PHP - Display different image based on mysql row result - php

Question

Sorry for my English.
I'm learning PHP and I try to create a small member area, I think it's a good way to learn.
In my member area, some members are "verified" and some are not.
How can I display different pics based on data stored in Mysql using PHP?
What I want is to display "Picture1" if "1" is the value stored in a MySQL column and "Picture2" if "2" is the value, etc... You are "unverified member" so you see picture 1, verified member see picture 2...
I know how to SELECT data using MySQLi but I can't find what I have to do next?
Thank you.

Answer 1

Lets assume that you have a database named as db and table as tb with columns name and verify. Name is string and verify is Boolean. Also verify stored 1 if user is verified and 0 it isn't.
so, you can do it by iteration statements (either by using if else or by switch statements)
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query="SELECT * FROM `tb`";
$sql=$conn->query($query);
if($sql->num_rows>0)
{
while($row=$sql->fetch_assoc())
{
//display name
echo $row['name'];
//check if user is verified or not
//using if else statement
if($row['verify']=="0")
{
echo '<img src="./picture1.jpg">';
}
else
{
echo '<img src="./picture2.jpg">';
}
//using switch case
switch($row['verify'])
{
case(0):
{
echo '<img src="./picture1.jpg">';
}
case(1);
{
echo '<img src="./picture2.jpg">';
}
}
//remember i have used both methods hence it will show the image twice , i did it just you possible ways , you choose any one from it.(if else or switch case)
}
}
$conn->close();
?>

Answer 2

There's a few ways. If it will always follow a number system... use this:
$picture = "Picture" . $verified;
// example: $picture = "Picture1", if $verified = 1
Lots of ways to concatenate...
$pic = "Picture$verified.png";
$picture = "Picture" . $verified . ".jpeg";
It will just depends on how you have it setup..
switch($verified){
case 1:
$picture = "Picture1";
break;
case 2:
$picture = "Picture2";
break;
case default:
$picture = "Picture1";
break;
}
Another way...
if($verified == 1){ $picture = "Picture1"; } else { $picture = "Picture2"; }
The reason for the switch statement, or this if/else statement is that it can account for verified being equal to 0, null, or another non 1 or 2 value that may be stored in the database for whatever reason. With the first example, if $verified (verified column = to 1 or 2) has an odd value, the picture# may not actually be real.
Now, of course this is just to handle the different ways. You would have to make sure it lines up properly with your actual image...
$picture = "Picture1"; // would be dynamic to code above
$ext = ".png"; // optional, could be added into the $picture variable above easily too
$filePath = "/images/user_uploads/"; // make sure path is correct
$filePathName = $filePath . $picture . $ext;
// see? We combine path/pictureName/extension to create an image URL
if(file_exists($filePathName) !== false){ // make sure exists
$imageCode = "<img src='$filePathName' alt='Not Found'>";
} else {
// file does not exist! ut oh!
}

Answer 3

It depends on how your images map to the values in your database, but assuming you have a simple relationship like:
1: /images/1.jpg
2: /images/2.jpg
Simply grab the data from your database and map it to a variable (in my example it's $image), and then concatenate it with the filepath inside an echo statement inside of an <img src> attribute:
<img src="<?php echo '/images/' $image . '.jpg'; ?>"/>
This will output:
<img src="/images/1.jpg"/>
<img src="/images/2.jpg"/>
Based on the chosen image.
Now you just have to set that variable to the right value based on your conditional:
<?php
if ($authenticated) {
$image = 1;
}
else {
$image = 2;
}
?>
<img src="<?php echo '/images/' $image . '.jpg'; ?>"/>
Or more succinctly with a ternary:
<?php ($authenticated ? $image = 1 : $image = 2) ?>
<img src="<?php echo '/images/' $image . '.jpg'; ?>"/>

Answer 4

Try this:
<img src="<?php echo "/pictures/Picture{$number}.jpg"; ?>" />