Replacing one specific line in a textfile in c# [duplicate]

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Modify the content of specific line in text file (1 answer)

Closed 8 years ago.

I'm having a textfile say something like this:

#TITLE:What's Up
#ARTIST:4 Non Blondes - Whats Up
#MP3:Four Non Blondes - Whats Up.mp3
#COVER:4 Non Blondes - Whats Up [CO].jpg
#BACKGROUND:4 Non Blondes - Whats Up [CO].jpg
#BPM:135
#GAP:32100

it's saved as 4 Non Blondes - Whats Up.txt In the same folder there's a MP3 file which is in this example: 4 Non Blondes - Whats Up.mp3

What i want is to replace the line:

#MP3:Four Non Blondes - Whats Up.mp3

into this line:

#MP3:4 Non Blondes - Whats Up.mp3

Every MP3 line has infront of the line this:

#MP3:[Songname].mp3

I know i can do this manually but i have like 2k files like this, and they all need to link to the correct mp3 file. I'm trying this in C#, but without luck.

This is what i've tried so far:

private static void testMethod(string path)
    {
        var x = System.IO.Directory.GetDirectories(path);
        foreach (var directory in x)
        {
            string[] mp3Files = System.IO.Directory.GetFiles(directory, "*.mp3");
            string[] txtFiles = System.IO.Directory.GetFiles(directory, "*.txt");
            string MP3FileNameWithExtensions = System.IO.Path.GetFileName(mp3Files[0]);
            Console.WriteLine(txtFiles[0]);
            var lines = System.IO.File.ReadAllLines(txtFiles[0]);
            for (int i = 0; i < lines.Length; i++)
            {
                if(lines[i].Contains("#MP3")){
                    Console.WriteLine("Jeeeej working");
                    lines[i] = "#MP3:"+MP3FileNameWithExtensions;
                    System.IO.File.WriteAllLines(txtFiles[0], lines);
                }
            }
        }
    }

Answer 1

As the filename of the .txt file is [Songname].txt, you can use Path.GetFilenameWithoutExtension(files[i]) to get [Songname]. Then replace the #MP3 line with the filename + ".mp3". Now write out the file.

N.B. You will probably want to make a copy of the directory you are working on just in case something goes wrong.

Answer 2

I know I am late to the party but here is an example.

public void FixTheFiles(String startFolderPath)
        {
            foreach (String dirName in Directory.GetDirectories(startFolderPath))
            {
                FixTheFiles(dirName);
            }
            foreach (String fileName in Directory.GetFiles(startFolderPath))
            {
                FileInfo fi = new FileInfo(fileName);
                if (fi.Extension.Equals("MP3"))
                {
                    String fileContents = "";
                    using (StreamReader sr = new StreamReader(File.Open(fileName.Replace(".mp3",".txt"),FileMode.Open)))
                    {
                        String currentLine = sr.ReadLine();
                        if (currentLine.StartsWith("#MP3:"))
                        {
                            currentLine = "#MP3:" + fileName.Substring(fileName.LastIndexOf('\\')+1);
                        }
                        fileContents += currentLine;
                    }
                    using (StreamWriter sw = new StreamWriter(File.Open(fileName.Replace(".mp3",".txt"),FileMode.Open)))
                    {
                        sw.Write(fileContents);
                    }
                }
            }
        }

Answer 3

I would simutaneously open a stream reader and a stream writer (with a different file name) and go through each file one line at a time searching for whatever changes you need to make. You can select your file names with an openfiledialog with Multiselect = true or run this command in a command prompt window to generate a textfile to paste in your code with quotation marks around them as initial values for a string array instantiation.

dir *.mp3 /b > filenames.txt

    string[] array = new string[] {"file0.mp3",
                                        "file1.mp3",
                                        "file2.mp3",
                                        "file3.mp3",
                                        "file4.mp3",
                                        "file5.mp3"};