abstract                                                            long         addFeature(Feature feature)

Adds the given feature to the table.
            abstract                                                            long[]         addFeatures(List<Feature> features)

Adds the given features to the table.
            abstract                                                            void         deleteFeature(long featureId)

Deletes the specified feature from the table.
            abstract                                                            void         deleteFeatures(long[] featureId)

Deletes the specified features from the table.
            abstract                                                            String         getCopyright()

Gets the copyright information.
            abstract                                                             Envelope         getExtent()

Returns the full extent associated with this feature table.
            abstract                                                             Feature         getFeature(long id)

Retrieves the feature with the given ID from the table.
            abstract                                                             FeatureResult         getFeatures(long[] ids)

Retrieves the features with the given IDs from the table.
            abstract                                                             Field         getField(String fieldName)

Returns the Field object with the specified field name.
            abstract                                                            List<Field>         getFields()

Returns all the Fields in the table.
            abstract                                                            long         getNumberOfFeatures()

Gets the number of features in the table.
            abstract                                                             SpatialReference         getSpatialReference()

Returns the spatial reference that the geometries in this table are in.
            abstract                                                            String         getTableName()

This returns the table name as defined by the table type.
abstract boolean         hasGeometry()

This returns true if this feature table has a geometry column.
            abstract                                                            boolean         isEditable()

Checks if the table can be edited.
            abstract                                                            Future<FeatureResult>         queryFeatures(QueryParameters query, CallbackListener<FeatureResult> callback)

Queries the table using the given query parameters, and returns an iterator  of features found by the query.
            abstract                                                            Future<long[]>         queryIds(QueryParameters query, CallbackListener<long[]> callback)

Queries the table using the given query parameters, and returns an array  of feature IDs found by the query.
                                                                        String         toString()          
            abstract                                                            void         updateFeature(long featureId, Feature feature)

Updates the feature specified by the unique feature ID.
            abstract                                                            void         updateFeatures(long[] featureIds, List<Feature> features)

Updates the features specified by the array of feature IDs passed in.

FeatureTable()的更多相关文章

  1. 京东评论情感分类器(基于bag-of-words模型)

    京东评论情感分类器(基于bag-of-words模型) 近期在本来在研究paraVector模型,想拿bag-of-words来做对照. 数据集是京东的评论,经过人工挑选,选出一批正面和负面的评论. ...

  2. Cesium原理篇:3D Tiles(2)数据结构

    上一节介绍3D Tiles渲染调度的时候,我们提到目前Cesium支持的Cesium3DTileContent目前支持如下类型: Batched3DModel3DTileContent Instanc ...

  3. Cesium原理篇:3D Tiles(3)个人总结

    个人结论:目前,在演示层面,3D Tiles问题不大,但项目应用上就不够成熟了,所以问问自己,你是想吃瓜呢还是想吃螃蟹? 好的方面 数据规范 我非常喜欢glTF的整体设计,概括有四点:第一,数据块(B ...

  4. ArcGIS API for JavaScript 4.2学习笔记[0] AJS4.2概述、新特性、未来产品线计划与AJS笔记目录

    放着好好的成熟的AJS 3.19不学,为什么要去碰乳臭未干的AJS 4.2? 4.2全线基础学习请点击[直达] 4.3及更高版本的补充学习请关注我的博客. ArcGIS API for JavaScr ...

  5. [置顶] ArcGIS Runtime SDKs 10.2 for iOS & Android& OS X发布

    我们高兴的宣布:ArcGISRuntime SDKs 10.2 for iOS & Android & OS X正式发布!在10.2版本中,你可以在iOS.Android和Mac设备上 ...

  6. 扩增子分析QIIME2. 1简介和安装

    原网站:https://blog.csdn.net/woodcorpse/article/details/75103929 声明:本文为QIIME2官方帮助文档的中文版,由中科院遗传发育所刘永鑫博士翻 ...

  7. arcgis andriod Edit features

    来自:https://developers.arcgis.com/android/guide/edit-features.htm#ESRI_SECTION1_56C60DB71AF941E98668A ...

  8. 扩增子分析QIIME2-3数据导出Exporting data

    # 激活工作环境 source activate qiime2-2017.8 # 建立工作目录 mkdir -p qiime2-exporting-tutorial cd qiime2-exporti ...

  9. 扩增子分析QIIME2-2数据导入Importing data

    # 激活工作环境 source activate qiime2-2017.8 # 建立工作目录 mkdir -p qiime2-importing-tutorial cd qiime2-importi ...

随机推荐

  1. C语言调用正则表达式

    标准的C和C++都不支持正则表达式,但有一些函数库可以辅助C/C++程序员完成这一功能,其中最著名的当数Philip Hazel的Perl-Compatible Regular Expression库 ...

  2. UVA 11624 Fire!(两次BFS+记录最小着火时间)

    题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...

  3. socket-----爬虫&&文件传输

    最近想着写几个小demo 写了一个爬虫,用的是C++,基本思想就是一层一层的找类似深搜吧,抓取的页面是www.cnblogs.com,从localhost发送request请求,给www.cnblog ...

  4. apache2.2控制目录文件访问设置

    1.apache2.2控制目录文件访问需求 红框可以访问,其他不能访问 2.apache2.2具体正则配置 <locationMatch ^/f/user/Panorama/81/581/(gr ...

  5. 洛谷 P2639 [USACO09OCT]Bessie的体重问题Bessie's We… 题解

    题目传送门 这也是个01背包,只是装的很... #include<bits/stdc++.h> #define MAXN 45010 using namespace std; int f[ ...

  6. c++ primer 6 语句

    没什么重要的东西,异常处理在17章再讲吧

  7. merc_timer_handle_t函数的使用

    merc_timer_handle_t,是定义一个时间类型,这个时间类型可以用来接收2个函数之间的wasted time 但是在项目中出现这个情况: 因为在脚本中添加了该函数:

  8. Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) E - Goods transportation 最大流转最小割转dp

    E - Goods transportation 思路:这个最大流-> 最小割->dp好巧妙哦. #include<bits/stdc++.h> #define LL long ...

  9. 【BZOJ 3622】3622: 已经没有什么好害怕的了(DP+容斥原理)

    3622: 已经没有什么好害怕的了 Time Limit: 10 Sec  Memory Limit: 256 MBSubmit: 683  Solved: 328 Description Input ...

  10. 【UOJ #206】【APIO 2016】Gap

    http://uoj.ac/problem/206 对于T=1,直接从两端往中间跳可以遍历所有的点. 对于T=2,先求出最小值a和最大值b,由鸽巢原理,答案一定不小于\(\frac{b-a}{N-1} ...