CODEWITHSUNDEEP
php
spraff
spraff

Reputation: 33395

PHP fopen returning TRUE

PHP's fopen is supposed to

Returns a file pointer resource on success, or FALSE on error. 

$f = fopen ($logfile, "a");

file_put_contents("/tmp/foo", gettype($f)."--".print_r ($f), TRUE);

The fopen is printing failed to open stream: Permission denied in the Apache log, which is what I expect in this particular case, however the error-handling logic that comes after isn't working because if ($f) succeeds.

The trace in /tmp/foo tells us

boolean--1

I guess I can use is_resource to make the error handling work, but this looks like a bug in PHP, no? Or is there something else that could bring about this situation?

Upvotes: 1

Views: 1459

Answers (2)

WEBjuju
WEBjuju

Reputation: 6581

I think a minor misuse of print_r is leaving confusion. Note that if you wish to use print_r "inline" then a second parameter is required.

echo print_r($f, 1).' '.TRUE;

That's going to print:

1

The reason it prints a space and then 1 is that $f is FALSE, which in print_r prints nothing. But if you ask a string to print TRUE directly, it converts it to an integer representation, or "1".

Try this:

if ($f) {
  echo 'f!';
} else {
  echo 'no f';
}

You'll find that there is "no f" if fopen fails. Try using this test:

if ($f !== false) {

That way you'll be sure it's a resource.

Upvotes: 1

OptimusCrime
OptimusCrime

Reputation: 14863

What system are you on? It works as expected here:

<?php
$f = fopen('/var/log/mail.err', "a");
var_dump($f);
if ($f === false) {
    echo 'Not a valid stream';
}

Outputs:

<br />
<b>Warning</b>:  fopen(/var/log/mail.err): failed to open stream: Permission denied in <b>/foobar.php</b> on line <b>2</b><br />
bool(false)
Not a valid stream

Tested on Debian. PHP7.0.

Upvotes: 0

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