Awk how to print the next line of a specific expression but only of it contains a number

Here is a test.txt file

sometext
someothertext
...
lol
74
hhhhh
lol
789AA

I want a bash command that only displays:

74

For this I have tried:

awk '$0 == "lol" {i=1;next};i && i++ <= 1' test.txt

which for now displays

74
789AA

So, how can I augment this command so that it only displays a line that only contains a number (74) ?

Thanks

>Solution :

This awk should work for you:

awk 'p == "lol" && $0+0 == $0; {p = $0}' file

74

$0+0 == $0 will return true only when $0 contains a number only.