小编Man*_*nki的帖子

Doctrine实体对象到数组

想要将doctrine entiry对象转换为普通数组,这是我的代码到目前为止,

 $demo = $this->doctrine->em->find('Entity\User',2);
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获取实体对象,

Entity\User Object
(
[id:Entity\User:private] => 2
[username:Entity\User:private] => TestUser
[password:Entity\User:private] => 950715f3f83e20ee154995cd5a89ac75
[email:Entity\User:private] => test@test.com
[firm_id:Entity\User:private] => Entity\Firm Object
    (
        [id:Entity\Firm:private] => 16
        [company_name:Entity\Firm:private] => TestFirm
        [company_detail:Entity\Firm:private] => India
        [created_at:Entity\Firm:private] => DateTime Object
            (
                [date] => 2014-08-01 18:16:08
                [timezone_type] => 3
                [timezone] => Europe/Paris
            )

        [user:Entity\Firm:private] => 
    )

[created_at:Entity\User:private] => DateTime Object
    (
        [date] => 2014-08-01 15:12:36
        [timezone_type] => 3
        [timezone] => Europe/Paris
    )

[updated_at:Entity\User:private] => DateTime Object
    (
        [date] => 2014-08-01 15:12:36
        [timezone_type] …
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php arrays doctrine doctrine-orm

15
推荐指数
2
解决办法
4万
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PHP - 从'Point'形状生成kml

我正在使用PHPShapefile库来生成KML并将数据显示到谷歌地图,但是当涉及到"点"形状时它不起作用而不生成KML.这是Polygon形状的代码片段,帮助我为Point形状创建.

//this shape data i'm fetching from shapefile library.        
$shp_data = $record->getShpData();
if (isset($shp_data['parts'])) {
  $counter1 = 0;
  if ($shp_data['numparts']) {
    $polygon_array['polygon']['status'] = 'multi-polygon';
  } else {
    $polygon_array['polygon']['status'] = 'single-polygon';
  }

  $polygon_array['polygon']['total_polygon'] = $shp_data['numparts'];

  foreach ($shp_data['parts'] as $polygon) {
    foreach ($polygon as $points) {
      $counter = 0;
      $polygon_string = '';

      while ($counter < count($points)) {
        if ($counter == 0) {
          $polygon_string = $points[count($points) - 1]['x'] . ',';
          $polygon_string .= $points[$counter]['y'] . ' ' . $points[$counter]['x'] . ',';
        } …
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php google-maps kml shapefile

6
推荐指数
1
解决办法
354
查看次数

使用 Instant.parse 解析 2018-05-01T00:00:00 日期时出错

这是我用来使用 Instant.parse 解析字符串的代码,

String date = "2018-05-01T00:00:00";
Instant.parse(date)
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并低于错误

java.time.format.DateTimeParseException: Text '2018-05-01T00:00:00' could not be parsed at index 19
        at java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1949)
        at java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1851)
        at java.time.Instant.parse(Instant.java:395)
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我不能使用其他然后Instant所以只为它寻找解决方案!

java datetime java.time.instant

3
推荐指数
1
解决办法
367
查看次数