2012-06-08 120 views
0

我正在尝试使用PHP/POST/etc来帮助我在当天的Webdev工作。从教程等,我已经看到在线,这应该工作。但是,当我点击提交时,页面被重定向到bagelBack.php,但是有一个空白页面,并且未提交推文。我在我自己的机器上使用XAMPP Apache。 (在HTML页面上有jQuery,如果有帮助的话)Form will not post

编辑:它在php($connection->request等)的第40行上失败。 var_dump的工作,但echo没有。我对这一切都很陌生。为什么这不会引发错误?

HTML:

<form id="bagelForm" action="bagelBack.php" method="POST"> 
    <label for="twitterName">Twitter Name: </label><input type="text" id="twitterName" name="twitterName"/><br /> 
    <label for="bagelType">Bagel Type: </label><input type="text" id="bagelType" name="bagelType"/><br /> 
    <input type="submit" /> 
</form> 

PHP(主要来自here):

<?php 
/** 
* bagelBack.php 
* Example of posting a tweet with OAuth 
* Latest copy of this code: 
* http://140dev.com/twitter-api-programming-tutorials/hello-twitter-oauth-php/ 
* @author Adam Green <[email protected]> 
* @license GNU Public License 
*/ 

$name = "@".$_POST['twitterName']; 
$type = $_POST['bagelType']; 

$tweet_text = $name.", your ".$type." bagel has finished toasting!"; 
$result = post_tweet($tweet_text); 
echo "Response code: " . $result . "\n"; 

function post_tweet($tweet_text) { 

    // Use Matt Harris' OAuth library to make the connection 
    // This lives at: https://github.com/themattharris/tmhOAuth 
    require_once('tmhOAuth.php'); 

    // Set the authorization values 
    // In keeping with the OAuth tradition of maximum confusion, 
    // the names of some of these values are different from the Twitter Dev interface 
    // user_token is called Access Token on the Dev site 
    // user_secret is called Access Token Secret on the Dev site 
    // The values here have asterisks to hide the true contents 
    // You need to use the actual values from Twitter 
    $connection = new tmhOAuth(array(
    'consumer_key' => '[redacted]', 
    'consumer_secret' => '[redacted]', 
    'user_token' => '[redacted]', 
    'user_secret' => '[redacted]' 
)); 

    // Make the API call 
    $connection->request('POST', 
    $connection->url('1/statuses/update'), 
    array('status' => $tweet_text)); 

    return $connection->response['code']; 
} 
?> 
+0

您是否尝试过使用var_dump($ name)变量来查看是否有任何内容?还有,你是否试图删除post_tweet函数,看看你是否可以让帖子进入下一页?文件名是否正确? – chadpeppers

+0

你有没有试过'echo $ name;'和'echo $ type;'?你看到了什么? – robonerd

+0

这些变量都很好。问题出现在'post_tweet'中,为什么它会像没有显示错误那样死去? – SomeKittens

回答