检查图像是否存在，如果不存在，请使用php选择另一个图像

Question

I want to display an image with php .

How can I check if it exists and if not, link to another one?

This is what i've got:

<img class="MyClass" src="Images/<?php echo $image1?>.png">

So if image1 doesn't exist it should link to image2

Answer 1

there are some way but this is the way that I use:

php documentation

<?php
  $filename="images/".$image_name;
  if(is_readable($filename)){
     $fileToShow=$filename;
  }else{
    $fileToShow="images/default.jpg";
  }
    echo '<img class="MyClass" src="<?php  echo $fileToShow; ?>"/>';
?>

Answer 2

First i need to know are you saving image files without extention. If so see the code below

$filename = 'Images/'.$image1.'.png';

 if (file_exists($filename)) {
echo '<img class="MyClass" src="Images/<?php echo $image1;?>.png">';
} 
else {
 echo '<img class="MyClass" src="Images/<?php echo $image2;?>.png">';;
 }

Answer 3

Try this

<?php
  $file_name="images/".$image_name;
  if(file_exists($file_name))
  {
     $path=$file_name;
  }
  else
  {
    $path="images/default.jpg";
  }
    echo '<img class="MyClass" src="<?php  echo $path; ?>"/>';
?>