检查文件是空的imagecreatefromgif - php

Question

我正在使用PHP显示文件中的图像。我的代码看起来像这样

$image = @imagecreatefromgif(../ProfileIcons/".$use);

$text_color = imagecolorallocate($image, $r, $g, $b);
imagestring($image,$font_size,14,4, $code, $text_color);

header('Content-Type: image/gif');
$kk = imagegif($image);

imagedestroy($image);

此处如何检查ProfileIcons/中的图片是否为空？

我必须提出一个条件

if (image is present in `ProfileIcons/`) 

   $image = @imagecreatefromgif(../ProfileIcons/".$use);

else (image is not present in the folder)

   $image = Show default image

如何实现这一目标？

Answer 1

尝试查看file_exists

if (file_exists($filePath)) {
// process image
} else {
// use default image
}

Answer 2

 // check if file exist 
  if(file_exists('../ProfileIcons/'.$use.''))
  {// do what you like }else{//}