PHP showing image is not working - php

Question

Hello i am having a site which shows users their own profile picture. But i am unable to do so, I have built a code but it is not working & giving an error as Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in /home/u522159750/public_html/users/myaccount.php on line 156 i.e. the line $sql = mysqli_query($link, "SELECT id, name, avatar FROM users WHERE id="'.$id.'" ") ;Can anyone suggest me what could be the possible reasons. The code for showing image is as follow---
<p>
<?php
//We check if the users ID is defined
$db_host = "xxxxxxxxxxxx";
$db_username = "xxxxxxxxxxxxxx";
$db_password = "xxxxxxxxx";
$db_database = "xxxxxxxxxxxxxx";
$link = mysqli_connect($db_host,$db_username,$db_password) or die("Cannot connect");
mysqli_select_db($link, $db_database) or die("Cannot select the database");
$sql = mysqli_query($link, "SELECT id, name, avatar FROM users WHERE id="'.$id.'" ") ;
while($result = mysqli_fetch_object($sql)):
<img src="<?php echo $result->avatar; ?>" alt="<?php echo $result->name; ?>" width="200" height="300" />
?> </p>
<?php endwhile; ?>

Answer 1

The part where you intend to output the link with the image, is html and should either be outside of the php part, or printed, as below.
<?php
//We check if the users ID is defined
$db_host = "xxxxxxxxxxxx";
$db_username = "xxxxxxxxxxxxxx";
$db_password = "xxxxxxxxx";
$db_database = "xxxxxxxxxxxxxx";
$link = mysqli_connect($db_host,$db_username,$db_password) or die("Cannot connect");
mysqli_select_db($link, $db_database) or die("Cannot select the database");
$sql = mysqli_query($link, "SELECT id, name, avatar FROM users WHERE id="'.$id.'" ") ;
while($result = mysqli_fetch_object($sql)) {
print ' <img src="'.$result->avatar.'" alt="'.$result->name.'" width="200" height="300" />';
}
?>
</p>

Answer 2

after viewing at the browser, check the url src of each image, is it at the right path? and make sure the image is existed at the src path.

Answer 3

i will not show you a code or an example but this sugestions might be useful :
what does the avatar column in db takes ? does it take the image name or name.extension
or location/name.extension
if it takes only the name : then you must define the location in php example
echo '<a href="#" > <IMG src="location/'.$avatar.'.jpg" alt="..."
hight="200"width="200"/></a>';
but note : in this case all the avatars must be the same extension ( *.jpg )
if it takes the name.extension then you have just to define the location only
if it takes location/name.extension then you problem is not the $avatar
but still can be the avatar's name itself be sure to do not leave avatars names contains
white spaces or non english alphabet characters. you have to renam them.