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Question

我有这种CSV文件，我必须用Java解析。

2012-11-01 00,  1106,   2194.1971066908
2012-11-01 01,  760,    1271.8460526316
.
.
.
2012-11-30 21,  1353,   1464.0014781966
2012-11-30 22,  1810,   1338.8331491713
2012-11-30 23,  1537,   1222.7826935589

720 rows selected.      

Elapsed: 00:37:00.23

这是我创建的Java代码，用于隔离每个列并将其存储在列表中。

public void extractFile(String fileName){
        try{
            BufferedReader bf = new BufferedReader(new FileReader(fileName));
            try {
                String readBuff = bf.readLine();

                while (readBuff!=null){

                    Pattern checkData = Pattern.compile("[a-zA-Z]");
                    Matcher match = checkData.matcher(readBuff);

                    if (match.find()){
                        readBuff = null;
                    }

                    else if (!match.find()){

                        String[] splitReadBuffByComma = new String[3];
                        splitReadBuffByComma = readBuff.split(",");

                            for (int x=0; x<splitReadBuffByComma.length; x++){

                                if (x==0){
                                    dHourList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==1){
                                    throughputList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==2){
                                    avgRespTimeList.add(splitReadBuffByComma[x]);
                                }
                            }
                    }

                    readBuff = bf.readLine();
                }
            }
            finally{
                bf.close();
            }
        }
        catch(FileNotFoundException e){
            System.out.println("File not found dude: "+ e);
        }
        catch(IOException e){
            System.out.println("Error Exception dude: "+e);
        }
    }

问题是我创建的正则表达式有点错误，因为它仍然包含“720行选择”文本并将它们存储在dHourList中。
dHourList应该只存储如此表示的日期列“2012-11-01 00 ... etc”
throughputList =“1106,760 ......等”
avgResponseTime =“2194.192， 1271.846 ...等“

这对于正确的reg表达式应该是什么？

更新

2012-11-30 21 2012-11-30 22 2012-11-30 23

选择了720行。

经历时间：00：37：00.23

日期 - 小时的大小：724吞吐量大小：720平均大小：720

我在checkData regex中使用了这个，因为如果我使用一个斜杠\ d编译会说无效的转义序列

Pattern checkData = Pattern.compile("^(19|20)\\d\\d([-/.])(0[1-9]|1[012])\2(0[1-9]|[12][0-9]|3[01])\b.+$");

但它仍然显示选择了720行，另一行不应该在那里。

更新2

工作代码：

while (readBuff!=null){


                    Pattern checkData = Pattern.compile("^(19|20)\\d\\d([-/.])(0[1-9]|1[012])\\2(0[1-9]|[12][0-9]|3[01])\\b.+$");

                    Matcher match = checkData.matcher(readBuff);

                    if (!match.find()){
                        readBuff = null;
                    }

                    else{

                        String[] splitReadBuffByComma = new String[3];
                        splitReadBuffByComma = readBuff.split(",");

                            for (int x=0; x<splitReadBuffByComma.length; x++){

                                if (x==0){
                                    dHourList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==1){
                                    throughputList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==2){
                                    avgRespTimeList.add(splitReadBuffByComma[x]);
                                }
                            }
                    }

                    readBuff = bf.readLine();
                }

我删除了else if条件并将其更改为else并使用了Cylian建议的正则表达式现在我有输出

2012-11-30 21
2012-11-30 22
2012-11-30 23

Size of date-hour: 720 size of throughput: 720 size of avg resp time: 720

非常感谢！

Answer 1

首先，在^正则表达式的开头插入checkData是有意义的。然后表达式只会在行的开头找到，而不是在整个字符串中查找，应该使它更快。

你可以让你的正则表达式以更多日期格式开始，比如表达式（比如4个数字和短划线），就像在最后一行一样，在行数之后永远不会有破折号。

也许是这样的：

Pattern checkData = Pattern.compile("^\\d\\d\\d\\d-");

