phpを使って他のURLを呼び出す方法..？

Question

私は

http://192.168.1.220/cgi-bin/handle_login.tcl

が、カールを試みたときには、以下

The requested URL was not found on this server. The link on the referring page seems to be wrong or outdated. Please inform the author of that page about the error. 

私のphpで私に次のエラーを与えてその作業罰金をアプリ配達員に次のURLを打ったときコード

<?php 
    $postData = array(
    'user' => 'admin', 
    'pw' => 'admin', 
    'submit' => 'Login' 
    ); 

    // Setup cURL 
    $ch = curl_init('http://192.168.1.220/cgi-bin/handle_login.tcl'); 
    curl_setopt_array($ch, array(
    CURLOPT_POST => TRUE, 
    CURLOPT_RETURNTRANSFER => TRUE, 
    CURLOPT_HTTPHEADER => array('Content-Type: application/x-www-form-urlencoded' 
), 
    CURLOPT_POSTFIELDS => json_encode($postData) 
)); 

// Send the request 
    $response = curl_exec($ch); 

var_dump($response); 

// Check for errors 
    if($response === FALSE){ 
    die(curl_error($ch)); 
    } 

    // Decode the response 
    $responseData = json_decode($response, TRUE); 

// Print the date from the response 
echo $responseData['published']; 
?> 

Answer 1

私は自分のコード内のエラーを整流してきたと私は答えを希望しまった...

は以下

$ch = curl_init(); 
    curl_setopt($ch, CURLOPT_URL,"URL"); 
    curl_setopt($ch, CURLOPT_POST, 1); 
    curl_setopt($ch, CURLOPT_POSTFIELDS, 
    "key1=value1&key2=value2&key3=value3");   
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 

    // Send the request 
    $response = curl_exec($ch); 
    curl_close ($ch); 
    echo "$response";