java - 在 Java 中替换字符串时出现问题

Question

我正在尝试用字符串内的缩短 URL 替换 URL:

public void shortenMessage()
    {
        String body = composeEditText.getText().toString();
        String shortenedBody = new String();
        String [] tokens = body.split("\\s");

        // Attempt to convert each item into an URL.  
        for( String token : tokens ) 
        {
            try
            {
                Url url = as("mycompany", "someapikey").call(shorten(token));
                Log.d("SHORTENED", token + " was shortened!");
                shortenedBody = body.replace(token, url.getShortUrl());
            }
            catch(BitlyException e)
            {
                //Log.d("BitlyException", token + " could not be shortened!");

            }
        }

        composeEditText.setText(shortenedBody);
        // url.getShortUrl() -> http://bit.ly/fB05
    }

链接缩短后，我想在 EditText 中打印修改后的字符串。我的 EditText 无法正确显示我的消息。

例如:

"I like www.google.com" should be "I like [some shortened url]" after my code executes.

Answer 1

在 Java 中，字符串是不可变的。 String.replace() 返回一个新字符串，它是替换的结果。因此，当您在循环中执行 shortenedBody = body.replace(token, url.getShortUrl()); 时，shortenedBody 将保存(仅是最后一个)的结果替换。

这是使用 StringBuilder 的修复。

public void shortenMessage()
{
    String body = composeEditText.getText().toString();
    StringBuilder shortenedBody = new StringBuilder();
    String [] tokens = body.split("\\s");

    // Attempt to convert each item into an URL.  
    for( String token : tokens ) 
    {
        try
        {
            Url url = as("mycompany", "someapikey").call(shorten(token));
            Log.d("SHORTENED", token + " was shortened!");
            shortenedBody.append(url.getShortUrl()).append(" ");
        }
        catch(BitlyException e)
        {
            //Log.d("BitlyException", token + " could not be shortened!");

        }
    }

    composeEditText.setText(shortenedBody.toString());
    // url.getShortUrl() -> http://bit.ly/fB05
}