如何使用Oauth，API，JSON和PHP将图片上传到Twitter，需要帮助

Question

我对twitter api非常陌生，今天我是我生命中第一次注册twitter api。 我的目的是通过Twitter在我的网站上发布图片和消息。 我已经从互联网上下载了当前的库，它提供了以下2个文件和一些示例。

我已经包含了这两个文件，并获得了此代码photo_tweet.php意味着与照片一起鸣叫

这是我的代码

<?php
require 'tmhOAuth.php';
require 'tmhUtilities.php';

$tmhOAuth = new tmhOAuth(array(
  'consumer_key'    => 'ZlhOWeCWG2MS5Wxxxxxx',
  'consumer_secret' => 'DIPjoKcIWjpGWmw5jGJSKGAOLxxxxxx',
  'user_token'      => 'xxxxx-OGtPyRSUOUaR6XQRLAFVuv14xxxxxx',
  'user_secret'     => 'xxxxxoimsrNFhlz7mPa9h5pyVSjxxxxxxx',
));

// we're using a hardcoded image path here. You can easily replace this with
// an uploaded image - see images.php in the examples folder for how to do this
// 'image = "@{$_FILES['image']['tmp_name']};type={$_FILES['image']['type']};filename={$_FILES['image']['name']}",

// this is the jpeg file to upload. It should be in the same directory as this file.
$image = 'a.jpg';

$code = $tmhOAuth->request(
  'POST',
  'https://upload.twitter.com/1/statuses/update_with_media.json',
  array(
    'media[]'  => "@{$image};type=image/jpeg;filename={$image}",
    'status'   => 'Picture time',
  ),
  true, // use auth
  true  // multipart
);

if ($code == 200) {
  tmhUtilities::pr(json_decode($tmhOAuth->response['response']));
} else {
  tmhUtilities::pr($tmhOAuth->response['response']);
}

?>

a.jpg在我的服务器上。

这是我在Twitter上的设置！

请求类型：GET请求URI：* https://api.twitter.com/1/

当我上传所有这些代码并访问我的代码时，当我回显$ code时，它将返回空白屏幕。 它返回0

这是我所观察和经历的一切。 我需要您的帮助。 我将非常非常感谢您

注意：事件删除我的使用者密钥和使用者密钥时，此代码不会显示任何错误。 只是一个空白的屏幕。 我希望这在出现任何问题的情况下显示错误，但没有错误

---

这是状态代码，仅对我有用

$consumerKey = $consumer_key;
$consumerSecret = $consumer_secret;
$OAuthToken = $user_token;
$OAuthSecret = $user_secret;

include "OAuth.php";
include "twitteroauth.php";

$tweet = new TwitterOAuth($consumerKey, $consumerSecret, $OAuthToken, $OAuthSecret);

$image = 'a.jpg';

$optsArray['status'] = 'Hi';



if (isset($_GET['msg'])) {
    $tweetmsg = $_GET['msg'];
    $tweet->post('statuses/update',$optsArray);

   if($tweet)
     echo "Your message has been sent to Twitter.";
   else
     echo "Your message has not been sent to Twitter.";

} else {
    echo "Your message has not been sent to Twitter.";
}

但它仅在我使用时发布状态

update_with_media.json

和

$ optsArray ['@ media []'] =“ @ {$ image}”;

它不起作用，请帮助我找出此问题

Answer 1

您可能想查看这篇文章。 它引导您完成所有步骤，并提供许多示例。

API提供了以下方法。 twitpic徽标使用PHP在twitpic API上发布图片

METHOD: uploadAndPost (http://twitpic.com/api/uploadAndPost)
METHOD: upload (http://twitpic.com/api/upload)
Error codes
    1001 – Invalid twitter username or password
    1002 – Image not found
    1003 – Invalid image type
    1004 – Image larger than 4MB

上面给出的是twitpic的API页面中的API描述。 但是您不必深入研究那些复杂的细节。 相反，您可以使用一个类，该类使您可以在twitpic上发布图片并在同一帐户上更新twitter状态。 这是简单易用。 查看下面的代码。

<?php
class twitpic
{
  /* 
   * variable declarations
   */
  var $post_url='http://twitpic.com/api/upload';
  var $post_tweet_url='http://twitpic.com/api/uploadAndPost';
  var $url='';
  var $post_data='';
  var $result='';
  var $tweet='';
  var $return='';

