为什么在我的方法中imagecreatefromjpeg($src）方法返回false - Why does he method imagecreatefromjpeg($src) return false in my method?

Why does he method imagecreatefromjpeg($src) return false in my method?

我是php的新手，我正在尝试制作缩略图

$src是图像的路径$thumbWidth是所需的宽度$imageName并不重要，它需要传递给为缩略图生成html代码的函数。

问题出现在第174行，我评论道，如果图像是jpeg文件，函数返回false，那么$source_image为false，有人能解释为什么吗？

这是我的方法：

 function makeThumb( $src, $thumbWidth, $imageName )
 {
$count = 0;
$len = strlen($src);
$indexlen = $len - 1;
$sourceArray = str_split($src);
for($i = $indexlen; $i > -1; $i--)
{
    if($sourceArray[$i] == '.')
    {
        $count = $count + 1;
        if($count == 1)
        {
            $hold = $i; 
        }
    }
}
$ending = substr($src, $hold, $len);
if($ending === '.gif') 
{
    $type = '.gif';
    $source_image = imagecreatefromgif($src);
}
if($ending === '.jpeg' || $ending === '.pjpeg' || $ending === '.jpg') 
{
    $type = '.jpg';
    $source_image = imagecreatefromjpeg($src);
}
if($ending === '.png')
{
    $type = '.png';
    $source_image = imagecreatefrompng($src);
}
else
{   
    //throw new Exception('This file is not in JPG, GIF, or PNG format!');
    $type = null;
}
/* read the source image */
if($ending = null)
{   return null;    } 
$width = imagesx($src);
$height = imagesy($src);
$newWidth = $thumbWidth;
/* find the "desired height" of this thumbnail, relative to the desired width  */
$newHeight = floor($height * ($newWidth / $width));
/* create a new, "virtual" image */
$tempImg = imagecreatetruecolor($desired_width, $desired_height);
$pic = formatImage($tempImg, $imageName);
return $pic;
/* copy source image at a resized size */
//imagecopyresampled($virtual_image, $source_image, 0, 0, 0, 0, $desired_width, $desired_height, $width, $height);
/* create the physical thumbnail image to its destination */
//imagejpeg($virtual_image, $dest);

}

您确定该图像是有效的jpeg图像吗？

您可以通过扩展来检查图像的类型。您可以用一种简单得多的方式检查是否获得扩展。您还应该检查文件是否存在：

function makeThumb( $src, $thumbWidth, $imageName )
{
    if(!file_exists($src))
        throw new Exception('The file '.$src.' does not exist.');
    $ext = pathinfo($src, PATHINFO_EXTENSION);

但是使用扩展来检查图像的类型是不可靠的。使用getimagesize函数还有另一种检查方法：

function makeThumb( $src, $thumbWidth, $imageName )
{
    if(!file_exists($src)) 
        throw new Exception('The file '.$src.' does not exist.');

    $info = getimagesize($src);
    if($info === false)
        throw new Exception('This file is not a valid image');
    $type    = $info[2];
    $width   = $info[0]; // you don't need to use the imagesx and imagesy functions
    $height  = $info[1];
    switch($type) {
        case IMAGETYPE_JPEG:
            $source_image = imagecreatefromjpeg($src);
            break;
        case IMAGETYPE_GIF:
            $source_image = imagecreatefromgif($src);
            break;
        case IMAGETYPE_PNG:
            $source_image = imagecreatefrompng($src);
            break;
        default:
            throw new Exception('This file is not in JPG, GIF, or PNG format!');
    }
    $newWidth = $thumbWidth;
    // ..and here goes the rest of your code

请注意-我没有检查你的代码的其余部分，因为你的问题与函数的第一部分有关。