2012-06-08 118 views
0

我正在嘗試使用PHP/POST/etc來幫助我在當天的Webdev工作。從教程等,我已經看到在線,這應該工作。但是,當我點擊提交時,頁面被重定向到bagelBack.php,但是有一個空白頁面,並且未提交推文。我在我自己的機器上使用XAMPP Apache。 (在HTML頁面上有jQuery,如果有幫助的話)Form will not post

編輯:它在php($connection->request等)的第40行上失敗。 var_dump的工作,但echo沒有。我對這一切都很陌生。爲什麼這不會引發錯誤?

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(主要來自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

您是否嘗試過使用var_dump($ name)變量來查看是否有任何內容?還有,你是否試圖刪除post_tweet函數,看看你是否可以讓帖子進入下一頁?文件名是否正確? – chadpeppers

+0

你有沒有試過'echo $ name;'和'echo $ type;'?你看到了什麼? – robonerd

+0

這些變量都很好。問題出現在'post_tweet'中,爲什麼它會像沒有顯示錯誤那樣死去? – SomeKittens

回答