/*
* @param1 is the array of data which is to be uploaded
* @param2 if passed true will display result in the XML format, default is false
* @param3 if passed true will update status twitter,default is false
*/

  function __construct($data,$return=false,$tweet=false)
  {
    $this->post_data=$data;
    if(empty($this->post_data) || !is_array($this->post_data)) //validates the data
      $this->throw_error(0);
    $this->display=$return;
    $this->tweet=$tweet;

  }

  function post()
  {
    $this->url=($this->tweet)?$this->post_tweet_url:$this->post_url; //assigns URL for curl request based on the nature of request by user
    $this->makeCurl();
  }
  private function makeCurl()
  {
    $curl = curl_init();
    curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 2);
    curl_setopt($curl, CURLOPT_HEADER, false);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($curl, CURLOPT_BINARYTRANSFER, 1);
    curl_setopt($curl, CURLOPT_URL, $this->url);
    curl_setopt($curl, CURLOPT_POST, 3);
    curl_setopt($curl, CURLOPT_POSTFIELDS, $this->post_data);
    $this->result = curl_exec($curl);
    curl_close($curl);
    if($this->display)
    {
      header ("content-type: text/xml");
      echo $this->result ;
    }

  }
  private function throw_error($code) //handles few errors, you can add more

  {
    switch($code)
    {
      case 0:
        echo 'Think, you forgot to pass the data';
        break;
      default:
        echo 'Something just broke !!';
        break;
    }
    exit;
  }
} //class ends here

?>

上面的PHP类完成了所有技巧，可以在twitpic上上传图片并在twitter上发布状态。 您可以将此类与HTML表单或PHP脚本一起使用，以自动上传图像并发布推文。 您可以结合使用以下HTML表单和PHP脚本，通过从用户处获取输入来上传图片。

// This block of code should be written above the HTML and it will exit after the picture has been uploaded. If //you have turned display on (by passing 3rd param as true) then it will display the success message.
if($_POST)
{

  $file=$_FILES['media'];
  $postfields = array();

  $postfields['username'] = $_POST['username'];

  $postfields['password'] = $_POST['password'];
  $postfields['message'] = $_POST['message'];
  $postfields['media'] = "@$file[tmp_name]";

  $t=new twitpic($postfields,true,true);
  $t->post();
  exit;
}

<style type="text/javascript">
  *{font-family:verdana;}
  span{font-size:12px;color:#393939;}
  h3{font-size:14px;color:#5AAAF7;}
</style>
<body>

  <h3>Upload your pic to twitpic, and post status on twitter</h3>
  <form method="post"  enctype="multipart/form-data" action="<?= $_SERVER[PHP_SELF] ?>"   >
    <p><span style="height:40px;font-weight:bold;margin-right:56px;">Twitter Username :</span><input type="text" name="username" /></p>
    <p><span style="height:40px;font-weight:bold;margin-right:61px;">Twitter Password:</span><input type="password" name="password" /></p>

    <p><span style="vertical-align:text-top;height:40px;font-weight:bold;margin-right:28px;">Message to be posted :</span> <textarea cols="35" rows="2" name="message"></textarea></p>
    <p><span style="vertical-align:text-top;height:40px;font-weight:bold;">Choose an image to upload: </span><input type="file" name="media" /></p>
    <p style="width:250px;text-align:right;margin-top:50px;"><input type="submit" value="Upload&nbsp;&raquo;" /> </p>
  </form>
  <sup>Script powered by <a href="http://www.digimantra.com/">www.digimantra.com</a></sup>
</body>

You can skip posting update to twitter by passing the third argument as false or just by skipping it. If you want to upload image programmatically, without the user input or the form then you can do it using the following code. Make sure the image path is correctly mention, else it will throw an error.

<?php
$file='file_to_be_uploaded.gif';
$postfields = array();

$postfields['username'] = 'twitter_username';

$postfields['password'] = 'twitter_password';
$postfields['message'] = 'Message to be posted' ;
$postfields['media'] = "@$file"; //Be sure to prefix @, else it wont upload

$t=new twitpic($postfields,true,true);
$t->post();
?>

上面的此类以二进制格式发布图像，如使用curl PHP发送二进制数据中所述。 您可以根据需要更改班级，也可以根据自己的喜好对其进行增强。 希望这篇文章对您有所帮助。