Remove markdown code block from python string

Question

I'm trying to remove markdown code blocks from the following string:

Problem with encoding a very large BigInteger or BigDecimal without fraction

We have an issue to encode a very large BigDecimal. For example, when I tried to encode _7.533938258014959827307132527342E+545_ I got the following error:

``` java
Error com.n1analytics.paillier.EncodeException: Input value cannot be encoded.
at com.n1analytics.paillier.StandardEncodingScheme.encode(StandardEncodingScheme.java:115)
at com.n1analytics.paillier.StandardEncodingScheme.encode(StandardEncodingScheme.java:239)
...
```

The reason why I got the error is as follows:

I've tried:

txt = re.sub('(```[a-z]*\n[\s\S]*?\n```)', '', txt) # https://regex101.com/r/aA5bI3/3
txt = re.sub('(```.+?```)', '', txt)  # https://regex101.com/r/aA5bI3/3

Even though I saw some of these regex at https://coderwall.com/p/r6b4xg/regex-to-match-github-s-markdown-code-blocks and https://regex101.com/r/aA5bI3/3, I had no success

Kirill Polishchuk · Accepted Answer · 2020-09-29 00:51:10Z

2

I believe you can simply use this regex

```.*?```

with single-line option enabled

Example: https://regex101.com/r/fubH7e/1

answered Sep 29, 2020 at 0:51

Kirill Polishchuk

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The fourth bird · Accepted Answer · 2020-09-29 09:45:06Z

The first pattern almost works, but in the example data there is a space before java

If the spaces and the chars a-z are the only acceptable characters after the backticks and they are optional, you could match optional whitespace chars without a newline [^\S\r\n]*

The pattern could look like

```[^\S\r\n]*[a-z]*\n.*?\n```

Instead of using [\s\S]*? you can use the re.DOTALL flag to make the dot match a newline.

Regex demo | Python demo

For example

txt = re.sub(r"```[^\S\r\n]*[a-z]*\n.*?\n```", '', txt, 0, re.DOTALL)

If the backticks always begin at the start of the string, and also end at the start of the string, you could make the pattern a bit more efficient matching all the lines in between that do not start with 3 backticks using a negative lookahead to prevent unnecessary backtracking.

As the pattern uses an anchor, you should use the re.MULTILINE flag.

^```[^\S\r\n]*[a-z]*(?:\n(?!```$).*)*\n```

Explanation

^ Start of string
``` Match 3 backticks
[^\S\r\n]*[a-z]* Match optional spaces without a newline and optional chars a-z
(?: Non capture group
- \n(?!```$) Match a newline and assert that the line does not start with 3 backticks
- .* If that is the case, match the whole line
)* Close non capture group and repeat 0+ times to match all lines
\n``` Match a newline and 3 backticks

Regex demo | Python demo

For example

txt = re.sub(r"^```[^\S\r\n]*[a-z]*(?:\n(?!```$).*)*\n```", '', txt, 0, re.MULTILINE)

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