How to create a regex pattern for a specific pattern in json file in scala

Question

I am trying to create a regex pattern to match a pattern in a Json file. Json file consists of the following kind of patterns - Examples

“raw”: “”\\""
“raw”: “true”
“raw”: “”’""

raw attribute can have any type of value between double quotes. I want to match all the patterns like this and replace it with “*” of the same length at that place.

I created a pattern “”"“raw”:(".*")""".r which works fine normally in editor but not in scala and it gives the complete string after raw.

How can I achieve this?

Wiktor Stribiżew · Accepted Answer · 2019-03-17 11:26:49Z

1

Replacing a part of a matched pattern with asterisks of the same legnth as the pattern part, you may use a solution like

val s = "Text here, \"raw\": \"Remove\" and here"
val rx = "(\"raw\":\\s*\")([^\"]+)(\")".r
val res = rx.replaceAllIn(s, m => m.group(1) + ("*" * m.group(2).length) + m.group(3))
println( res ) // => Text here, "raw": "******" and here

The regex is

(\"raw\":\\s*\")([^\"]+)(\")
|___ Group1 ___||_ G2 _||G3|

It matches and captures into Group 1 (thanks to the capturing parentheses) "raw": and then 0+ whitespaces (with \s*), then captures into Group 2 any one or more chars other than ", and then captures into Group 3 a double quotation mark.

With the help of ReplaceAllIn, you can pass the match data into a lambda expression where you may manipulate the match before the replacement occurs. So, m is the match object, m.group(1) is Group 1 value, m.group(2).length is the length of Group 2 value and m.group(3) here holds the " char, Group 3 value.

answered Mar 17, 2019 at 11:26

Wiktor Stribiżew

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1 Comment

ankur gupta Over a year ago

Thanks for informing me about this. I rarely post questions on stack overflow , therefore was not aware of it. I will take care of these things now onwards.

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