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0

0

I'm trying to print out a users image using PHP and HTML.
The user's image is saved inside the $_SESSION['usericon']

The error is on line 3.

I've tried removing the PHP tags and switched print_r out etc.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else {
?>

php html

share|improve this question

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You are in a PHP code section already, so using <?php inside of that again makes absolutely no sense. Please go read up on some basics of string concatenation. (We should not have to explain those here, this site is not a replacement for tutorials etc.)
  
  – 04FS
  Mar 28 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  You could just concat the variable, not sure if this is still the syntax (not done php since v4): echo '<img class="img-responsive" src="img/' . $_SESSION['userIcon']) . '">';
  
  – Pete
  Mar 28 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @executable I have assigned $_SESSION['userIcon'] to a variable but, when i look at the page source now i just do this: <img class="img-responsive" src="img/$userIcon">
  
  – Mikkel Hebel
  Mar 28 at 13:54
  

  
  
  
  

add a comment
 | 

0

0

I'm trying to print out a users image using PHP and HTML.
The user's image is saved inside the $_SESSION['usericon']

The error is on line 3.

I've tried removing the PHP tags and switched print_r out etc.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else {
?>

php html

share|improve this question

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You are in a PHP code section already, so using <?php inside of that again makes absolutely no sense. Please go read up on some basics of string concatenation. (We should not have to explain those here, this site is not a replacement for tutorials etc.)
  
  – 04FS
  Mar 28 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  You could just concat the variable, not sure if this is still the syntax (not done php since v4): echo '<img class="img-responsive" src="img/' . $_SESSION['userIcon']) . '">';
  
  – Pete
  Mar 28 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @executable I have assigned $_SESSION['userIcon'] to a variable but, when i look at the page source now i just do this: <img class="img-responsive" src="img/$userIcon">
  
  – Mikkel Hebel
  Mar 28 at 13:54
  

  
  
  
  

add a comment
 | 

I'm trying to print out a users image using PHP and HTML.
The user's image is saved inside the $_SESSION['usericon']

The error is on line 3.

I've tried removing the PHP tags and switched print_r out etc.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else {
?>

php html

share|improve this question

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else {
?>

share|improve this question

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

share|improve this question

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

edited Mar 28 at 14:06

executable

2,06822 gold badges1111 silver badges2727 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

asked Mar 28 at 13:43

Mikkel HebelMikkel Hebel

455 bronze badges

4 Answers
4

active

oldest

votes

1

Here is a simple solution to your problem. You need to remove the php tags and use concatenation.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon'].'">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 


?>

share|improve this answer

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @executeable Worked, thank you! :D
  
  – Mikkel Hebel
  Mar 28 at 14:03
  

  
  
  
  

add a comment
 | 

1

change

 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">

to

 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon']).'">';

share|improve this answer

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

add a comment
 | 

0

You can use something like this:

<?php
 if (isset($_SESSION['userId'])) 
 $img = sprintf('<img class="img-responsive" src="img/%s" />', $_SESSION['userIcon']);
 echo $img . ' <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 
 // @todo
 
?>

rsrs

share|improve this answer

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

add a comment
 | 

0

You have started a PHP code, within a PHP code...
This is the correct img src:

src="img/' . print_r($_SESSION['userIcon']) . '"

I would suggest you store all (temporary) images on your server, and not in a session file :) Have a nice day.

share|improve this answer

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

add a comment
 | 

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1

Here is a simple solution to your problem. You need to remove the php tags and use concatenation.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon'].'">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 


?>

share|improve this answer

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @executeable Worked, thank you! :D
  
  – Mikkel Hebel
  Mar 28 at 14:03
  

  
  
  
  

add a comment
 | 

1

Here is a simple solution to your problem. You need to remove the php tags and use concatenation.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon'].'">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 


?>

share|improve this answer

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @executeable Worked, thank you! :D
  
  – Mikkel Hebel
  Mar 28 at 14:03
  

  
  
  
  

add a comment
 | 

Here is a simple solution to your problem. You need to remove the php tags and use concatenation.

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon'].'">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 


?>

share|improve this answer

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

<?php
if (isset($_SESSION['userId'])) 
 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon'].'">
 <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 


?>

share|improve this answer

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

answered Mar 28 at 13:58

executableexecutable

2,06822 gold badges1111 silver badges2727 bronze badges

1

change

 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">

to

 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon']).'">';

share|improve this answer

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

add a comment
 | 

1

change

 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">

to

 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon']).'">';

share|improve this answer

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

add a comment
 | 

change

 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">

to

 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon']).'">';

share|improve this answer

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

 echo '<img class="img-responsive" src="img/<?php print_r($_SESSION['userIcon']); ?>">

 echo '<img class="img-responsive" src="img/'.$_SESSION['userIcon']).'">';

share|improve this answer

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

answered Mar 28 at 14:04

Danyal SandeeloDanyal Sandeelo

9,03633 gold badges2626 silver badges4646 bronze badges

0

You can use something like this:

<?php
 if (isset($_SESSION['userId'])) 
 $img = sprintf('<img class="img-responsive" src="img/%s" />', $_SESSION['userIcon']);
 echo $img . ' <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 
 // @todo
 
?>

rsrs

share|improve this answer

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

add a comment
 | 

0

You can use something like this:

<?php
 if (isset($_SESSION['userId'])) 
 $img = sprintf('<img class="img-responsive" src="img/%s" />', $_SESSION['userIcon']);
 echo $img . ' <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 
 // @todo
 
?>

rsrs

share|improve this answer

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

add a comment
 | 

You can use something like this:

<?php
 if (isset($_SESSION['userId'])) 
 $img = sprintf('<img class="img-responsive" src="img/%s" />', $_SESSION['userIcon']);
 echo $img . ' <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 
 // @todo
 
?>

rsrs

share|improve this answer

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

<?php
 if (isset($_SESSION['userId'])) 
 $img = sprintf('<img class="img-responsive" src="img/%s" />', $_SESSION['userIcon']);
 echo $img . ' <a href="settings.php">Settings</a>
 <form action="lib/logout.php" method="POST">
 <button type="submit" name="logout-submit">Logout</button>
 </form>';
 else 
 // @todo
 
?>

share|improve this answer

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

answered Mar 28 at 14:01

Lucas CampeloLucas Campelo

122 bronze badges

0

You have started a PHP code, within a PHP code...
This is the correct img src:

src="img/' . print_r($_SESSION['userIcon']) . '"

I would suggest you store all (temporary) images on your server, and not in a session file :) Have a nice day.

share|improve this answer

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

add a comment
 | 

0

You have started a PHP code, within a PHP code...
This is the correct img src:

src="img/' . print_r($_SESSION['userIcon']) . '"

I would suggest you store all (temporary) images on your server, and not in a session file :) Have a nice day.

share|improve this answer

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

add a comment
 | 

You have started a PHP code, within a PHP code...
This is the correct img src:

src="img/' . print_r($_SESSION['userIcon']) . '"

I would suggest you store all (temporary) images on your server, and not in a session file :) Have a nice day.

share|improve this answer

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

src="img/' . print_r($_SESSION['userIcon']) . '"

share|improve this answer

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges

answered Mar 28 at 14:01

Hans SchiesslHans Schiessl

922 bronze badges