如果您确定没有获得意外数据，这应该足够了 - 如果您想确保您的程序即使您的csv数据格式错误也能正常工作，只需扩展正则表达式以捕获整行并改用matches()。

Answer 2

试试这个[您的代码，但有点修改]：

public void extractFile(String fileName){
        try{
            BufferedReader bf = new BufferedReader(new FileReader(fileName));
            try {
                String readBuff = bf.readLine();

                while (readBuff!=null){

                    Pattern checkData = Pattern.compile("^(19|20)\\d\\d([-/.])(0[1-9]|1[012])\\2(0[1-9]|[12][0-9]|3[01])\\b.+$");
                    Matcher match = checkData.matcher(readBuff);

                    if (!match.find()){
                        readBuff = null;
                    }

                    else if (match.find()){

                        String[] splitReadBuffByComma = new String[3];
                        splitReadBuffByComma = readBuff.split(",");

                            for (int x=0; x<splitReadBuffByComma.length; x++){

                                if (x==0){
                                    dHourList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==1){
                                    throughputList.add(splitReadBuffByComma[x]);
                                }
                                else if (x==2){
                                    avgRespTimeList.add(splitReadBuffByComma[x]);
                                }
                            }
                    }

                    readBuff = bf.readLine();
                }
            }
            finally{
                bf.close();
            }
        }
        catch(FileNotFoundException e){
            System.out.println("File not found dude: "+ e);
        }
        catch(IOException e){
            System.out.println("Error Exception dude: "+e);
        }
    }

正则表达式解剖

# ^(19|20)\d\d([-/.])(0[1-9]|1[012])\2(0[1-9]|[12][0-9]|3[01])\b.+$
# 
# Options: ^ and $ match at line breaks
# 
# Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
# Match the regular expression below and capture its match into backreference number 1 «(19|20)»
#    Match either the regular expression below (attempting the next alternative only if this one fails) «19»
#       Match the characters “19” literally «19»
#    Or match regular expression number 2 below (the entire group fails if this one fails to match) «20»
#       Match the characters “20” literally «20»
# Match a single digit 0..9 «\d»
# Match a single digit 0..9 «\d»
# Match the regular expression below and capture its match into backreference number 2 «([-/.])»
#    Match a single character present in the list “-/.” «[-/.]»
# Match the regular expression below and capture its match into backreference number 3 «(0[1-9]|1[012])»
#    Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
#       Match the character “0” literally «0»
#       Match a single character in the range between “1” and “9” «[1-9]»
#    Or match regular expression number 2 below (the entire group fails if this one fails to match) «1[012]»
#       Match the character “1” literally «1»
#       Match a single character present in the list “012” «[012]»
# Match the same text as most recently matched by capturing group number 2 «\2»
# Match the regular expression below and capture its match into backreference number 4 «(0[1-9]|[12][0-9]|3[01])»
#    Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
#       Match the character “0” literally «0»
#       Match a single character in the range between “1” and “9” «[1-9]»
#    Or match regular expression number 2 below (attempting the next alternative only if this one fails) «[12][0-9]»
#       Match a single character present in the list “12” «[12]»
#       Match a single character in the range between “0” and “9” «[0-9]»
#    Or match regular expression number 3 below (the entire group fails if this one fails to match) «3[01]»
#       Match the character “3” literally «3»
#       Match a single character present in the list “01” «[01]»
# Assert position at a word boundary «\b»
# Match any single character that is not a line break character «.+»
#    Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
# Assert position at the end of a line (at the end of the string or before a line break character) «$»

<强>更新

据我所知，您的输入字符串包含许多以日期开头但不包含逗号的行。为此，将以前的模式更改为以下内容：

^(19|20)\d\d([-/.])(0[1-9]|1[012])\2(0[1-9]|[12][0-9]|3[01])\s+\d+,[^,]+,[^,]+$

或 escaped

^(19|20)\\d\\d([-/.])(0[1-9]|1[012])\\2(0[1-9]|[12][0-9]|3[01])\\s+\\d+,[^,]+,[^,]+$

Answer 3

你不必用正则表达式来做。（ 如果它显示为您的示例 ）

你可以检查

如果该行包含逗号“,”或
如果分割数组的长度为3或
在while条件下更改一下，如果该行以“selected。”结尾，则跳出。

如何使用Java Regex丢弃CSV文件的某些部分

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更新2

3 个答案: