return curl code in php

Question

I am trying to find out if a curl call was successful or not in php, the equivalent in bash would be:

`eval curl -f --data $payload http://localhost/service`
if [ $? -ne 0 ]; then
    echo passed
else
    echo failed
fi

so far I have this in php but it does not work:

    `eval curl -f --data $PAYLOAD "http://localhost/service"`;
    if ( $? == 0 )
        echo fail;
    else
        echo passed;

however I get the following error PHP Parse error: syntax error, unexpected '?', expecting variable (T_VARIABLE) or '$'

the payload looks like the folioing:

    $PAYLOAD="'payload={ 
    \"firstname\":\"$firstname\",
    \"id\":\"$id\", 
    \"author_lastname\":\"$lastname\"
    }' "

thanks in advance!

`$?` is not a valid variable identifier. Your entire PHP code is everything but valid PHP code. — N.B., Jul 29 '15 at 10:20

score 4 · Answer 1 · answered Jul 29 '15 at 10:17

4

Why not just use PHP's curl library?

https://secure.php.net/manual/en/ref.curl.php

Then you can use the built in error functions to catch exceptions.

answered Jul 29 '15 at 10:17

DrJohnZoidberg

91
1
8

score 1 · Accepted Answer · edited May 23 '17 at 11:43

1

As user2405162 mentioned it'd be better to use PHP's curl library, but if you really want to use shell instead, do it like this:

exec("curl -f --data $PAYLOAD \"http://localhost/service\"", $output, $status);
// constant on left side, to catch assignment 
// instead of comparison typos
if ( 0 == $status )
    echo "fail";
else
    echo "passed";

Backquote syntax only returns output in PHP, no status code is provided. Yet you can try some workarounds:

$status = `curl -f --data $payload http://localhost/service > /dev/null; echo $?`
if ( $status == 0 )
    echo "fail";
else
    echo "passed";

As mentioned in this question: getting output and exit status from shell_exec()

edited May 23 '17 at 11:43

Community

1
1

answered Jul 29 '15 at 10:23

ptkoz

2,388
1
20
28

this almost works! But I am seeing errors like : eval: id:40,: not found, sh: 21: eval: type:11,: not found - these are fields in the payload, the payload worked fine in my example above – godzilla Jul 29 '15 at 10:46
Can you post contents of the $payload variable? – ptkoz Jul 29 '15 at 10:47
sure, i have added it to the question above – godzilla Jul 29 '15 at 10:49
1

it's the eval causing the issue, removing it fixes the problem, but i do need the status of that curl call ... – godzilla Jul 29 '15 at 10:57
So use exec without eval - command status is saved to it's third argument. – ptkoz Jul 29 '15 at 11:00
working now, thanks!! – godzilla Jul 29 '15 at 11:01

return curl code in php

2 Answers2