result of "less than" operator in bash script - bash

Question

I'm trying to do a script which downloads a file and if it takes less than 'x' seconds it's a pass but from some reason I can't get it to work properly, i've tried '<=' '-lt' in the below example it's always 'FAST' i'e 2.09 FAST
#!/bin/bash
file=&(time -p wget -O /dev/null -q http://site/file.iso ) 2>&1 | grep real | sed -e s/real//g -e s/' '//g
if [ $file <= 1 ]
then
echo "FAST"
else
echo "SLOW"
fi

Answer 1

The bash builtin time outputs directly to your terminal, not to a stdio channel.
You'll need to use /bin/time which uses stderr:
$ time -p sleep 1 >/dev/null 2>&1
real 1.00
user 0.00
sys 0.00
$ /bin/time -p sleep 1
real 1.00
user 0.00
sys 0.00
$ /bin/time -p sleep 1 2>/dev/null
$
So:
$ command time -p sleep 1 2>&1 | awk -v limit=0.5 '$1 == "real" {exit ($2 <= limit)}'
$ echo $?
0
$ command time -p sleep 1 2>&1 | awk -v limit=1.5 '$1 == "real" {exit ($2 <= limit)}'
$ echo $?
1
and then
limit=1 # 1 second
if command time -p wget -O /dev/null -q http://site/file.iso |
awk -v lmt=$limit '$1 == "real" {exit ($2 <= lmt)}'
then
echo "FAST"
else
echo "SLOW"
fi

Answer 2

To capture the output of time you need curly braces:
file=$( { time -p wget -O /dev/null -q http://site/file.iso ; } 2>&1 | grep real | sed -e s/real//g -e s/' '//g )
Bash can't do floating-point calculations, so you can use bc for that:
[[ $( echo $file' <= 1.0' | bc ) == 1 ]] && echo FAST || echo SLOW