Question

我想通过PHP添加post to a Blogger blog。谷歌提供了以下示例。如何在PHP中使用它？

您可以通过向帖子发送POST请求来为博客添加帖子带有帖子JSON正文的集合URI：

POST https://www.googleapis.com/blogger/v3/blogs/8070105920543249955/posts/
Authorization: /* OAuth 2.0 token here */
Content-Type: application/json

{
  "kind": "blogger#post",
  "blog": {
    "id": "8070105920543249955"
  },
  "title": "A new post",
  "content": "With <b>exciting</b> content..."
}

Answer 1

您需要使用cURL library发送此请求。

<?php
// Your ID and token
$blogID = '8070105920543249955';
$authToken = 'OAuth 2.0 token here';

// The data to send to the API
$postData = array(
    'kind' => 'blogger#post',
    'blog' => array('id' => $blogID),
    'title' => 'A new post',
    'content' => 'With <b>exciting</b> content...'
);

// Setup cURL
$ch = curl_init('https://www.googleapis.com/blogger/v3/blogs/'.$blogID.'/posts/');
curl_setopt_array($ch, array(
    CURLOPT_POST => TRUE,
    CURLOPT_RETURNTRANSFER => TRUE,
    CURLOPT_HTTPHEADER => array(
        'Authorization: '.$authToken,
        'Content-Type: application/json'
    ),
    CURLOPT_POSTFIELDS => json_encode($postData)
));

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

// 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'];

如果由于某种原因，你不能/不想使用cURL，你可以这样做：

<?php
// Your ID and token
$blogID = '8070105920543249955';
$authToken = 'OAuth 2.0 token here';

// The data to send to the API
$postData = array(
    'kind' => 'blogger#post',
    'blog' => array('id' => $blogID),
    'title' => 'A new post',
    'content' => 'With <b>exciting</b> content...'
);

// Create the context for the request
$context = stream_context_create(array(
    'http' => array(
        // http://www.php.net/manual/en/context.http.php
        'method' => 'POST',
        'header' => "Authorization: {$authToken}\r\n".
            "Content-Type: application/json\r\n",
        'content' => json_encode($postData)
    )
));

// Send the request
$response = file_get_contents('https://www.googleapis.com/blogger/v3/blogs/'.$blogID.'/posts/', FALSE, $context);

// Check for errors
if($response === FALSE){
    die('Error');
}

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

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

Answer 2

我认为cURL是一个很好的解决方案。这没有经过测试，但您可以尝试这样的事情：

$body = '{
  "kind": "blogger#post",
  "blog": {
    "id": "8070105920543249955"
  },
  "title": "A new post",
  "content": "With <b>exciting</b> content..."
}';
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "https://www.googleapis.com/blogger/v3/blogs/8070105920543249955/posts/");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json","Authorization: OAuth 2.0 token here"));
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $body);
$result = curl_exec($ch);

Answer 3

如果你不想使用CURL，你可以在stackoverflow上找到一些例子，就像这里的一样：How do I send a POST request with PHP?。我建议你观看一些关于如何在PHP中使用GET和POST方法的教程，或者在这里查看php.net手册：httprequest::send。你可以找到很多教程：HTTP POST from PHP, without cURL等等......

Answer 4

我通过网站上的表格将数据发送到基于@Rocket Hazmat，@ dbau和@maraca代码的prosperworks。我希望，它会对某人有所帮助：

<?php

if(isset($_POST['submit'])) {
    //form's fields name:
    $name = $_POST['nameField'];
    $email = $_POST['emailField'];

    //API url:
    $url = 'https://api.prosperworks.com/developer_api/v1/leads';

    //JSON data(not exact, but will be compiled to JSON) file:
    //add as many data as you need (according to prosperworks doc):
    $data = array(
                            'name' => $name,
                            'email' => array('email' => $email)
                        );

    //sending request (according to prosperworks documentation):
    // use key 'http' even if you send the request to https://...
    $options = array(
        'http' => array(
            'header'  => "Content-Type: application/json\r\n".
             "X-PW-AccessToken: YOUR_TOKEN_HERE\r\n".
             "X-PW-Application:developer_api\r\n".
             "X-PW-UserEmail: YOUR_EMAIL_HERE\r\n",
            'method'  => 'POST',
            'content' => json_encode($data)
        )
    );

    //engine:
    $context  = stream_context_create($options);
    $result = file_get_contents($url, false, $context);
    if ($result === FALSE) { /* Handle error */ }
    //compiling to JSON (as wrote above):
    $resultData = json_decode($result, TRUE);
    //display what was sent:
    echo '<h2>Sent: </h2>';
    echo $resultData['published'];
    //dump var:
    var_dump($result);

}
?>
<html>
    <head>
    </head>

    <body>

        <form action="" method="POST">
            <h1><?php echo $msg; ?></h1>
            Name: <input type="text" name="nameField"/>
            <br>
            Email: <input type="text" name="emailField"/>
            <input type="submit" name="submit" value="Send"/>
        </form>

    </body>
</html>

Answer 5

<?php
// Example API call
$data = array(array (
    "REGION" => "MUMBAI",
    "LOCATION" => "NA",
    "STORE" => "AMAZON"));
// json encode data
$authToken = "xxxxxxxxxx";
$data_string = json_encode($data); 
// set up the curl resource
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "http://domainyouhaveapi.com");   
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_HEADER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
    'Content-Type:application/json',       
    'Content-Length: ' . strlen($data_string) ,
    'API-TOKEN-KEY:'.$authToken ));   // API-TOKEN-KEY is keyword so change according to ur key word. like authorization 
// execute the request
$output = curl_exec($ch);
//echo $output;
// Check for errors
if($output === FALSE){
    die(curl_error($ch));
}
echo($output) . PHP_EOL;
// close curl resource to free up system resources
curl_close($ch);

使用JSON正文的POST请求

5 个答案: