECK 的行为。

Answer 5

您在chartTicks[i] 中的数据点似乎需要与chartData[i].data 中的刻度顺序相匹配。 确保这种匹配的一种方法是在 sql 中按名称对数据进行排序，并在 php 中按客户第一和标签第二来堆叠结果。

$query = $this->db->query("call GetAllCustomersV2($id, $year, $month, $day)");
$customers = $query->result_array(); //should be sorted by name
$results = array();

foreach ($customers as $customer) {
    $i = is_array($results[$customer['name']][$customer['innumber']]) 
        ? count($results[$customer['name']][$customer['innumber']])
        : 0;

    //stack data points by customer name first and label second
    $results[$customer['name']][$customer['innumber']][] = array($i,$customer['count']);
}

$chartData = array();
$chartTicks = array();
$i=0;

foreach($results as $name => $labels) {
    $chartTicks[] = array($i++,$name);
    foreach($labels as $label => $data) {
        $chartData[] = array(
            'label' => $label,
            'data' => $data,
        );
    }
}

print json_encode($chartData);
print json_encode($chartTicks);

Answer 6

这是将当前 JSON 数据结构转换为所需输出的简洁方法：

var reduced;
var chartData = Object.keys(reduced = data.reduce(function(a, b) {
  if(a[b.label]) {
    a[b.label].push([a[b.label].length, parseInt(b.count, 10)]);
  } else {
    a[b.label] = [[0, parseInt(b.count, 10)]];
  }
  return a;
}, {})).map(function(key) {
  return {
    label: key,
    data: reduced[key]
  };
});

小提琴：http://jsfiddle.net/rdkgbteq/1/

如果您想在服务器上转换数据，PHP 中也是这样：

$reduced = array_reduce($data, function($result, $current) {
    if(array_key_exists($current['label'], $result)) {
        array_push($result, [count($result[$current['label']]), $current['count']]);
    } else {
        $result[$current['label']] = [[0, $current['count']]];
    }
    return $result;
}, array());

$formatted = array_map(function($key) {
    return array(
        'label' => $key,
        'data'  => $reduced[$key]
    ); 
}, array_keys($reduced));

echo json_encode($formatted);

如果您希望我详细说明这里发生的事情，请告诉我。