是否可以将数据发布到curl url？

Question

所以我想从以下形式的值张贴到卷曲URL（脚本名称是pagination.php）就像这样：

<form action="pagination.php" method="post"> 
<input style="width:50%" type="text" name="search_term"> 
<input type="submit" value="Submit"> 
</form> 

$search_term = $_POST['search_term']; 

echo $search_term; 

$post_data = array('q' => $search_term); 

$ch = curl_init(); 

curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data); 
curl_setopt($ch, CURLOPT_URL, 'http://gdata.youtube.com/feeds 
/api/videos?q='.urlencode($search_term).'&safeSearch=none&orderby=viewCount& 
v=2&alt=json&start-index=1&max-results=50'); 
curl_setopt($ch, CURLOPT_HEADER, 0); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
curl_setopt($ch, CURLOPT_TIMEOUT, 10); 
$output = curl_exec($ch); 
curl_close($ch); 

$data = json_decode($output,true); 

$info = $data["feed"]; 
$video = $info["entry"]; 
$nVideo = count($video); 



echo "<ul style='float:right'>"; 
foreach($video as $video) { 
echo '<img src="'.$video['media$group']['media$thumbnail'][0]['url'].'"><br><br>'; 
$title = $video['title']['$t']; 
$video_id = $video['media$group']['yt$videoid']['$t']; 
echo '<a href="search_4.php?video_get_id='.$video_id.'">'.$title.'</a>'; 
echo '<br>'; 
} 

当我运行像这样的事情的发生代码（是的我附和了$ SEARCH_TERM，以确保它不为空），但是如果我手动分配值$ SEARCH_TERM这样

$search term = 'baseball'; 

一切工作正常。是否有关于cURL的内容不允许发布数据？谢谢！

Answer 1

你看上去mising

curl_setopt($ch, CURLOPT_POST, 1); 
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data); 

编辑 现在回答这个问题没有提出的“如何使这项工作？”

像@ Kostanos说，看起来问题是在你的网址断行。这工作：

<form action="pagination.php" method="post"> 
<input style="width:50%" type="text" name="search_term"> 
<input type="submit" value="Submit"> 
</form> 
<?php 
if($_POST) 
    { 
    $search_term = $_POST['search_term']; 

    echo $search_term; 

    $post_data = array('q' => $search_term); 

    $ch = curl_init(); 
    curl_setopt($ch, CURLOPT_URL, 'http://gdata.youtube.com/feeds/api/videos?q='.urlencode($search_term).'&safeSearch=none&orderby=viewCount&v=2&alt=json&start-index=1&max-results=50'); 
    curl_setopt($ch, CURLOPT_HEADER, 0); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
    curl_setopt($ch, CURLOPT_TIMEOUT, 10); 
    $output = curl_exec($ch); 
    curl_close($ch); 

    $data = json_decode($output,true); 

    $info = $data["feed"]; 
    $video = $info["entry"]; 
    $nVideo = count($video); 



    echo "<ul style='float:right'>"; 
    foreach($video as $video) 
    { 
     echo '<img src="'.$video['media$group']['media$thumbnail'][0]['url'].'"><br><br>'; 
     $title = $video['title']['$t']; 
     $video_id = $video['media$group']['yt$videoid']['$t']; 
     echo '<a href="search_4.php?video_get_id='.$video_id.'">'.$title.'</a>'; 
     echo '<br>'; 
    } 

    } 

Answer 2

根据您的代码：

... 
$search_term = $_POST['search_term']; 
... 
curl_setopt($ch, CURLOPT_POSTFIELDS, $search_term); 
... 

$ search_form是不正确的POSTFIELDS值，它应该是key=>value配对数组或字符串像这样：&key=value&key1=value1..

尝试使用此：

... 
$search_term = $_POST['search_term']; 
$post_data = array('q' => $search_term); 
... 
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data); 
... 

顺便说一句：For youtube API

您不需要POST请求，只需GET请求。从代码中删除以下两行：

curl_setopt($ch, CURLOPT_POST, 1); 
curl_setopt($ch, CURLOPT_POSTFIELDS, $search_term); 

而替换该行（确保它是一条线，没有任何字符串中进入，似乎你的编辑器插件进入到长字符串！）：

curl_setopt($ch, CURLOPT_URL, 'http://gdata.youtube.com/feeds/api/videos?q=' . urlencode($search_term) . '&safeSearch=none&orderby=viewCount&v=2&alt=json&start-index=1&max-results=50'); 

PS。通过将urlencode添加到$ search_term