英文:
Split string and extract text and number
问题
以下是您要的翻译内容：
我必须将地址分割为街道和编号。示例：
Lievensberg 31DJablunkovska 21/2Weimarstraat 113 APastoor Baltesenstraat 22Van Musschenboek strasse 84
我需要这样分割：
街道1：Lievensberg编号1：31D街道2：Jablunkovska编号2：21/2街道3：Weimarstraat编号3：113 A街道4：Pastoor Baltesenstraat编号4：22街道5：Van Musschenboek strasse编号5：84
我使用了以下代码，但不起作用，因为我只需要在空格后面的字符是数字时才进行分割：
String[] arrSplit = address_line.split("\\s");for (int i = 0; i < arrSplit.length; i++) {System.out.println(arrSplit[i]);}
但我不知道如何做到满足我所有的要求。有任何想法吗？
英文:
I have to divide an address into street and number. Examples
Lievensberg 31DJablunkovska 21/2Weimarstraat 113 APastoor Baltesenstraat 22Van Musschenboek strasse 84
I need to split like this:
Street1: LievensbergNumber1: 31DStreet2: JablunkovskaNumber2: 21/2Street3: WeimarstraatNumber3: 113 AStreet4: Pastoor BaltesenstraatNumber4: 22Street5: Van Musschenboek strasseNumber5: 84
I used this code but not working, because I need to split only when the character after the white space will be a number:
String[] arrSplit = address_line.split("\\s");for (int i = 0; i < arrSplit.length; i++) {System.out.println(arrSplit[i]);}
But I don't know how to do it so that all my requirements are met. Any idea?
答案1
得分: 2
如果数字是可选的，可以使用两个捕获组，其中第二个组是可选的。
^([^\d\r\n]+?)(?:\h*(\d.*)|$)
解释
^字符串开始([^\d\r\n]+?)匹配1个或多个字符，但不包括数字或换行符，非贪婪模式(?:非捕获组\h*(\d.*)匹配0个或多个水平空白字符|或者$字符串结束
)关闭非捕获组
示例代码
String regex = "^([^\\d\\r\\n]+?)(?:\\h*(\\d.*)|$);"String string = "Lievensberg 31D\n"+ "Jablunkovska 21/2\n"+ "Weimarstraat 113 A\n"+ "Pastoor Baltesenstraat 22\n"+ "Van Musschenboek strasse 84\n"+ "Lievensberg";Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);Matcher matcher = pattern.matcher(string);while (matcher.find()) {System.out.println("Street: " + matcher.group(1));if (matcher.group(2) != null) {System.out.println("Number: " + matcher.group(2));}System.out.println("------------------");}
输出
Street: LievensbergNumber: 31D------------------Street: JablunkovskaNumber: 21/2------------------Street: WeimarstraatNumber: 113 A------------------Street: Pastoor BaltesenstraatNumber: 22------------------Street: Van Musschenboek strasseNumber: 84------------------Street: Lievensberg------------------
英文:
If the number can be optional, instead of using split, you could use 2 capturing groups where the second group is optional.
^([^\d\r\n]+?)(?:\h*(\d.*)|$)
Explanation
^Start of string([^\d\r\n]+?)Match 1+ times any char except a digit or newline non greedy(?:Non capture group\h*(\d.*)Match 0+ horizontal whitespace chars|Or$End of string
)Close non capture group
Example code
String regex = "^([^\\d\\r\\n]+?)(?:\\h*(\\d.*)|$)";String string = "Lievensberg 31D\n"+ "Jablunkovska 21/2\n"+ "Weimarstraat 113 A\n"+ "Pastoor Baltesenstraat 22\n"+ "Van Musschenboek strasse 84\n"+ "Lievensberg";Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);Matcher matcher = pattern.matcher(string);while (matcher.find()) {System.out.println("Street: " + matcher.group(1));if (matcher.group(2) != null) {System.out.println("Number: " + matcher.group(2));}System.out.println("------------------");}
Output
Street: LievensbergNumber: 31D------------------Street: JablunkovskaNumber: 21/2------------------Street: WeimarstraatNumber: 113 A------------------Street: Pastoor BaltesenstraatNumber: 22------------------Street: Van Musschenboek strasseNumber: 84------------------Street: Lievensberg------------------
答案2
得分: 1
你可以使用正则表达式首先验证是否匹配，然后再进行处理。
String str1 = "Lievensberg 31D"; // 街道 = Lievensberg，号码 = 31DString str2 = "Lievensberg NN31D"; // 不匹配String str3 = "Lievensberg"; // 街道 = Lievensberg，号码 = nullString str4 = "Pastoor Baltesenstraat 22"; // 街道 = Pastoor Baltesenstraat，号码 = 22Pattern pattern = Pattern.compile("([a-zA-Z ]+?)(\\s(\\d+)(.*))?");Matcher matcher = pattern.matcher(str1);if (matcher.matches()) {String street = matcher.group(1);String number = matcher.group(2) != null ? matcher.group(3) + matcher.group(4) : null;System.out.println("街道 = " + street);System.out.println("号码 = " + number);}
英文:
You can use regex to verify first whether it matches or not, then only process it.
