我试图用curl替换使用url(例如...... /?id = 123)的传递参数。如果不能使用curl则忽略。
我尝试在线查找示例,但我只能看到如何发布数据但没有得到它。我在网站上找到了这个,但我没有参考。
$post_data['ID'] = '1';
//traverse array and prepare data for posting (key1=value1)
foreach ( $post_data as $key => $value) {
$post_items[] = $key . '=' . $value;
}
//create the final string to be posted using implode()
$post_string = implode ('&', $post_items);
//create cURL connection
$curl_connection = curl_init('posturl.php');
//set options
curl_setopt($curl_connection, CURLOPT_CONNECTTIMEOUT, 30);
curl_setopt($curl_connection, CURLOPT_USERAGENT, "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1)");
curl_setopt($curl_connection, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curl_connection, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($curl_connection, CURLOPT_FOLLOWLOCATION, 1);
//set data to be posted
curl_setopt($curl_connection, CURLOPT_POSTFIELDS, $post_string);
//perform our request8
$curlresult = curl_exec($curl_connection);
//show information regarding the request
// print_r(curl_getinfo($curl_connection));
echo curl_getinfo($ch, CURLINFO_HTTP_CODE);
//close the connection
curl_close($curl_connection);
如何从posturl.php获取参数
有没有人有一个发布和使用curl的例子?
答案 0 :(得分:0)
cURL
主要用于检索从HTTP POST
方法渲染的HTML。你做得很好。那究竟你想要检索什么?来自posturl.php
的参数?然后最后你可以正确使用这个代码:
foreach($post_items as $parameter => $value)
echo "The value of $parameter is $value.\n";
我不确定我是否理解这个问题,但这是我可以帮助你的方式。希望这会有所帮助。
答案 1 :(得分:0)
尝试替换
//create cURL connection
$curl_connection = curl_init('posturl.php');
与
//create cURL connection
$curl_connection = curl_init("posturl.php?$post_string");
和评论
//curl_setopt($curl_connection, CURLOPT_POSTFIELDS, $post_string);