Question

我需要向字符串'SUCCESS'发送HTTP 200响应，但我的php服务器版本是5.2.17！

在我的情况下，webhook将数据发送到捕获到名为notification.php的文件，我读取了contentes，保存在数据库中并需要发送响应但不知道如何执行此操作！

有谁知道如何在php 5.2.17中执行此操作？

我尝试了以下方法但没有成功：

// error 1
header("Content-Type: text/plain");
echo "SUCCESS";

// error 2
$httpStatusCode = 200;
$httpStatusMsg  = 'SUCCESS';
$protocol = isset($_SERVER['SERVER_PROTOCOL']) ? $_SERVER['SERVER_PROTOCOL'] : 'HTTP/1.0';
header($protocol.' '.$httpStatusCode.' '.$httpStatusMsg);

// error 3
header("200 SUCCESS");
return "200 SUCCESS";

// error 4
header('Content-Type: application/json');
echo 'SUCCESS';

// error 5
header('Content-Type: application/json');
return 'SUCCESS';

//error 6
header('Content-Type: application/json');
header('SUCCESS');

///error 7
header('Content-Type: application/json');
$success =json_encode('SUCCESS');
header($success);


///error 8
header("HTTP/1.1 200 SUCCESS");
header("Content-Type:application/json; charset=utf-8");

///error 9
header("HTTP/1.1 200");
header("Content-Type:application/json; charset=utf-8");
result 'SUCCESS';

//error 10
header("Content-Type:application/json;");
header('HTTP/1.0 200 SUCCESS');

// error 11
$code = 200;
$text = 'OK';
$protocol = (isset($_SERVER['SERVER_PROTOCOL']) ? $_SERVER['SERVER_PROTOCOL'] : 'HTTP/1.0');
header($protocol . ' ' . $code . ' ' . $text);
$GLOBALS['http_response_code'] = $code;
echo 'SUCCESS';

我的PhpCode

<?php
 $src_data = $_REQUEST["data"];           
 $une_data = stripslashes($src_data);
 $data = json_decode($une_data);

 //get payment data
 $id_cob  = $data->payment->id;
 $id_cus  = $data->payment->customer;
 $status  = $data->payment->status;
 $dtsts   = date("Y-m-d");

 if ($data->event == 'PAYMENT_RECEIVED') {
// post client
include('dbconnection.php');
$qryn  = "UPDATE PAYMENTS SET STATUS='$status', DATASTATUS='$dtsts' WHERE ID_COB_ASAAS='$id_cob' AND ID_CLI_ASAAS='$id_cus'";
    mysql_query($qryn,$cnx);
 } 

// webhook return
// my solution was:
header('HTTP/1.1 200 OK');
echo 'SUCCESS';
return;

?>

Answer 1

200是标准返回码。

所以

.introduction {
  background-image: url('../images/karma-background.jpg'); 
}

基本上应该足够了。

Answer 2

我使用了以下内容：

echo "SUCCESS";
http_response_code(200); // PHP 5.4 or greater

这似乎有效。

Answer 3

尝试：

ARRAY start{*} col1 col5 col8 var1;
DO OVER start;

在PHP中发送响应HTTP 200 SUCCESS

3 个答案: