if image is → on-error fetch image on the base of gender value

December 30, 2022

I’m trying to fetch an image, and if on-error, get the default image based on the gender value.
pls advice How can I fetch an on-error default image on base of gender value ?

echo"
   
<img  src=".'imagefolder/'.$_row['img']."
   
<?php if (".$_row['gender']." != 'Male'):?>

onerror = this.src='female-default.jpg'

<?php else: ?>

onerror = this.src='male-default.jpg'

<?php endif; ?> 
   
>";

>Solution :

Your quotes mean that none of the PHP inside is being evaluated, and you can’t use if/else blocks inside an echo anyway. You can break it apart

echo "<img  src='imagefolder/'" . $_row['img'] . "' ";

if ( $_row['gender'] != 'Male'):

echo "onerror = this.src='female-default.jpg'";

else:

echo "onerror = this.src='male-default.jpg'";

 endif;
   
echo ">";

or you can use a ternary

echo "<img src='imagefolder/{$_row['img']}'" . ($_row['gender'] != 'Male' ? " onerror = this.src='female-default.jpg'" : " onerror = this.src='male-default.jpg'") . ">";