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PO
 

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    <p>There is something fishy here; either <code>text</code> is not a <code>String</code>, or you don't use <code>.replace()</code> but something else (<code>.replaceAll()</code>?), or Android's <code>.replace()</code> is buggy.</p> <p>And I frankly doubt that Android devs would have had such an overlook.</p> <p>The <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replace%28java.lang.CharSequence,%20java.lang.CharSequence%29" rel="nofollow">Javadoc for <code>String#replace()</code></a> says:</p> <blockquote> <p>Replaces each substring of this string that matches the <strong>literal target sequence</strong> with the specified <strong>literal replacement sequence.</strong> [emphasis mine]</p> </blockquote> <p>Unlike its sibling methods (<code>.replaceFirst()</code> and <code>.replaceAll()</code>) which do use regexes, <code>.replace()</code> doesn't (and the fact that internally it does use <code>Pattern</code>, at least in Oracle's JDK [*], is not the problem).</p> <p>Therefore, if you actually use <code>.replace()</code> and gain the result you say, this is a bug in Android. If this is the case, try an alternative, like so (UNTESTED):</p> <pre><code>public static String realStringReplace(final String victim, final String target, final String replacement) { final int skip = target.length(); final StringBuilder sb = new StringBuilder(victim.length()); String tmp = victim; int index; while (!tmp.isEmpty()) { index = tmp.indexOf(target); if (index == -1) break; sb.append(tmp.subString(0, index)).append(replacement); tmp = tmp.subString(index + skip); } return sb.append(tmp).toString(); } </code></pre>
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    1. POreplace ". " with "\n\n" within a string in Android
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    1. COThere is ZERO probability of a bug that major still existing in the Android code-base. In fact, almost ZERO probability that it ever existed in the first place.
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    2. CO@StephenC I agree; then what?
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    3. COIt is a bug in the Question. Either he's given us the wrong code, or it doesn't behave how he thinks/claims it does. Most likely the former.
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