Java从多行字符串获取子字符串或拆分

Question

I want to get one String from a String composed of many lines

Example String:

Date: May 12, 20003
ID: 54076
Content: Test
filename: Testing.fileextension
folder: Test Folder
endDate: May 13, 2003

The output I want is "Testing.filextension" I tried a couple methods, but none return just the filename.

Method Attempt 1:

String[] files = example.substring(example.indexOf("filename: ")
for (String filename : files){
    System.out.printlin(filename);
}

Method Attempt 2 was to try to split on a newline, but right now it just returns all the lines. My main issue is that the .split method takes (String, int) , however, I don't have the int.

String[] files = example.split("\\r?\\n");
for (String filename : files){
    System.out.println(filename)
}

Answer 1

What about using streams? You should consider filtering lines you cannot split, eg when the splitting character is missing.

So try this one:

Map<String, String> myMap = Arrays.stream(myString.split("\\r?\\n"))
            .map(string -> string.split(": "))
            .collect(Collectors.toMap(stringArray -> stringArray[0], 
                    stringArray -> stringArray[1]));

    System.out.println(myMap.get("filename"));

Answer 2

Is there another limitation? If not just combine with an if:

String keyword = "filename: ";
String[] files = example.split("\\r?\\n");
for(String file : files) {
    if(file.startsWith(keyword)){
        System.out.println(file.substring(keyword.length()));
    }
}

Answer 3

You can filter it this way:

String[] files = example.split("\\r?\\n");
for (String filename : files){
    if (filename.startsWith("filename"))
        System.out.println(filename);
}

Answer 4

use

String[] files = example.split("\\r?\\n");
for (String filename : files){
    if (filename.startsWith("filename"))
        System.out.println(filename);
}

and if you wan't filename without fileextension

String[] files = example.split("\\r?\\n");
for (String filename : files){
    if (filename.startsWith("filename"))
        System.out.println(filename.split(".")[0]);
}

Answer 5

try this:

String searchStr = "filename:";
example.substring(example.indexOf(searchStr) + searchStr.length()).split("\n")[0].trim();

Though I would suggest you that to structure the input into a HashMap, which would be easier for later use.

Answer 6

/**
 * @Author Jack <J@ck>
 * https://stackoverflow.com/questions/49264343/java-get-substring-or-split-from-multiline-string
 */
public class StringFromStringComposedOfManyLines {
    public static void main(String[] args) {
        String manyLinesString =
                "Date: May 12, 20003\n" +
                        "ID: 54076\n" +
                        "Content: Test\n" +
                        "filename: Testing.fileextension\n" +
                        "folder: Test Folder\n" +
                        "endDate: May 13, 2003\n";
        System.out.println(manyLinesString);

        // Replace all \n \t \r occurrences to "" and gets one line string
        String inLineString = manyLinesString.replaceAll("\\n|\\t||\\r", "");
        System.out.println(inLineString);
    }
}

Out: Date: May 12, 20003ID: 54076Content: Testfilename: Testing.fileextensionfolder: Test FolderendDate: May 13, 2003

Answer 7

I'd suggest using regular expression and named group like this:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class RegExTest {

    public static void main(String[] args) {

        // Your example string
        String exampleStr = "Date: May 12, 20003\nID: 54076\nContent: Test\nfilename: Testing.fileextension\nfolder: Test Folder\nendDate: May 13, 2003.";

        /* create the regex:
         * you are providing a regex group named fname
         * This code assumes that the file name ends with a newline/eof.
         * the name of the group that will match the filename is "fname".
         * The brackets() indicate that this is a named group.
         */
        String regex = "filename:\\s*(?<fname>[^\\r\\n]+)";

        // create a pattern object by compiling the regex
        Pattern pattern = Pattern.compile(regex);

        // create a matcher object by applying the pattern on the example string:
        Matcher matcher = pattern.matcher(exampleStr);

        // if a match is found
        if (matcher.find()) {
            System.out.println("Found a match:");
            // output the content of the group that matched the filename:
            System.out.println(matcher.group("fname"));
        }
        else {
            System.out.println("No match found");
        }
    }
}