if/else to display an image

In my sql database, I have a number stored, between 1 and 10.

On the page I am working on, the number in the database is displayed on the screen.

I would like to display a different image depending on whatever number is in the database.

ie.- if 1, display 1.jpg, if 2, display 2.jpg and so on (instead of just displaying the actual number.

Would some sort of if/else statement be used for this? If so, how should it be written? Thank you!

 

if($number==1)
  echo "<img src=\"the source here\">";

else if($number==2)
 echo "<img src=\"the source here\">";

else if($number==3)
 echo "<img src=\"the source here\">";

else if($number==4)
 echo "<img src=\"the source here\">";

else if($number==5)
 echo "<img src=\"the source here\">";

PHP:

in this way

 

better a switch - case

 

much simpler would be:

if($number >= 1 && $number <= 10){
    echo "<img src=\"$number.jpg\">";
}

PHP:

 

Code is poetry. redtrim got it better than 100 lines of coderbari

 

Looks great guys, thank you so much for your help.

 

I would have an array of image paths and then have number indicate what the index is.

$image[0] = ...
$image[1] = ...
.
.
.
$image[10] = ...

echo "<img src=\"" . $image[$number] . "\">";

 

sitefever, there is no need for an if statement or case statement. To answer exactly to you example you could do like this, if $number would be your variable holding the number from the DB

 echo '<img src="'.$number.'display.jpg">';

PHP:

 

printf( '<img src="%d.jpg" />', file_exists( sprintf( '%d.jpg', $number ) ) ? $number : 'not_found' );

PHP:

make a not_found.jpg or something that is relevant to whatever you're doing.

 

Instead of using sprintf, you can use the intval function.

~
Thomas

 

I'm not evaluating the number, there is no point if they are all in a database and you know what the results are, I just prefer to use sprintf and printf over

echo "this" . $sort . " of " . $business ;

PHP:

it just reads easier and you're less likely to make errors when writing code