无法发送使用CURLOPT_POST到其他PHP文件

Question

我是新来的PHP，我试图让使用curl从网站的数据（刮）数据，

无法从index.php来data.php获取数据，使用CURLOPT_POST ..我做错了什么..？

• followed this tutorial on youtube

的index.php

<?php 
$data = array("name"=>"john","age"=>31); 
$string = http_build_query($data); 

echo $string; 

$ch = curl_init("http://localhost/scrap_practise/data.php"); 
curl_setopt($ch, CURLOPT_POST,true); 
curl_setopt($ch, CURLOPT_POSTFIELDS,$string); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
curl_exec($ch); 
curl_close($ch); 

?> 

data.php

<?php 
echo 'finlly in'; // this never echos 
if(isset($_POST['name'],$_POST['age'])){ 
    $db = new Mysqli("localhost","root","","mydb"); 
    $name = $db->real_escape_string($_POST['name']); 
     $age = (int) $_POST['age']; 
     $query = "INSERT INTO data SET data='$name,$age'"; 
     $db->query($query); 

} 

?> 

Answer 1

只要你需要更新你index.php脚本与这些代码行。

的index.php

<?php 
$data = array("name"=>"john","age"=>31); 
$string = http_build_query($data); 

echo $string; 

$ch = curl_init("http://localhost/scrap_practise/data.php"); 
curl_setopt($ch, CURLOPT_POST,true); 
curl_setopt($ch, CURLOPT_POSTFIELDS,$string); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
if(! $result = curl_exec($ch)) 
{ 
    trigger_error(curl_error($ch)); 
} 
curl_close($ch); 

// to see the return result uncomment the below line code. 
//print_r($result); 
?> 

更多选项或函数引用看到这个链接 - http://php.net/manual/en/function.curl-exec.php

希望这将有助于解决您的问题！

Answer 2

试试这个代码，我希望它会为你工作...

<?php 
$url = 'http://localhost/scrap_practise/data.php'; 

$data = array("name"=>"john","age"=>31); 

$string = http_build_query($data); 

echo $string; 

$ch = curl_init(); 
curl_setopt($ch,CURLOPT_URL, $url); 
curl_setopt($ch, CURLOPT_POST,true); 
curl_setopt($ch, CURLOPT_POSTFIELDS,$string); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 

$result = curl_exec($ch); 
print_r($result) ; 

curl_close($ch); 

?> 

在data.php文件

<?php 
echo 'finlly in'; // this never echos 
if(isset($_POST['name'],$_POST['age'])) 
{ 
    echo "<pre>"; print_r($_POST); 
} 
?> 

这将输出：

finlly in 
Array 
(
    [name] => john 
    [age] => 31 
)