Java String replace not working [duplicate]

2019-01-01 02:29发布

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String delimiter = "\\*\\*";
String html = "<html><head></head><body>**USERNAME** AND **PASSWORD**</body></html>";
Map<String, String> mp = new HashMap<String, String>();
mp.put("USERNAME", "User A");
mp.put("PASSWORD", "B");
for (Entry<String, String> entry : mp.entrySet()) {
  html.replace(delimiter + entry.getKey()+ delimiter, entry.getValue());
}

That should usually replace those both strings, but it does not. Does anyone has an idea?

6条回答
妖精总统
2楼-- · 2019-01-01 03:04

You don't need to escape * character. Difference between replace and replaceAll is that replace escapes any regex metacharacters for us automatically:

String delimiter = "**";
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人气声优
3楼-- · 2019-01-01 03:04

Analogy:

If you have an immutable file on your desktop. To make some changes you do a copy and replace. This leads to a state wherein you can not access the old file unless you have a backup.

In the same way in most computer languages like Java, JavaScript, python and C# the replace method does not replace/modify the String. It only operates over the former String and returns a new String with all the changes.

Now if you really want to store the changes you'll need to get the returned String in the same variable (if your situation permits) or in a new variable.

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余欢
4楼-- · 2019-01-01 03:08

String is immutable, which means that the html reference doesn't change, rather the replace method returns a new String object that you have to assign.

html = html.replace(delimiter + entry.getKey()+ delimiter, entry.getValue());
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大哥的爱人
5楼-- · 2019-01-01 03:16

as said you are discarding the results and replace doesn't take a regex only a literal char sequence to be replaced so you don't need to escape in the delimiter

but replaceAll and replaceFirst do take a regex string (bad design that)

and as an aside it's advisable to use Patter.quote(String) and Matcher.quoteReplacement(String) to ensure no weird things are happening when using regex (it's a bit easier and ensures there's no error in escaping the chars)

here's for when only one occurrence must be replaced

String delimiter = "**";
String html = "<html><head></head><body>**USERNAME** AND **PASSWORD**</body></html>";
Map<String, String> mp = new HashMap<String, String>();
mp.put("USERNAME", "User A");
mp.put("PASSWORD", "B");
for (Map.Entry<String, String> entry : mp.entrySet()) {
  html = html.replace(delimiter + entry.getKey()+ delimiter, entry.getValue());
}

and here's for when multiple occurrences must be replaced

String delimiter = "**";//unescaped because I'm handling that in my replace 
String html = "<html><head></head><body>**USERNAME** AND **PASSWORD**</body></html>";
Map<String, String> mp = new HashMap<String, String>();
mp.put("USERNAME", "User A");
mp.put("PASSWORD", "B");
for (Map.Entry<String, String> entry : mp.entrySet()) {
  html = html.replaceAll(Pattern.quote(delimiter + entry.getKey()+ delimiter), Matcher.quoteReplacement(entry.getValue()));
}
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人气声优
6楼-- · 2019-01-01 03:19

The replace method returns its result, which you're discarding.

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心情的温度
7楼-- · 2019-01-01 03:23

String is an immutable type,so we should new a String object to save the new String returned by the replace Method

html = html.replace(delimiter + entry.getKey()+ delimiter, entry.getValue());
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