Bash脚本无效的空间符号

Question

My goal is it is to gate time in second for last modified file in catalog.

I have catalog with space in name and want to stat it with script. With a terminal I simply use:

sudo stat -c %Y /var/www/pre/data/jko/files/My\ Files

and works perfectly. In my bash script i read files one by one from directory /var/www/pre/data/jko/files/ with (this is inside while loop)

touch tempFile.txt
#ls sort by last modified date
sudo ls -rt /var/www/pre/data/$line/files/ > tempFile.txt 
# read newest file
outputFile=$(tail -1 tempFile.txt) 
# replace all spaces with "\ " sign
outputFile=$(echo ${outputFile// /"\ "})
outputDirectoryToFile=/var/www/pre/data/$line/files/$outputFile
echo $outputDirectoryToFile
expr `sudo date +%s` - `sudo stat -c %Y $outputDirectoryToFile`

And if i fire this script with bash script.sh i got

/var/www/pre/data/jko/files/My\ Files //line from echo
stat: cannot stat '/var/www/pre/data/jko/files/My\': No such file or directory
stat: cannot stat 'Files': No such file or directory
expr: syntax error

Or maybe there is simpler solution

Answer 1

Correctly, and efficiently, finding the latest file in a directory

{ read -r -d ' ' mtime && IFS= read -r -d '' filename; } \
  < <(find /directory -type f -printf '%T@ %p\0' | sort -z -r -n)

...will put the time in seconds for the most-recently-modified file in the shell variable mtime , and the name of this file in the shell variable filename .

Moreover, it will work even with files with surprising or intentionally malicious names -- filenames with newline literals in their names, filenames with glob characters, etc. I have a more complete explanation of this idiom and why it works here .

---

Why your original code didn't work

Now, what was wrong with your original code? In short: Literal quotes do not substitute for syntactic quotes . Let's go into what this means.

In /var/www/pre/data/jko/files/My\\ Files , the backslash is syntactic : It's shell syntax. When you run stat /var/www/pre/data/jko/files/My\\ Files , the result is a syscall that looks like the following:

execv("/usr/bin/stat", "stat", "/var/www/pre/data/jko/files/My Files")

Note how the backslash is gone? That's because the shell consumed it when it was parsing the string.

The following are all exactly identical:

# each of these assigns the same string to the variable named s
s=Hello\ World
s=Hello" "World
s='Hello World'

...and they can be expanded as follows:

# this passes the *exact* contents of that variable as an argument to stat, and thus tries
# to stat a file named Hello World (no literal quotes in the name):
stat "$s"

In the above, the quotes are again syntactic -- they're telling the shell not to split the result of the variable expansion into multiple separate words or evaluate it as a glob -- not literal , in which case they would be passed to stat as part of its argument.

---

So, what happens when you run s=${s// /'\\ '} ? You're putting literal backslashes into the data.

Thereafter:

s="Hello World"
s=${s// /'\ '}    # change string to 'Hello\ World'
stat "$s"         # try to stat a file named 'Hello\ World', with the backslash 

What happens if you leave out the syntactic double quotes on the expansion? It's even uglier:

s="Hello World"
s=${s// /'\ '}    # change string to 'Hello\ World'
stat $s           # try to stat a file named 'Hello\', and a second file named 'World'

That's because unquoted expansion doesn't run through all shell parsing steps; it only does field splitting and glob expansion.

Answer 2

If you paths contain spaces, quote them all to avoid problems:

touch tempFile.txt
sudo ls -rt "/var/www/pre/data/$line/files/" > tempFile.txt //ls sort by last modified date
outputFile=$(tail -1 tempFile.txt) // read newest file
outputDirectoryToFile="/var/www/pre/data/$line/files/$outputFile"
echo $outputDirectoryToFile
expr `sudo date +%s` - `sudo stat -c %Y "$outputDirectoryToFile"`