AE中如何由IFeature 如何获取所对应的FeatureClass

Posted 2023-04-26

参考技术A private void OnDeleteFeatureMethod(object o) IFeature pFeature = o as IFeature; IFeatureClass pFeatureClass = pFeature.Class as IFeatureClass; for (int i = 0; i < axMapControl1.Map.LayerCount;i++ ) IFeatureLayer iFeatureLayer = axMapControl1.get_Layer(i) as IFeatureLayer; IFeatureClass iFeatureCla = iFeatureLayer.FeatureClass; if (iFeatureCla == pFeatureClass) IWorkspace pWorkSpace = m_EngineEditor.EditWorkspace; textBox3.Text += "操作的文件全路径:" + pWorkSpace.PathName + "\\" + axMapControl1.get_Layer(i).Name + ".shp " + "\r\n"; break; if (pFeature.Shape.GeometryType == esriGeometryType.esriGeometryPoint) IGeometry iGe = pFeature.Shape; IPoint ipo = new PointClass(); ipo = iGe as IPoint; int a = 0; int b = 0; axMapControl1.FromMapPoint(ipo, ref a, ref b); textBox3.Text += "删除的点的ID号：" + pFeature.OID + "，坐标：(" + a + "," + b + ")" + "\r\n"; else if (pFeature.Shape.GeometryType == esriGeometryType.esriGeometryPolygon) textBox3.Text += "删除的多边形对象的ID号：" + pFeature.OID + "，坐标："; IPolygon pPolygon = (IPolygon)pFeature.Shape; int a = 0; int b = 0; //把该feature强制转换为一个点的集合，再取点的坐标 IPointCollection pPointCollection = pPolygon as IPointCollection; for (int i = 0; i < pPointCollection.PointCount - 1; i++) IPoint ipo = pPointCollection.get_Point(i); axMapControl1.FromMapPoint(ipo, ref a, ref b); textBox3.Text += "(" + a + "," + b + ")" + "\t"; textBox3.Text += "\r\n"; else if (pFeature.Shape.GeometryType == esriGeometryType.esriGeometryPolyline) textBox3.Text += "删除的线对象的ID号：" + pFeature.OID + "，其坐标："; IPolyline pPolygon = (IPolyline)pFeature.Shape; int a = 0; int b = 0; //把该feature强制转换为一个点的集合，再取点的坐标 IPointCollection pPointCollection = pPolygon as IPointCollection; for (int i = 0; i < pPointCollection.PointCount; i++) IPoint ipo = pPointCollection.get_Point(i); axMapControl1.FromMapPoint(ipo, ref a, ref b); textBox3.Text += "(" + a + "," + b + ")" + "\t"; textBox3.Text += "\r\n"; axMapControl1.ActiveView.PartialRefresh(esriViewDrawPhase.esriViewGeography, null, null); 本回答被提问者采纳

PHP-CodeIgniter：如何通过Javascript获取要删除的对应html表格行值

【中文标题】PHP-CodeIgniter：如何通过Javascript获取要删除的对应html表格行值

【英文标题】：PHP-CodeIgniter: How to get the corresponding html table row values to be deleted by Javascript

【发布时间】：2015-02-17 06:56:49

【问题描述】：

在我的 codeigniter 项目中，我使用了一个由 Ajax 动态生成的表。在每一行上都有一个按钮，可以从 Html 表和 Mysql 表中删除相应的行。

我已经试过了。我得到了删除 Html 表格行的代码，它遵循

$(document).on('click', '#deleteRow', function() 

     $(this).parent().parent().remove();

);

它奏效了。但我也想从 Mysql 中删除相应的行。所以首先，它需要从javascript传递相应的行信息。然后通过 URL 将其传递给控制器​​。？

window.location.href = "<?php echo base_url("settings/remove_company"); ?>?id="+current;

我如何获取对应的行信息，例如 company_id、lic_id（html 表的字段名称）。？ 任何帮助将不胜感激。

【问题讨论】：

使每一行都是唯一的，然后在您的删除按钮上调用一个函数。make 函数发出 ajax 请求以删除该行，当响应是使用您的唯一 ID 接收删除元素时

【参考方案1】：

为<tr>添加属性

<tr data-companyId="<?php echo $companyId;?>" data-licId="<?php echo $licId;?>">

在您的 jQuery 中，通过单击删除链接获取这些属性：

$(document).on('click', '#deleteRow', function()      
  var companyId = $(this).parent().parent().attr('data-companyId');
  var licId = $(this).parent().parent().attr('data-licId');
  $(this).parent().parent().remove();
);

甚至，您可以进行对象缓存（使用变量而不是对象来提高性能。

$(document).on('click', '#deleteRow', function() 
  var obj = $(this).parent().parent();
  var companyId = obj.attr('data-companyId');
  var licId = obj.attr('data-licId');
  obj.remove();
);