Check wget's return value

Question

I'm writing a script to download a bunch of files, and I want it to inform when a particular file doesn't exist.

r=`wget -q www.someurl.com`
if [ $r -ne 0 ]
  then echo "Not there"
  else echo "OK"
fi

But it gives the following error on execution:

./file: line 2: [: -ne: unary operator expected

What's wrong?

score 81 · Answer 1 · answered Apr 26 '10 at 22:32

81

Others have correctly posted that you can use $? to get the most recent exit code:

wget_output=$(wget -q "$URL")
if [ $? -ne 0 ]; then
    ...

This lets you capture both the stdout and the exit code. If you don't actually care what it prints, you can just test it directly:

if wget -q "$URL"; then
    ...

And if you want to suppress the output:

if wget -q "$URL" > /dev/null; then
    ...

answered Apr 26 '10 at 22:32

Cascabel

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Is it standard that `if` will get the exit code of the program when this form is used? (`if ; then...`) – Vassilis Nov 08 '13 at 16:53

nobody · Accepted Answer · 2010-04-26T23:24:58.223

46

$r is the text output of wget (which you've captured with backticks). To access the return code, use the $? variable.

edited Apr 26 '10 at 23:24

answered Apr 26 '10 at 22:25

nobody

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While this is correct, a better explanation would be why `$r` is blank and why the error message occurs. – Engineer2021 Apr 03 '14 at 12:06

Bolo · Answer 3 · 2010-04-26T22:33:21.103

18

$r is empty, and therefore your condition becomes if [ -ne 0 ] and it seems as if -ne is used as a unary operator. Try this instead:

wget -q www.someurl.com
if [ $? -ne 0 ]
  ...

EDIT As Andrew explained before me, backticks return standard output, while $? returns the exit code of the last operation.

edited Apr 26 '10 at 22:33

answered Apr 26 '10 at 22:26

Bolo

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this is the only answer that explains why the asker is getting that error message. +1. – Lesmana Jan 02 '13 at 21:25

score 12 · Answer 4 · answered Apr 27 '10 at 12:31

you could just

wget ruffingthewitness.com && echo "WE GOT IT" || echo "Failure"

-(~)----------------------------------------------------------(07:30 Tue Apr 27)
risk@DockMaster [2024] --> wget ruffingthewitness.com && echo "WE GOT IT" || echo "Failure" 
--2010-04-27 07:30:56--  http://ruffingthewitness.com/
Resolving ruffingthewitness.com... 69.56.251.239
Connecting to ruffingthewitness.com|69.56.251.239|:80... connected.
HTTP request sent, awaiting response... 200 OK
Length: unspecified [text/html]
Saving to: `index.html.1'

    [ <=>                                                                                    ] 14,252      72.7K/s   in 0.2s    

2010-04-27 07:30:58 (72.7 KB/s) - `index.html.1' saved [14252]

WE GOT IT
-(~)-----------------------------------------------------------------------------------------------------------(07:30 Tue Apr 27)
risk@DockMaster [2025] --> wget ruffingthewitness.biz && echo "WE GOT IT" || echo "Failure"
--2010-04-27 07:31:05--  http://ruffingthewitness.biz/
Resolving ruffingthewitness.biz... failed: Name or service not known.
wget: unable to resolve host address `ruffingthewitness.biz'
zsh: exit 1     wget ruffingthewitness.biz
Failure
-(~)-----------------------------------------------------------------------------------------------------------(07:31 Tue Apr 27)
risk@DockMaster [2026] -->

Code only answer are discouraged. Please write some explanation on how it works? — foobar, Nov 29 '19 at 12:45

score 6 · Answer 5 · answered Aug 24 '16 at 11:49

Best way to capture the result from wget and also check the call status

wget -O filename URL
if [[ $? -ne 0 ]]; then
    echo "wget failed"
    exit 1; 
fi

This way you can check the status of wget as well as store the output data.

If call is successful use the output stored
Otherwise it will exit with the error wget failed

score 3 · Answer 6 · edited Feb 08 '21 at 16:38

I been trying all the solutions without lucky.

wget executes in non-interactive way. This means that wget work in the background and you can't catch the return code with $?.

One solution it's to handle the "--server-response" property, searching http 200 status code Example:

wget --server-response -q -o wgetOut http://www.someurl.com
sleep 5
_wgetHttpCode=`cat wgetOut | gawk '/HTTP/{ print $2 }'`
if [ "$_wgetHttpCode" != "200" ]; then
    echo "[Error] `cat wgetOut`"
fi

Note: wget need some time to finish his work, for that reason I put "sleep 5". This is not the best way to do but worked ok for test the solution.

Check wget's return value

6 Answers6

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