Sorting a file containing line delimiters - bash

Question

I have a text file looking like
text_a_3 xxx yyy
- - - - - - - - - - -
text_b_2 xyx zyz
- - - - - - - - - - -
text_b_3 xxy zyy
- - - - - - - - - - -
text_a_2 foo bar
- - - - - - - - - - -
text_a_1 foo bla
- - - - - - - - - - -
text_b_1 bla bla
I want to sort this file numerically, based on the first field, so that my output would look like:
text_a_1 foo bla
- - - - - - - - - - -
text_a_2 foo bar
- - - - - - - - - - -
text_a_3 xxx yyy
- - - - - - - - - - -
text_b_1 bla bla
- - - - - - - - - - -
text_b_2 xyx zyz
- - - - - - - - - - -
text_b_3 xxy zyy
I thought sort would do the job. I thus tried
sort -n name_of_my_file
sort -k1 -n name_of_my_file
But it gives
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - -
text_a_1 foo bla
text_a_2 foo bar
text_a_3 xxx yyy
text_b_1 bla bla
text_b_2 xyx zyz
text_b_3 xxy zyy
The option --field-separator is not of any help.
Is there any way to achieve this with a one-line, sort based command ?
Or is the only solution to extract text containing lines, sort them, and insert line delimiters afterwards ?

Answer 1

Using GNU awk only, and relying with internal sort function asort() and record separator set to dashes line:
awk -v RS='- - - - - - - - - - -\n' '
{a[++c]=$0}
END{
asort(a)
for(i=1;i<=c;i++)
printf "%s%s",a[i],(i==c?"":RS)
}' name_of_my_file
The script first fills the content of the input file into the array a. When the file is read, the array is sorted and then printed with the same input record separator.

Answer 2

When the line delimiters are all on the even lines, you can use
paste -d'\r' - - < yourfile | sort -n | tr '\r' '\n'
I actually prefer removing the delimiters in front, sort and add them afterwards, so please reconsider your requirements:
grep -Ev "(- )*-" yourfile | sort -n | sed 's/$/\n- - - - - - - - - - -/'

Answer 3

Following sort + awk may help you.
sort -t"_" -k2 -k3 Input_file | awk '/^-/ && !val{val=$0} !/^-/{if(prev){print prev ORS val};prev=$0} END{print prev}'
Adding a non-one liner form of solution too now.
sort -t"_" -k2 -k3 Input_file |
awk '
/^-/ && !val{
val=$0}
!/^-/{
if(prev){
print prev ORS val};
prev=$0
}
END{
print prev
}'