Output showing too early - bash

Question

I made a small Bash-script to make my life easier. But I encountered a problem which I can't fix.
What I want
I've made a small script which will check for php-errors in a file each time that file gets saved/changed. This is done without me needing to run a command each time. So I run the Bash-script once on my second screen, and than each time when I save my PHP-file on screen one; I get the eventual errors shown on screen two automatically.
Basic algorithm
Get the hash of the file
Compare it to it's previous hash
If it differs, the file is changed/saved: Check if there are errors using the php -l command
Print out the result from php -l
Problem:
The result from php -l gets printed out before my code asked for it.
Code
#!/bin/bash
#Declaring basic variables here
fileToCheck="$1"
oldHash=("")
checksum=("")
#Function to get a striped line as long as the terminal width
function lineAmount {
cwidth=`tput cols`
lines=""
for i in $(seq $(expr $cwidth - 33)); do lines="$lines-";done
echo $lines
}
#Function to show the actual error
function showError {
msg=$1
time=`date +"%c"`
l=$(lineAmount)
if [ "$msg" == "No" ]
then
msg="No errors detected."
fi
printf "\n\n$time $l \n$msg\n"
}
#Start-screen------------------------------------------------
printf "Starting session for $1 at $time \n$(lineAmount)\n"
if [ ! -f $1 ]
then
echo "[Error] File $1 not found."
exit
fi
printf "\n\n\n"
#------------------------------------------
#Printing out error when file changed
while true
do
sleep 0.6
checksum=($(sha256sum $fileToCheck))
checksum=${checksum[0]}
if [ "$checksum" != "$oldHash" ]
then
error=$(php -l $fileToCheck)
oldHash=$checksum
showError $error
fi
done
Test file (test.php):
<?php
function foo() {
}
?>
Output of script:
Starting session for /home/name/Desktop/test.php at
-----------------------------------------------
Thu 11 Aug 2016 08:16:15 PM CEST -----------------------------------------------
No errors detected.
Now, in test.php I delete line 4:
<?php
function foo() {
?>
This will of course give an error, and my script shows that error:
Starting session for /home/name/Desktop/test.php at
-----------------------------------------------
Thu 11 Aug 2016 08:16:15 PM CEST -----------------------------------------------
No errors detected.
PHP Parse error: syntax error, unexpected end of file in /home/name/Desktop/test.php on line 6
Thu 11 Aug 2016 08:19:37 PM CEST ----------------------------------------------------------------------------------
Parse
But like you can see, this is not a nice output.
PHP Parse error: syntax error, unexpected end of file in /home/name/Desktop/test.php on line 6 should be printed below the second dotted line. Not below "No errors found." (The first output).
Expected output:
Starting session for /home/name/Desktop/test.php at
-----------------------------------------------
Thu 11 Aug 2016 08:16:15 PM CEST -----------------------------------------------
No errors detected.
Thu 11 Aug 2016 08:19:37 PM CEST ----------------------------------------------------------------------------------
PHP Parse error: syntax error, unexpected end of file in /home/name/Desktop/test.php on line 6
I tried a lot, I tried to change my algorithm a bit, searched up a lot; but it ain't working.
I guess the problem is somewhere on line 51, or 29. But I really can't see what's wrong.
Thanks!

Answer 1

Here's a stripped down and simplified version of your problem:
Why does this print an error message immediately instead of assigning it to the variable?
$ error=$(php -l test.php)
PHP Parse error: syntax error, unexpected end of file in test.php on line 5
php -l prints error messages to stderr like a good Unix citizen should. $(..) only captures stdout.
If you want to capture stdout and stderr together, you can use:
error=$(php -l $fileToCheck 2>&1)
You should also quote your variables so that the message is passed as a single parameter, since you're currently throwing away most of it (shellcheck is helpful):
showError "$error"
Being a good citizen, php also returns a useful exit code, so instead of trying to match a "No" to see if it's successful, you can just check the status directly:
if error=$(php -l $fileToCheck 2>&1)
then
echo "No problems"
else
echo "It failed with these messages: $error"
fi