Question

我已经解决了这个问题好几天了，但是找不到我做错的解释。我希望你能伸出援助之手。

我有一组UTM坐标filter { grok { match => { "message" => [ "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} fd=%{NUMBER:file_descriptor} slot=%{NUMBER} %{WORD:connection_method} connection from %{IP:source} to %{IP:destination}$", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} %{NOTSPACE:ssl_version} (?<encryption_method>%{NOTSPACE} %{NOTSPACE})$", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} op=%{NUMBER:op_number} %{WORD:ldap_operation} dn=%{QUOTEDSTRING:user_dn} method=%{NOTSPACE:bind_method} version=%{NUMBER:ldap_version}($)?(mech=%{NOTSPACE:auth_mechanism}$)?", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} op=%{NUMBER:op_number} %{WORD:ldap_operation} err=%{NUMBER:error_code} tag=%{NUMBER:tag_number} nentries=%{NUMBER:number_of_entries} etime=%{NUMBER:operation_time}($)?(dn=%{QUOTEDSTRING}$)?", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} op=%{NUMBER:op_number} %{WORD:ldap_operation} base=%{QUOTEDSTRING:search_base} scope=%{NUMBER:search_scope} filter=%{QUOTEDSTRING:search_filter} attrs=%{QUOTEDSTRING:search_attributes}$", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} op=%{NUMBER:op_number} %{WORD:ldap_operation}$", "^(\s)?\[%{HTTPDATE:timestamp}\] conn=%{NUMBER:connection_id} op=%{NUMBER:op_number} fd=%{NUMBER:file_descriptor} %{WORD:connection_result} - %{WORD:connection_code}$" ] } } if "" in [connection_method] { aggregate { task_id => "%{connection_id}" code => " map['timestamp'] = event['@timestamp'] map['tags'] ||= ['aggregated'] map['source'] = event['source'] map['destination'] = event['destination'] map['file_descriptor'] = event['file_descriptor'] map['connection_method'] = event['connection_method'] " map_action => "create" } } else if "" in [connection_code] { mutate { add_tag => [ "map_finished" ] } aggregate { task_id => "%{connection_id}" code => " map['operations'][event['op_number']]['connection_code'] = event['connection_code'] map['operations'][event['op_number']]['connection_result'] = event['connection_result'] " map_action => "update" } } else { aggregate { task_id => "%{connection_id}" code => " map['@timestamp'] = map['timestamp'] " timeout => 0 push_map_as_event_on_timeout => true } } }，我希望通过使用适用于C ++ {0}的proj4库将其转换为LongLat坐标(epsg:23030)。我的代码如下：

(epsg:4326)

我对函数(libproj-dev)的调用：

#include "proj_api.h
#include <geos/geom/Coordinate.h>

geos::geom::Coordinate utm2longlat(double x, double y){ 

    // Initialize LONGLAT projection with epsg:4326
    if ( !( pj_longlat = pj_init_plus("+init=epsg:4326" ) ) ){
       qDebug() << "pj_init_plus error: longlat";
    }

    // Initialize UTM projection with epsg:23030
    if ( ! (pj_utm = pj_init_plus("+init=epsg:23030" ) ) ){
       qDebug() << "pj_init_plus error: utm";
    }

    // Transform UTM projection into LONGLAT projection
    int p = pj_transform( pj_utm, pj_longlat, 1, 1, &x, &y, NULL );

    // Check for errors
    qDebug() << "Error message" << pj_strerrno( p ) ;

    // Return values as coordinate
    return geos::geom::Coordinate(x, y)
}

在我的例子中：

我知道 UTM 坐标utm2longlat指的是 LongLat 坐标... // UTM coordinates double x = 585363.1; double y = 4796767.1; geos::geom::Coordinate coord = utm2longlat( x, y ); qDebug() << coord.x << coord.y; /* Result is -0.0340087 0.756025 <-- WRONG */。
但是，在调用时，该函数返回一组错误的坐标：(585363.1 4796767.1)。

我想知道在初始化投影时是否有任何错误配置，所以我决定测试Proj4 Python绑定(-1.94725 43.3189)，只是为了测试我是否得到了相同的错误坐标...而且奇怪的是，我得到了好的。

(-0.0340087 0.756025 )

据我所知，(pyproj)是from pyproj import Proj, transform // Initialize UTM projection proj_utm = Proj(init='epsg:23030') // Initialize LongLat projection proj_lonlat = Proj(init='epsg:4326') x_utm, y_utm = 585363.1, 4796767.1 x_longlat, y_longlat = transform(proj_utm, proj_lonlat, x_utm, y_utm) // Print results print "original", x_utm, y_utm print "utm2lonlat", x_longlat, y_longlat /* Result is -1.94725 43.3189 <-- CORRECT */库上的一组Cython绑定，所以我在两种编程语言中使用相同的核心。

你有什么可能出错的线索吗？我在C ++函数中缺少某种类型的转换吗？

提前致谢。

Answer 1

结果似乎对我来说是正确的，但它以弧度而不是度数返回。将结果转换为度数并再次检查。

使用C ++中的Proj4从UTM转换为LongLat

1 个答案: