replace multi-line string in java

Question

Trying to replace multi line string in java using replaceAll method but it's not working. Is there anything wrong with below logic?

    String content="      \"get\" : {\n" + 
    "        \"name\" : [ \"Test\" ],\n" + 
    "        \"description\" : \"Test description to replace\",\n" + 
    "        \"details\" : \"Test details\"";


    String searchString="        \"name\" : [ \"Test\" ],\n" + 
"        \"description\" : \"Test description to replace\",";


String replaceString="        \"name\" : [ \"Actual\" ],\n" + 
"        \"description\" : \"Replaced description\",";

Tried below options and none of them worked-

Pattern.compile(searchString, Pattern.MULTILINE).matcher(content).replaceAll(replaceString);

Pattern.compile(searchString, Pattern.DOTALL).matcher(content).replaceAll(replaceString);

content = content.replaceAll(searchString, replaceString);

I have no way to test this at the moment, but you definitely need to escape the `[` brackets in your search string (change them to `\\[`). Maybe that will already fix it. — Tim Pietzcker, Apr 02 '19 at 06:46
you can quote the whole search string with `\Q` at the beginning or call `Pattern.quote()` — Sharon Ben Asher, Apr 02 '19 at 06:47
Use Something like this : String str = content.replaceAll("\\n",""); — Coder, Apr 02 '19 at 07:15
Since you are not using regular expressions, use either the simple `content = content.replace(searchString, replaceString);` or the more complex `content = Pattern .compile(searchString, Pattern.LITERAL).matcher(content).replaceAll(replaceString);`. The latter makes sense if you want to keep the result of `compile` and use it multiple times. Then, you get a benefit from the preparation, e.g. the Boyer–Moore algorithm used behind the scenes. — Holger, Apr 02 '19 at 10:06

Allan · Answer 1 · 2019-04-02T07:24:19.553

DISCLAIMER: You should not use regex to manipulate JSON or XML that have infinite nested content. Finite automate are not adapted to manipulate those data structures and you should use a JSON/XML parser instead.

This being said, purely for learning purpose, I will quickly fix your code.

1) Use either replace instead of replaceAll to avoid that your searchString is interpreted as a regex:

String content="      \"get\" : {\n" + 
            "        \"name\" : [ \"Test\" ],\n" + 
            "        \"description\" : \"Test description to replace\",\n" + 
            "        \"details\" : \"Test details\"";


String searchString="        \"name\" : [ \"Test\" ],\n" + 
        "        \"description\" : \"Test description to replace\",";


String replaceString="        \"name\" : [ \"Actual\" ],\n" + 
        "        \"description\" : \"Replaced description\",";

System.out.println(content.replace(searchString, replaceString));

output:

  "get" : {
    "name" : [ "Actual" ],
    "description" : "Replaced description",
    "details" : "Test details"

2) Or use replaceAll but escape the brackets to avoid them being interpreted as a character class definition attempt.

String searchString="        \"name\" : \\[ \"Test\" \\],\n" + 
        "        \"description\" : \"Test description to replace\",";


String replaceString="        \"name\" : [ \"Actual\" ],\n" + 
        "        \"description\" : \"Replaced description\",";

System.out.println(content.replaceAll(searchString, replaceString));

output:

  "get" : {
    "name" : [ "Actual" ],
    "description" : "Replaced description",
    "details" : "Test details"

Link: How to parse JSON in Java

You should load your json structure in an object
Change the values of the attributes of that object to the new values
Export it again in json format

replace multi-line string in java

1 Answers1