如何在字符串中查找最后一个字母表，它是java中数字和字符的组合(how to find last alphabet in string which is combination of number

如何在字符串中查找最后一个字母表，它是java中数字和字符的组合(how to find last alphabet in string which is combination of number and character in java)

我有以下字符串我想基于字符串中的最后一个字母表来spilts字符串，例如考虑

String str = '02191204E8DA4459.jpg' how to extract 4459 from str ?

我尝试下面的代码，但但在上面的字符串字母表不是常量actullally我想在数据库中添加图像sequeunce为例如

'02191204E8DA4459.jpg' to '02191204E8DA4465.jpg' i.e 6 images String sec1 = null, sec2 = null, result = null, initfilename = null; int lowerlimit, upperlimit; String realimg1, realimg2; String str1 = "120550DA121.jpg"; // here DA is constant String str2 = "120550DA130.jpg"; // String[] parts1; String[] parts2; realimg1 = str1.substring(0, str1.length() - 4); // remove .jpg from image name realimg2 = str2.substring(0, str2.length() - 4); // if (realimg1.matches(".*[a-zA-Z]+.*")) { // checking whether imagename has alphabets parts1 = realimg1.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)"); // It matches positions between a number and not-a-number (in any order). parts2 = realimg2.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)"); // It matches positions between a number and not-a-number (in any order). sec1 = parts1[0]; sec2 = parts1[1]; result = sec1.concat(sec2); lowerlimit = Integer.parseInt(parts1[parts1.length - 1]); upperlimit = Integer.parseInt(parts2[parts2.length - 1]); for (int j = lowerlimit; j <= upperlimit; j++) { initfilename = result + j + ".jpg"; System.out.println(initfilename); } } else { for (int j = Integer.parseInt(realimg1); j <= Integer.parseInt(realimg2); j++) { initfilename = j + ".jpg"; System.out.println(initfilename); } }

I have the following string I want to spilts string based on last alphabets in string for e.g consider

String str = '02191204E8DA4459.jpg' how to extract 4459 from str ?

i try following code but that but in above string alphabets are not constant actullally i want to add images sequeunce in database for e.g

'02191204E8DA4459.jpg' to '02191204E8DA4465.jpg' i.e 6 images String sec1 = null, sec2 = null, result = null, initfilename = null; int lowerlimit, upperlimit; String realimg1, realimg2; String str1 = "120550DA121.jpg"; // here DA is constant String str2 = "120550DA130.jpg"; // String[] parts1; String[] parts2; realimg1 = str1.substring(0, str1.length() - 4); // remove .jpg from image name realimg2 = str2.substring(0, str2.length() - 4); // if (realimg1.matches(".*[a-zA-Z]+.*")) { // checking whether imagename has alphabets parts1 = realimg1.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)"); // It matches positions between a number and not-a-number (in any order). parts2 = realimg2.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)"); // It matches positions between a number and not-a-number (in any order). sec1 = parts1[0]; sec2 = parts1[1]; result = sec1.concat(sec2); lowerlimit = Integer.parseInt(parts1[parts1.length - 1]); upperlimit = Integer.parseInt(parts2[parts2.length - 1]); for (int j = lowerlimit; j <= upperlimit; j++) { initfilename = result + j + ".jpg"; System.out.println(initfilename); } } else { for (int j = Integer.parseInt(realimg1); j <= Integer.parseInt(realimg2); j++) { initfilename = j + ".jpg"; System.out.println(initfilename); } }

最满意答案

也许不是最干净的方式，因为它不适用于前面没有非数字的字符串：

String str = "02191204E8DA4459.jpg"; String lastNumber = str.replaceAll("^.*[^0-9](\\d+).*?$", "$1"); System.out.println(lastNumber);

Maybe not the cleanest way, since it wouldn't work for strings not preceded by non-digits:

String str = "02191204E8DA4459.jpg"; String lastNumber = str.replaceAll("^.*[^0-9](\\d+).*?$", "$1"); System.out.println(lastNumber);