why two error reports when include() and require() failed in php

Question

i have a script like this :

<html>
<body>

<?php
require("wrongFile.php");
echo "Hello World!";
?>

</body>
</html>

now if wrongfile not exists the two below error will report:

Warning: require(wrongFile.php) [function.require]: failed to open stream: No such file or directory in C:\home\website\test.php on line 5

Fatal error: require() [function.require]:Failed opening required 'wrongFile.php' (include_path='.;C:\php5\pear') in C:\home\website\test.php on line 5

i know that if not exists file in require() function a fatal error displayed that cause script execution stop but do not understand what is first error

the first error display in using of include() function too

You can also turn off warnings all together, stackoverflow.com/questions/1645661/… — locrizak
– locrizak, Commented Nov 7, 2011 at 15:57
@everyone: do not format error messages as a code. you're making them worse than just plain text. — Your Common Sense
– Your Common Sense, Commented Nov 7, 2011 at 16:18

CodeCaster · Accepted Answer · 2011-11-07 15:57:00Z

6

It's pretty obvious, isn't it? :-)

The first warning is thrown from within the require() function, when it tries to open the file you specified. It can't open a stream, which is used to read the file, because the file doesn't exist.

Because of that, the function call itself also fails, which explains the 'Fatal error'.

answered Nov 7, 2011 at 15:57

CodeCaster

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Michael Krelin - hacker · Accepted Answer · 2011-11-07 15:56:10Z

0

The first is warning, not error. You can try using @include to suppress it. (naturally, it's of no great use with require).

answered Nov 7, 2011 at 15:56

Michael Krelin - hacker

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1 Comment

Your Common Sense Over a year ago

why, why this answer was ever upvoted? What can be even a slightest reason?

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