RT.
我只要让PHP模拟HTTP POST发送NVP,到本地的其他文件。
而不用返回任何值,
怎么配置参数呢?
$ch = curl_init();
curl_setopt($ch,CURL_URL,$url);
curl_setopt($ch,CURL_POST,true);
curl_setopt($ch,CURL_POSTFIELDS,$postfields);
...
除了这几个之外还需要配置什么参数呢?
求大神
回复讨论(解决方案)
不返回值?你不管他返回不就行了吗?
不返回值?你不管他返回不就行了吗?
是啊,我就是这样的:
function httppost($url,$postvalue)
$ch = curl_init();
curl_setopt($ch,CURL_URL,$url);
curl_setopt($ch,CURL_POST,true);
curl_setopt($ch,CURL_POSTFIELDS,$postfields);
curl_setopt($ch,CURL_RETURNTRANSFER, 0);
curl_exec($ch);
curl_close($ch);
}
然后在a.php里面
$url="http://mysite.com/b.php";
$postvalue="name1=one&name2=two";
httppost($url,$postvalue);
然后在b.php里面,
print_r($_POST);
结果在b.php里面输出的数组什么都没有。
你看还有没有漏掉什么环节啊?
接受了也不出来就行。
接受了也不出来就行。
可是我上面那段代码就是差不多那样的啊,
但测试结果收不到传递的值,
麻烦能帮我看下 有没有缺少什么东西吗
错误也太多了点吧?
function httppost($url,$postvalue) { $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $postvalue); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 0); //不要返回数据应改设为1 curl_exec($ch); curl_close($ch);}
在A.php里面不输出,你在B.php输出什么东西,都不会看到的……
在A.php里面不输出,你在B.php输出什么东西,都不会看到的……
为什么啊,能说具体点吗
错误也太多了点吧?PHP code
function httppost($url,$postvalue) {
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, $pos……
啊除了那个returntransfer还有哪里不对呢?
谢谢指出,
还有,我在一个脚本里用了httppost()函数,然后接着js转跳到目标url的页面,
目标URL里有用来测试print_r($POST),
可是转跳过去以后页面里什么都没输出,$post数组是空的
你知道是怎么回事呢
其实我的意思是想在服务器端两个文件之间传递数据,但不想使用cookie/session或者memcache之类的方式,
最好是想在curl发送POST的同时转跳到另一个文件,同时实现数据的传送,就像HTML POST表单那样,
比如,
if ($verify) {
//验证成功
post($url,$postvalue);
?//发送Post信息的同时,页面转跳到目的$url,
}
我希望是这样的,
如果实现不了的话 就只能用php呼出

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