我尝试了这些建议，但最终执行如下

editView.setFilters(new InputFilter[]{new InputFilter()
        {
            @Override
            public CharSequence filter(CharSequence source, int start, int end,
                                       Spanned dest, int dstart, int dend)
            {
                String prefix = "http://";

                //make sure our prefix is visible
                String destination = dest.toString();

                //Check If we already have our prefix - make sure it doesn't
                //get deleted
                if (destination.startsWith(prefix) && (dstart <= prefix.length() - 1))
                {
                    //Yep - our prefix gets modified - try preventing it.
                    int newEnd = (dend >= prefix.length()) ? dend : prefix.length();

                    SpannableStringBuilder builder = new SpannableStringBuilder(
                            destination.substring(dstart, newEnd));
                    builder.append(source);
                    if (source instanceof Spanned)
                    {
                        TextUtils.copySpansFrom(
                                (Spanned) source, 0, source.length(), null, builder, newEnd);
                    }

                    return builder;
                }
                else
                {
                    //Accept original replacement (by returning null)
                    return null;
                }
            }
        }});

使用以下任何一种方法:

String str = String.valueOf('c');
String str = Character.toString('c');
String str = 'c' + "";

你可以使用Character.toString(char)。注意，这个方法只是返回一个对String.valueOf(char)的调用，这也是有效的。

正如其他人所注意到的，字符串连接也可以作为一种快捷方式:

String s = "" + 's';

但这可以归结为:

String s = new StringBuilder().append("").append('s').toString();

这是较低的效率，因为StringBuilder是由char[]支持的(由StringBuilder()过度分配到16)，只有该数组才会被结果String防御性地复制。

String.valueOf(字符)通过将字符包装在单元素数组中并将其传递给包私有构造函数String(char[]， boolean)来“进入后门”，这避免了数组复制。

下面是将char c转换为String s的各种方法(以速度和效率为递减顺序)

char c = 'a';
String s = String.valueOf(c);             // fastest + memory efficient
String s = Character.toString(c);
String s = new String(new char[]{c});
String s = String.valueOf(new char[]{c});
String s = new Character(c).toString();
String s = "" + c;                        // slowest + memory inefficient

我尝试了这些建议，但最终执行如下

editView.setFilters(new InputFilter[]{new InputFilter()
        {
            @Override
            public CharSequence filter(CharSequence source, int start, int end,
                                       Spanned dest, int dstart, int dend)
            {
                String prefix = "http://";

                //make sure our prefix is visible
                String destination = dest.toString();

                //Check If we already have our prefix - make sure it doesn't
                //get deleted
                if (destination.startsWith(prefix) && (dstart <= prefix.length() - 1))
                {
                    //Yep - our prefix gets modified - try preventing it.
                    int newEnd = (dend >= prefix.length()) ? dend : prefix.length();

                    SpannableStringBuilder builder = new SpannableStringBuilder(
                            destination.substring(dstart, newEnd));
                    builder.append(source);
                    if (source instanceof Spanned)
                    {
                        TextUtils.copySpansFrom(
                                (Spanned) source, 0, source.length(), null, builder, newEnd);
                    }

                    return builder;
                }
                else
                {
                    //Accept original replacement (by returning null)
                    return null;
                }
            }
        }});

  char vIn = 'A';
  String vOut = Character.toString(vIn);

对于这些类型的转换，我有一个名为https://www.converttypes.com/的站点 它帮助我快速获得我使用的大多数语言的转换代码。