PHP新手CURL和数组

Question

更新：我简化了代码（尝试）

我正在尝试下载数组中设置的一系列图像，但是显然不正确：

function savePhoto($remoteImage,$fname) {
    $ch = curl_init();
    curl_setopt ($ch, CURLOPT_NOBODY, true);
    curl_setopt ($ch, CURLOPT_URL, $remoteImage);
    curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, 0);
    $fileContents = curl_exec($ch);
    $retcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
    curl_close($ch);
    if($retcode==200) {
        $newImg = imagecreatefromstring($fileContents);
        imagejpeg($newImg, ".{$fname}.jpg",100);
    }
    return $retcode;
}

$filesToGet = array('009');
$filesToPrint = array();

foreach ($filesToGet as $file) {
        if(savePhoto('http://pimpin.dk/jpeg/'.$file.'.jpg',$file)==200) {
            $size = getimagesize(".".$file.".jpg");
            echo $size[0] . " * " . $size[1] . "<br />";
        }
}

我收到以下错误：

警告：imagecreatefromstring（）[function.imagecreatefromstring]：空字符串或第15行的C：\\ inetpub \\ vhosts \\ dehold.net \\ httpdocs \\ ripdw \\ index.php中的图像无效

警告：imagejpeg（）：提供的参数在第16行的C：\\ inetpub \\ vhosts \\ dehold.net \\ httpdocs \\ ripdw \\ index.php中不是有效的图像资源

警告：getimagesize（.009.jpg）[function.getimagesize]：无法打开流：第26行的C：\\ inetpub \\ vhosts \\ dehold.net \\ httpdocs \\ ripdw \\ index.php中没有此类文件或目录*

Answer 1

试试这个代替：

function get_file1($file, $local_path, $newfilename)
{
    $err_msg = '';
    echo "<br>Attempting message download for $file<br>";
    $out = fopen($newfilename, 'wb');
    if ($out == FALSE){
      print "File not opened<br>";
      exit;
    }

    $ch = curl_init();

    curl_setopt($ch, CURLOPT_FILE, $out);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_URL, $file);

    curl_exec($ch);
    echo "<br>Error is : ".curl_error ( $ch);

    curl_close($ch);
    //fclose($handle);

}//end function 

//摘自： http ： //www.weberdev.com/get_example-4009.html

或file_get_contents

Answer 2

您应该尝试用file_get_contents代替CURL（虽然简单，但是可以完成工作）：

 function savePhoto($remoteImage,$fname) {        
      $fileContents = file_get_contents($remoteImage);        
      try {        
        $newImg = imagecreatefromstring($fileContents);
        imagejpeg($newImg, ".{$fname}.jpg",100);        
      } catch (Exception $e) {        
        //what to do if the url is invalid        
      } 
}

Answer 3

在大家的帮助下，我终于找到了工作的机会：-)

我最终使用了CURL：

function savePhoto($remoteImage,$fname) {
    $ch = curl_init();

    curl_setopt ($ch, CURLOPT_URL, $remoteImage);
    curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, 0);
    $fileContents = curl_exec($ch);

    $retcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);

    curl_close($ch);
    if($retcode == 200) {
        $newImg = imagecreatefromstring($fileContents);
        imagejpeg($newImg, $fname.".jpg",100);
    }

    return $retcode;
}


$website = "http://www.pimpin.dk/jpeg";
$filesToGet = array('009');
$filesToPrint = array();

foreach ($filesToGet as $file) {
        if(savePhoto("$website/$file.jpg",$file)==200) {
            $size = getimagesize($file.".jpg");
            echo $size[0] . " * " . $size[1] . "<br />";
        } else {
            echo "File wasn't found";
        }
}