PHP: syntax error, unexpected 'else'

Question

I got a syntax error when trying to set an username in my header, handling the exception when no first name and last name is seated, using the username to fill the label, surely is a stupid error. I'm using Quadodo Login Script in my project by the easily implementing.

Here is the code:

if(isset($qls->user_info['firstname'] { 
  $realname = "$qls->user_info['firstname'] . ' ' . $qls->user_info['lastname']"
} else { 
  $realname = "$qls->user_info['username']"
};

Thanks guys for all your answers, as expected was a distraction error! — andreaem, Jan 30 '16 at 18:05

score 1 · Accepted Answer · edited Jan 30 '16 at 18:06

1

You have forget semicolon as well as ending bracket in your code.

 // add two ending brackets in below line.
    if(isset($qls->user_info['firstname'])) { 
      // add semicolon in this line
      $realname = "$qls->user_info['firstname'] . ' ' . $qls->user_info['lastname']";
    } else { 
       // add semicolon in this line
      $realname = "$qls->user_info['username']";
    } //removed semicolon

edited Jan 30 '16 at 18:06

Charlotte Dunois

4,638
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answered Jan 30 '16 at 17:56

Ravi Hirani

6,511
1
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42

1

You missed two closing braces for isset – Gagan Upadhyay Jan 30 '16 at 17:58
@GaganUpadhyay: Thanks brother. I have edited my answer with comments. – Ravi Hirani Jan 30 '16 at 18:01
Thanks, i'm feeling stupid! maybe i must to do a pause when coding a lot! – andreaem Jan 30 '16 at 18:02
@Ravi - remove semicolon after else – Gagan Upadhyay Jan 30 '16 at 18:04
@charlotte: Thanks for your edit. – Ravi Hirani Jan 30 '16 at 18:09

score 0 · Answer 2 · answered Jan 30 '16 at 17:56

0

if(isset($qls->user_info['firstname'])){ 
  $realname = "$qls->user_info['firstname'] . ' ' . $qls->user_info['lastname']";
} else { 
  $realname = "$qls->user_info['username']";
}

try this

answered Jan 30 '16 at 17:56

Gagan Upadhyay

255
2
11

score 0 · Answer 3 · answered Jan 30 '16 at 17:57

0

Well this is because you realy messed up this code. Here is good version:

if(isset($qls->user_info['firstname'])) {
$realname = $qls->user_info['firstname'] . ' ' . $qls->user_info['lastname'];
} else {
$realname = $qls->user_info['username'];
};

I would suggest to use some IDE which will tell you when you have syntax error

answered Jan 30 '16 at 17:57

Radek Adamiec

502
3
10

I'm using C9IDE, but i'm coding for 5 hours whit out a pause! thanks – andreaem Jan 30 '16 at 18:03

PHP: syntax error, unexpected 'else'

3 Answers3