线性输出格式由lat lon

时间:2012-07-02 18:12:22

标签: postgis

有没有办法以这种格式从postgis数据库返回线串?

线:1,“路” 35.62200200,-88.98259200 35.62203500,-88.98240800 35.62202700,-88.98231000 35.62180000,-88.98163400 35.62175700,-88.98149000 35.62172500,-88.97881200 35.62172000,-88.97798500 35.62169800,-88.97752400 35.62170300,-88.97736200 35.62173900,-88.97723200 35.62180600,-88.97715200 35.62292600,-88.97608000 35.62467700,-88.97441600 35.62482100,-88.97429800 端:

3 个答案:

答案 0 :(得分:3)

你应该连接xy的

LINESTRING('|| s。“X”||''|| s。“Y”||','|| s。“X”||''|| s。“Y”||' ) '',2309)

希望对你有用:)

答案 1 :(得分:0)

您有自定义格式,还是已知?有几个geometry output functions,您可以随时将其输出解析为新的输出,例如:

SELECT 'Line: ' || id || ', ' ||
  trim(translate(ST_AsText(ST_FlipCoordinates(ST_Transform(the_geom, 4326))),
                 'MULTILINESTRINGZM()', '')) || ' End:' AS output
FROM nqatrkl;

或者你可以通过其他方式转储积分并修补它们:

SELECT id, time, (d).path, (d).path[1] AS part,
       ST_Y((d).geom) AS lat, ST_X((d).geom) AS lon
FROM (SELECT id, time, ST_DumpPoints(ST_Transform(the_geom, 4326)) AS d
      FROM nqatrkl) AS foo
ORDER BY (d).path;

答案 2 :(得分:0)

这对我有用 它是一种自定义格式,还是已知?有几个geometry output functions,您可以随时将其输出解析为新的输出,例如:

SELECT 'Line: ' || id || ', ' ||
  trim(translate(ST_AsText(ST_FlipCoordinates(ST_Transform(the_geom, 4326))),
                 'MULTILINESTRINGZM()', '')) || ' End:' AS output
FROM nqatrkl;

或者你可以通过其他方式转储积分并修补它们:

SELECT id, time, (d).path, (d).path[1] AS part,
       ST_Y((d).geom) AS lat, ST_X((d).geom) AS lon
FROM (SELECT id, time, ST_DumpPoints(ST_Transform(the_geom, 4326)) AS d
      FROM nqatrkl) AS foo
ORDER BY (d).path;