String str1 = "Lievensberg 31D"; // street = Lievensberg, number = 31DString str2 = "Lievensberg NN31D"; // doesn't matchesString str3 = "Lievensberg"; // street = Lievensberg, number = nullString str4 = "Pastoor Baltesenstraat 22"; // street = Pastoor Baltesenstraat, number = 22Pattern pattern = Pattern.compile("([a-zA-Z ]+?)(\\s(\\d+)(.*))?");Matcher matcher = pattern.matcher(str1);if(matcher.matches()) {String street = matcher.group(1);String number = matcher.group(2) != null ? matcher.group(3) + matcher.group(4) : null;System.out.println("street = " + street);System.out.println("number = " + number);}
答案3
得分: 1
ArrayList<String> list = new ArrayList();list.add("Lievensberg 31D");list.add("Jablunkovska 21/2");list.add("Weimarstraat 113 A");list.add("Pastoor Baltesenstraat 22");list.add("Van Musschenboek strasse 84");for(int i=0;i<list.size();i++){System.out.println("Street"+(i+1)+": "+ list.get(i).split("\\s+(?=\\d)")[0]);System.out.println("Number"+(i+1)+": "+ list.get(i).split("\\s+(?=\\d)")[1]);}
英文:
Something like this:
ArrayList<String> list = new ArrayList();list.add("Lievensberg 31D");list.add("Jablunkovska 21/2");list.add("Weimarstraat 113 A");list.add("Pastoor Baltesenstraat 22");list.add("Van Musschenboek strasse 84");for(int i=0;i<list.size();i++){System.out.println("Street"+(i+1)+": "+ list.get(i).split("\\s+(?=\\d)")[0]);System.out.println("Number"+(i+1)+": "+ list.get(i).split("\\s+(?=\\d)")[1]);}
答案4
得分: 0
你可以使用这个逻辑：
- 找到第一个数字的索引
- 根据这个索引来分割字符串
为了更好地理解，使用以下代码：
public static void main(String[] args) {String address_line = "Weimarstraat 113 A";// 找到第一个数字的索引Matcher matcher = Pattern.compile("\\d+").matcher(address_line);int i = -1;for(char c: address_line.toCharArray() ){if('0'<=c && c<='9')break;i++;}// 使用索引分割字符串System.out.println(address_line.substring(0, i));System.out.println(address_line.substring(i+1));}
其输出将是：
Weimarstraat113 A
英文:
You can use this logic:
- Find the index of the first number
- Split the string based on this index
For better understanding use below code
public static void main(String[] args) {
String address_line = "Weimarstraat 113 A";// Find index of first numberMatcher matcher = Pattern.compile("\\d+").matcher(address_line);int i = -1;for(char c: address_line.toCharArray() ){if('0'<=c && c<='9')break;i++;}//Split string using indexSystem.out.println(address_line.substring(0, i));System.out.println(address_line.substring(i+1));}
Its output will be:
Weimarstraat113 A
答案5
得分: -1
这是使用正则表达式和分割的简单解决方案：
String str = "Jablunkovska 21/2";String[] split = str.split("\\s(?=\\d)", 2);System.out.println(Arrays.toString(split));
输出：
[Jablunkovska, 21/2]
表达式 (?=\\d) 是一个数字的前瞻，因此它不会在分割时被移除。
英文:
Here's a simple solution using regex and split:
String str = "Jablunkovska 21/2";String[] split = str.split("\\s(?=\\d)", 2);System.out.println(Arrays.toString(split));
Output:
[Jablunkovska, 21/2]
The expression (?=\\d) is a lookahead for a digit, so it doesn't get removed with the split.
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