CODEWITHSUNDEEP
phpparsing
DLA
DLA

Reputation: 79

What does this T_IF error mean?

<?php
$to = "[email protected]";
$subject = "Reparasjon av " . $_REQUEST['type'] . " fra mysite.no";
$types = if(!empty($_REQUEST['type'])) {echo($_REQUEST['type'] . ". ");};
$reps = if(!empty($_REQUEST['rep'])) {echo($_REQUEST['rep']);};
$message =  $types . . $reps . "\n\nKommentarer:\n" . $_REQUEST['kommentarer'] . "\n\nFra:\n" . $_REQUEST['navn'] . "\nTelefon: " . $_REQUEST['telefon'] . "\nEmail: " . $_REQUEST['email'] . "\nBosted: " . $_REQUEST['bosted'];
$headers =  "From: " . $_REQUEST['email'] . "\r\n" . 'MIME-Version: 1.0' . "\r\n" . 'Content-type: text/plain; charset=UTF-8' . "\r\n";
if (mail($to, '=?UTF-8?B?'.base64_encode($subject).'?=', $message, $headers)) {
   header( 'Location: http://www.mysite.no/' );
  } else {
   header( 'Location: http://www.mysite.no/' );
  }
?>

It says there's a T_IF error in line 4. What's the problem?

Upvotes: 0

Views: 279

Answers (5)

user229044
user229044

Reputation: 239402

The lines

$types = if(!empty($_REQUEST['type'])) {echo($_REQUEST['type'] . ". ");};
$reps = if(!empty($_REQUEST['rep'])) {echo($_REQUEST['rep']);};

are invalid. if statements are not expressions in PHP; they don't evaluate to a value which can be assigned to a variable. You're also not "returning" anything from the if; echo writes to the screen, it doesn't "echo" a value out of an if statement to the calling scope.

You want the following:

if(!empty($_REQUEST['type'])) {
  $types = ($_REQUEST['type'] . ". ");
}

Upvotes: 0

knittl
knittl

Reputation: 265574

The first thing I can see is in line $message = … there is a double concatenation operator, which is clearly a syntax error. Should be (and should use escaped output):

$message =  $types . $reps . "\n\nKommentarer:\n" . $_REQUEST['kommentarer'] . "\n\nFra:\n" . $_REQUEST['navn'] . "\nTelefon: " . $_REQUEST['telefon'] . "\nEmail: " . $_REQUEST['email'] . "\nBosted: " . $_REQUEST['bosted'];

ps. god, so much wrong with this code (still no escaping/sanitazation in place) …

<?php
$to = "[email protected]";
$subject = "Reparasjon av " . $_REQUEST['type'] . " fra mysite.no";
$types = !empty($_REQUEST['type']) ? $_REQUEST['type'] . ". " : '';
$reps = !empty($_REQUEST['rep']) ? $_REQUEST['rep'] : '' ;
$message =  $types . $reps . "\n\nKommentarer:\n" . $_REQUEST['kommentarer'] . "\n\nFra:\n" . $_REQUEST['navn'] . "\nTelefon: " . $_REQUEST['telefon'] . "\nEmail: " . $_REQUEST['email'] . "\nBosted: " . $_REQUEST['bosted'];
$headers =  "From: " . $_REQUEST['email'] . "\r\n" . 'MIME-Version: 1.0' . "\r\n" . 'Content-type: text/plain; charset=UTF-8' . "\r\n";
if (mail($to, '=?UTF-8?B?'.base64_encode($subject).'?=', $message, $headers)) {
   header( 'Location: http://www.mysite.no/' );
  } else {
   header( 'Location: http://www.mysite.no/' );
  }
?>

Upvotes: 0

Marc B
Marc B

Reputation: 360762

if() is a language construct, not a function. It does not return anything, and cannot be assigned to a variable.

$types = if(!empty($_REQUEST['type'])) {echo($_REQUEST['type'] . ". ");};
^^^^^^^^--- not allowed

Try:

if (!empty($_REQUEST['type']) {
   $types = $_REQUEST['type'];
}

As well, echo cause direct output to the client. it does not "return" anything that can be assigned.

Upvotes: 2

deceze
deceze

Reputation: 522412

You can't use if there, it's a syntax error. Technically if is a statement, not an expression. That means you can't use it in an assignment like $types = if (...).

Upvotes: 3

therin
therin

Reputation: 1468

An IF statement does not return a value, so assigning it to a variable does nothing (and might even be causing your error!) Take off the semicolons from the end of the if statements, too.

Try this:

if (!empty($some_variable)) {
  $my_var = $some_variable;
}

Upvotes: 1

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