CODEWITHSUNDEEP
phpif-statementecho
Paul
Paul

Reputation: 3368

Setting the result of an if statement

I am displaying a user's profile image. I have created an if statement to post a default profile image if a user has not updated their own. This is all working, but what I cannot figure out is how to echo or call for each without getting an error for the one not set.

For instance, if they do have a profile picture set, it posts fine, but then I get an error that the other variable is not defined and vise versa.

How should I be calling for this or what changes should I make in my code?

        $pics = array();
        while ($stmt->fetch()) {
            $pics[] = $profilePic;
        }
        if ($profilePic === NULL) {
            $default_profile_img = '<img class="welcome-pic" src="profile_images/default.jpg">';
        } else {
            $set_profile_img = '<img class="welcome-pic" src=" '.$profilePic.'">';
        }
  }
?>
<nav id="nav-panel">
    <div id="nav-container">
            <div id="welcome">
                <?php echo $default_profile_img; 
                echo $set_profile_img; ?>

EDIT:

How profilepic gets defined:

$sql = "
  SELECT *
  FROM profile_img
  WHERE user_id = ?
  ORDER BY id DESC LIMIT 1
  ";
if ($stmt = $con->prepare($sql)) {
        $stmt->bind_param("s", $user_id);
        $stmt->execute();
        if (!$stmt->errno) {
            // Handle error here
        }
        $stmt->bind_result($id, $user_id, $profilePic);

Upvotes: 1

Views: 41

Answers (3)

Eyy
Eyy

Reputation: 135

Just add $default_profile_img = null; and $set_profile_img = null; at the top of php code. 

$default_profile_img = null;
$set_profile_img = null;
$pics = array();
    while ($stmt->fetch()) {
        $pics[] = $profilePic;
    }
    if ($profilePic === NULL) {
        $default_profile_img = '<img class="welcome-pic" src="profile_images/default.jpg">';
    } else {
        $set_profile_img = '<img class="welcome-pic" src=" '.$profilePic.'">';
    }
}
?>
<nav id="nav-panel">
<div id="nav-container">
        <div id="welcome">
            <?php echo $default_profile_img; 
            echo $set_profile_img; ?>

Upvotes: 1

Rax Weber
Rax Weber

Reputation: 3780

You just need to initialize the variables before the if..else statements so that it won't be undefined when you try to echo both of them.

$profile_img = "";
$default_profile_img = "";

if (...

Upvotes: 1

Jose Luis T.R.
Jose Luis T.R.

Reputation: 31

Try this code. There is no need to use two different variables. This way, you won't get the warning.

$pics = array();
while ($stmt->fetch()) {
    $pics[] = $profilePic;
}
if (!isset($profilePic) OR $profilePic === NULL) {
    $profile_img = '<img class="welcome-pic" src="profile_images/default.jpg">';
} else {
    $profile_img = '<img class="welcome-pic" src=" '.$profilePic.'">';
}
}
?>
<nav id="nav-panel">
    <div id="nav-container">
        <div id="welcome">
            <?php echo $profile_img; ?>

Upvotes: 0

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