Reverse order of html with awk via line swapping - bash

Question

Basically every week I have to reverse the following snippet
<!-- Homepage Slider Begin -->
<div class="container-fluid">
<div class="single-item-home hidden-xs">
<div class="slide slide--has-caption">
<a href="/1">
<img src="/sliders/1_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/2">
<img src="/sliders/2_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/3">
<img src="/sliders/3_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/4">
<img src="/sliders/4_example.jpg">
</a>
</div>
</div>
</div>
<!-- Homepage Slider End -->
Basically I'm wanting to make awk script and have a cron job to essentially take lines 4-8 to swap with lines 22-26 and lines 10-14 swap with lines 16-20 however I can only seem to find a way to swap one line and not line blocks.
Is this even possible with awk or just silly?

Answer 1

You may use awk . Below script
awk 'NR==FNR{line[i++]=$0}
END{
for(j=0;j<i;j++){
if(j>=3 && j<=7){
print line[j+18];
continue;
}
else if(j>=21 && j<=25){
print line[j-18];
continue;
}
else if(j>=9 && j<=13){
print line[j+6];
continue;
}
else if(j>=15 && j<=19){
print line[j-6];
continue;
}
print line[j];
}
}' file
will do what you want.
Sample Output
<!-- Homepage Slider Begin -->
<div class="container-fluid">
<div class="single-item-home hidden-xs">
<div class="slide slide--has-caption">
<a href="/4">
<img src="/sliders/4_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/3">
<img src="/sliders/3_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/2">
<img src="/sliders/2_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/1">
<img src="/sliders/1_example.jpg">
</a>
</div>
</div>
</div>
<!-- Homepage Slider End -->
Note: I leave the array-bounds check up to you. If the content of the file is static, you may not need this

Answer 2

This doesn't care how many lines are in each block or where they start/end in the file and it doesn't require you to store the whole file in memory (though most of the file is the "slides" which DO need to be stored so that's probably a non-issue):
$ cat tst.awk
/<div class="slide/ { inSlide=1; slide="" }
inSlide {
slide = slide $0 ORS
if ( /<\/div>/ ) {
slides[++numSlides] = slide
inSlide = 0
}
next
}
/<\/div>/ {
for (slideNr=numSlides; slideNr>=1; slideNr--) {
printf "%s", slides[slideNr]
}
numSlides = 0
}
NF
.
$ awk -f tst.awk file
<!-- Homepage Slider Begin -->
<div class="container-fluid">
<div class="single-item-home hidden-xs">
<div class="slide slide--has-caption">
<a href="/4">
<img src="/sliders/4_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/3">
<img src="/sliders/3_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/2">
<img src="/sliders/2_example.jpg">
</a>
</div>
<div class="slide slide--has-caption">
<a href="/1">
<img src="/sliders/1_example.jpg">
</a>
</div>
</div>
</div>
<!-- Homepage Slider End -->

Answer 3

perl -e '#f=<>; print #f[0..2,21..25,8,15..19,14,9..13,20,3..7,26..$#f]' ip.html
-e option to pass Perl code from command line itself
#f=<> Reads the contents of file (passed as command line argument) into an array
and then print as per order required (index starts from 0, $#f gives last index of array #f)

Answer 4

This is a solution, where you define an order where to print in the BEGIN section and in that order it will print:
$ cat > preordered.awk
BEGIN {
split("1,2,3,22,23,24,25,26,9,16,17,18,19,20,15,10,11,12,13,14,21,4,5,6,7,8",a,",")
}
{
b[(NR in a?a[NR]:NR)]=$0
}
END {
PROCINFO["sorted_in"]="#ind_num_asc"
for(i in b)
print b[i]
}
Give it a go:
$ awk -f preordered.awk' file
<!-- Homepage Slider Begin -->
<div class="container-fluid">
<div class="single-item-home hidden-xs">
<div class="slide slide--has-caption">
<a href="/4">
<img src="/sliders/4_example.jpg">
</a>
</div>
...