Question

我正在尝试为我的公司制作一个程序，该程序接收图像并将它们发送到带有相关信息的API。

脚本获取标题，在我的系统上创建目录，将图像放在目录中，遍历目录，将每个图像转换为Base64Data，并将数据传递给将数据发布到API的python脚本。但由于某种原因，我的循环停在第一张图像上。

更新的代码：我试图使代码更容易理解。我从API获取了无效的base64代码。功能＆＃34;照片＆＃34;显示了XML必须构造的格式。

UPDATED - FIX：在PHP中使用cURL库，而不是将数据传递给python脚本。

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$count = 0;


function photo($base64, $caption, $description, $displayname, $credits){
    $data = "<image xmlns='http://apiapi.com/cms' xmlns:i='http://www.w3.org/2001/XMLSchema-instance'><caption>'";
    $data .= $caption;
    $data .= "'</caption>";
    $data .= "<credits>'";
    $data .= $credits;
    $data .= "'</credits><data>'";
    $data .= $base64;
    $data .= "'</data>";
    $data .="<description>'";
    $data .= $base64;
    $data .= "'</description><displayname>'";
    $data .= $displayname;
    $data .="'</displayname><extension>JPG</extension>";
    $data .= "<imageusage>SlideShow</imageusage><uploadmethod>Server</uploadmethod>";
    $data .= "</image>";
    return $data; }


function base64url_encode($s) {
    return str_replace(array('+', '/'), array('-', '_'), base64_encode($s));
    }

if ($_SERVER['REQUEST_METHOD'] == 'POST'){
    $dirty_title = $_POST['title'];
    $title = htmlspecialchars($dirty_title);
    chdir('uploads');
    mkdir($title, 0777, true);
    chdir($title);
    //echo getcwd();
    foreach ($_FILES['files']['name'] as $i => $name) {
       $type = pathinfo($name, PATHINFO_EXTENSION);

        $superstring = $title . $count .".". $type;
       //echo "SuperString = " . $superstring."<br />";
        if (strlen($_FILES['files']['name'][$i]) > 1) {
            if (move_uploaded_file($_FILES['files']['tmp_name'][$i], $superstring)) {
                $count++;
            }
        }
    }

}

function endsWith($haystack, $needle)
{
    return $needle === '' || substr_compare($haystack, $needle, -strlen($needle)) === 0;
}


if (isset($_POST['submit'])){
$curdir = getcwd();
//$exiftool = exec('exiftool -all= *.jpg');
//echo "exiftool = ".$exiftool."<br />";
$dir = new DirectoryIterator($curdir);
$count = 0;
foreach ($dir as $fileinfo){
   if ($dir == '.' || $dir == '..' || endsWith($fileinfo, ".jpg_original")){}
  else{
      $count ++;
      chmod($fileinfo, 0777);
      $caption = $_POST['caption'];
      $description = $_POST['description'];
      $displayname = $_POST['displayname'];
      $credit = $_POST['credits'];
      $final_info =  $curdir."/".$fileinfo;
      $type = pathinfo($fileinfo, PATHINFO_EXTENSION);
      $data = file_get_contents($final_info);
      $mybase = base64url_encode($data);
      echo "Function Worked";
      $photo = photo($mybase, $caption, $description, $displayname, $credit);
      $file_name = "data".$count.".txt";
      $file = fopen($file_name, "w");
      fwrite($file, $photo);
      fclose($file);
      $cwd = getcwd();
      echo "REAL PATH: ".realpath($cwd);



  } 
 }      
  $cwd = getcwd();
  $realpath = realpath($cwd);
  $test =exec("/usr/bin/python /var/www/html/helper_scripts/frankAPI.py '{$realpath}'");
  echo $test;
}
?>

这是python脚本，它将文件路径作为参数，读取PHP脚本编写的写入数据，并将其发送到API。现在我得到500错误，它说我发送的base64编码不被识别为base64。

#!/usr/bin/python
import sys
import os
import requests


if __name__ == "__main__":
    try:
        directory = str(sys.argv[1])

        headers = {'Authorization':'usernameusername',
           'Content-Type': 'application/xml'
        }
        os.chdir(directory)
        dirs = os.listdir('.')

        for file in dirs:
            if file.endswith(".txt"):
                print file
                with open(file, 'r') as content_file:
                    content = content_file.read()
                    r = requests.post('https://cms.worldnow.com/v1.0/images',
                        headers=headers,
                        data=content)
                    print r.text[0:100]

    except Exception, e:
        print e

使用Base64图像数据调用Python的PHP失败

0 个答案: