How can I grep for every uneven match? - regex

Question

Markdowns fenced code blocks look like this
```
Here is the code
in many lines
```
or like this:
```text
Here is the code
in many lines
```
The "text" specifies the language which should be used for highlighting.
I want to run over a flat directory and find all files which contain fenced code blocks without a specified language. How can I find fenced code blocks without a specified language?
What I tried
The following is a superset of what I want:
$ grep -rIE -m1 "\`\`\`[[:space:]]*$" *
The problem is the closing part. Essentially this finds all files which have a fenced code block at all. But how do I grep for every uneven triple backtick?
My guess is that I have to grep for the complete code block. It is guaranteed that there is either a newline after the triple backticks or a language.
So I tried the following two:
grep -rIzPo -m1 "\`\`\`\\n(.*?)\`\`\`" *
grep -rIzEo -m1 "\`\`\`\\n(.*?)\`\`\`" *
It found a couple of cases, but it missed at least one. I have no idea why.
Problem: Two codeblocks
I have many files with multiple code blocks, e.g:
```python
a = "Hello"
b = "Stackoverflow"
print(f"{a} {b}")
```
and
```python
print("foobar")
```
Please note that I don't want a file with this content to match! All regexes I tried so far match
```
and
```python
print("foobar")
```

Answer 1

I think that'd be easier with gawk.
awk 'BEGINFILE{f=0} /^```/{f=!f}
f&&/^```\s*$/{print FILENAME;nextfile}' *
f denotes whether last match was even or uneven. It is reset at the beginning of each file, and negated by each match. When f is 1 and exit condition (i.e current line is three backquotes followed by zero+ spaces) is met, the program prints filename and moves on to the next file.