这不是最简单的解决方案吗?

<string name="string">Please provide your <b>Name</b> properly</string>

只要使用你想要的字符串:)

结果如下所示:

请填写您的姓名

只需要在HTML中构建字符串并设置它:

String sourceString = "<b>" + id + "</b> " + name; 
mytextview.setText(Html.fromHtml(sourceString));

下面是我如何使用正则表达式和Kotlin来做到这一点

val BOLD_SPAN = StyleSpan(Typeface.BOLD)

    fun TextView.boldMatches(regexString: String) {
        this.applyStyleSpanToMatches(regexString, BOLD_SPAN)
    }
    

fun TextView.applyStyleSpanToMatches(regexString: String, span: StyleSpan){
        this.text = this.text.toString().applyStyleSpanToMatches(regexString, span)
        }

        fun String.applyStyleSpanToMatches(regexString: String, span: StyleSpan): Spannable {
            val result = SpannableString.valueOf(this)
            if(regexString.isEmpty()) return result
            val pattern = try{
                Pattern.compile(regexString)
            } catch (e: PatternSyntaxException){
                return result
            }
            val matcher = pattern.matcher(result)
            while (matcher.find()) {
                val start = matcher.start()
                val end = matcher.end()
                result.setSpan(span, start, end, Spannable.SPAN_EXCLUSIVE_EXCLUSIVE)
            }
            return result
        }

使用问句 它可以这样应用:

txtResult.boldMatches(id)

val phone = "+45xxxxxx"

val phoneText = "<font color=#757B7F><b>${phone}</b></font>"

val wholeString = requireActivity().resources.getString(R.string.loginwith)+" "+phoneText

当在list/recycler中搜索char时，使字符串的第一个char可扩展

拉维和阿杰

以前突出显示像这样，但我想像下面

ravi和ajay OR ravi和ajay

为此，我搜索单词长度，如果它等于1，我将主字符串分离成单词并计算单词开始位置，然后我搜索单词以char开头。

 public static SpannableString colorString(int color, String text, String... wordsToColor) {
    SpannableString coloredString = new SpannableString(text);

    for (String word : wordsToColor) {

        Log.e("tokentoken", "-wrd len-" + word.length());
        if (word.length() !=1) {
            int startColorIndex = text.toLowerCase().indexOf(word.toLowerCase());
            int endColorIndex = startColorIndex + word.length();
            try {
                coloredString.setSpan(new ForegroundColorSpan(color), startColorIndex, endColorIndex,
                        Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);

            } catch (Exception e) {
                e.getMessage();
            }
        } else {
            int start = 0;

            for (String token : text.split("[\u00A0 \n]")) {
                if (token.length() > 0) {
                    start = text.indexOf(token, start);
                   // Log.e("tokentoken", "-token-" + token + "   --start--" + start);
                    char x = token.toLowerCase().charAt(0);
                    char w = word.toLowerCase().charAt(0);
                   // Log.e("tokentoken", "-w-" + w + "   --x--" + x);

                    if (x == w) {
                        // int startColorIndex = text.toLowerCase().indexOf(word.toLowerCase());
                        int endColorIndex = start + word.length();
                        try {
                            coloredString.setSpan(new ForegroundColorSpan(color), start, endColorIndex,
                                    Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);

                        } catch (Exception e) {
                            e.getMessage();
                        }
                    }
                }
            }

        }

    }

    return coloredString;
}

如果你正在使用Kotlin和string资源，一个简单的解决方案是:

在strings.xml上创建你的字符串，使用<b> </b>加粗你想要的部分

<string name="my_message"> This is a very <b>important</b> message! </string>

在Kotlin代码中，你必须这样做

textView.setText(R.string.my_message)

就是这样!

重要提示!

使用属性语法将不起作用:

textView.text = resources.getString(R.string.my_message)

希望能有所帮助!