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我正在编写一个程序来解析一堆数据，（您可以在此处获取数据集本身的示例：https ://explore.data.gov/Geography-and-Environment/Worldwide-M1-Earthquakes-Past-7-天/7tag-iwnu）。

下面的类工作得很好，但我不知道为什么我需要在我的方法matcher.find()中的每个项目之间调用一个额外的时间。parseEarthquake()这是为什么？这是我必须处理的正常怪癖，还是我错误地设置了我的模式/匹配器？

该方法接受一个包含数据行之一的字符串（例如，nc,71958020,1,"Thursday, March 21, 2013 17:13:34 UTC",38.8367,-122.8298,1.4,2.60,28,"Northern California"），并为数据返回一个地震对象。

import java.text.DecimalFormat;
import java.text.FieldPosition;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.TimeZone;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class Earthquake {

    String src="xx";
    String eqid="00000000";
    short version;
    long dateTime;
    float lat, lon;
    float mag, dep;
    short nst;
    String region="Nowhere";

    private Earthquake(){
        date.setTimeZone(TimeZone.getTimeZone("UTC"));
    }



    private static DecimalFormat 
            coords      = new DecimalFormat( "##0.0000" ),
            magnitude   = new DecimalFormat( "###0.0" ),
            depth       = new DecimalFormat( "###0.00" );
    private static SimpleDateFormat date = new SimpleDateFormat("'\"'EEEE', 'MMMM' 'dd', 'yyyy' 'HH':'mm':'ss' 'zzz'\"'");


    // Src, Eqid, Version, Datetime, Lat, Lon, Magnitude, Depth, NST, Region;

    public static Earthquake parseEarthquake(String string){
        Earthquake result = new Earthquake();

        Matcher matcher = Pattern.compile("(\".*?\")|([^,]*)").matcher(string);


        try {

                                matcher.find(); result.src = matcher.group();
                matcher.find(); matcher.find(); result.eqid = matcher.group();
                matcher.find(); matcher.find(); result.version = Short.parseShort(matcher.group());
                matcher.find(); matcher.find(); result.dateTime = date.parse(matcher.group()).getTime();
                matcher.find(); matcher.find(); result.lat = coords.parse(matcher.group()).floatValue();
                matcher.find(); matcher.find(); result.lon = coords.parse(matcher.group()).floatValue();
                matcher.find(); matcher.find(); result.mag = magnitude.parse(matcher.group()).floatValue();
                matcher.find(); matcher.find(); result.dep = depth.parse(matcher.group()).floatValue();
                matcher.find(); matcher.find(); result.nst = Short.parseShort(matcher.group());
                matcher.find(); matcher.find(); result.region = matcher.group();    

        } catch (ParseException e) {
            e.printStackTrace();
        } catch (NumberFormatException e) {
            e.printStackTrace();
        }

        return result;
    }

    public String toString(){
        StringBuffer buf = new StringBuffer();

                            buf.append(src);
        buf.append(',');    buf.append(eqid);
        buf.append(',');    buf.append(version);
        buf.append(',');    date.format(dateTime, buf, new FieldPosition(0));
        buf.append(',');    coords.format(lat, buf, new FieldPosition(0));
        buf.append(',');    coords.format(lon, buf, new FieldPosition(0));
        buf.append(',');    magnitude.format(mag, buf, new FieldPosition(0));
        buf.append(',');    depth.format(dep, buf, new FieldPosition(0));
        buf.append(',');    buf.append(nst);
        buf.append(',');    buf.append('"'); buf.append(region); buf.append('"');

        return buf.toString();

    }
}
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1 回答 1

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更改([^,]*)为([^,]+)，因为前者将始终匹配 - 即使它只匹配任何内容。

于 2013-03-25T18:33:07.963 回答