Question

使用PHP版本5.4.28，我想要完成的是：

我的Facebook应用程序是脚本＆＃34; otherevent.php＆＃34;。它不应该对任何用户公开可见。它是一个位于Web服务器上的后端应用程序，因为Facebook要求它拥有一个Web地址，并且该应用程序由其他脚本运行和管理。
我必须运行和管理应用程序的第一个脚本叫做＃34; app.php＆＃34;。理想情况下，它应该执行的是执行＆＃34; otherevent.php＆＃34;，获取输出，并将输出回显到屏幕。
Craziness以＆＃34;意外T_STRING的形式出现，期待T_CONSTANT_ENCAPSED_STRING＆＃34;其中包括错误。

我在这里有三个测试文件的评论和代码，＆＃34; otherevent.php＆＃34;，＆＃34; app.php＆＃34;和＆＃34; test.php＆＃34;这是一个用于调试错误的测试文件。这是：

test.php的：

echo $argv[1] . " TADA!";
exit();

?>

otherevent.php（删除并审查，以获取敏感信息）：

require_once( 'Facebook/FacebookSession.php' );
require_once( 'Facebook/FacebookRedirectLoginHelper.php' );
require_once( 'Facebook/FacebookRequest.php' );
require_once( 'Facebook/FacebookResponse.php' );
require_once( 'Facebook/FacebookSDKException.php' );
require_once( 'Facebook/FacebookRequestException.php' );
require_once( 'Facebook/FacebookServerException.php' );
require_once( 'Facebook/FacebookOtherException.php' );
require_once( 'Facebook/FacebookAuthorizationException.php' );
require_once( 'Facebook/GraphObject.php' );
require_once( 'Facebook/GraphSessionInfo.php' );
require_once( 'Facebook/GraphUser.php' );

use Facebook\FacebookSession;
use Facebook\FacebookRedirectLoginHelper;
use Facebook\FacebookRequest;
use Facebook\FacebookResponse;
use Facebook\FacebookSDKException;
use Facebook\FacebookRequestException;
use Facebook\FacebookServerException;
use Facebook\FacebookOtherException;
use Facebook\FacebookAuthorizationException;
use Facebook\GraphObject;
use Facebook\GraphSessionInfo;
use Facebook\GraphUser;


  // Get session variable!
session_start();
$output = array(
);
// Initialize the app with the app's "secret" and ID. These are private values.
FacebookSession::setDefaultApplication( 'xxxxxxxxxxxx','xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx' );

$arg1 = $_SESSION['arg1']; //This is a password variable that tells the script it's being run by an authorized script on the server, not by an outside source.
$arg2 = $_SESSION['arg2']; //This is an argument that tells the script what it's supposed to be trying to do.
if($arg1 != "whatever"){
    echo"Something went wrong; the password variable didn't work.";
    //The following line is what is -supposed- to happen when this codeblock executes, but we're debugging right now, so we're providing output instead to see what happens in the code.
    //header("Location: http://www.fantasycs.com/app.php");
}elseif($arg2 == "loginurl"){
    echo "It worked! Login URL will be displayed here.";
    unset($_SESSION['arg2']);
    unset($_SESSION['arg1']);
    exit();
}
?>

最后，app.php：

<?php

session_start();
$_SESSION['arg1'] = "whatever";
$_SESSION['arg2'] = "loginurl";

$output = shell_exec('php-cli test.php Thisworks'); //This runs test.php and $output captures the output properly just fine. No worries.

echo $output;


$output = shell_exec('php test.php Thisworks'); //This keeps running indefinitely so far as I can tell, and I can't see what's happening because no output is produced. It simply keeps trying to load the page forever. No output is seemingly capture. Bizarre.

echo $output;


$output = shell_exec('php otherevent.php'); //This has the same problem as when you try to run test.php with the "php" command. It stalls.

echo $output;


$output = shell_exec('php-cli otherevent.php Thisworks'); //This produces an error on the "use" lines in otherevent.php. It says there's a syntax error on those lines. This has never happened or appeared before; these lines, simply put, do NOT have any errors in them. They've run fine before and have never been changed. As a result of the error, the script is aborted, and no output is captured.

echo $output;

?>

在这个例子中，我想要的是让app.php能够执行otherevent.php，获取登录URL（或输出的任何测试输出），并将登录URL打印到屏幕上。怎么做到这一点？显然它并不像看起来那么简单。

Answer 1

使用php-cli围绕'use'关键字的错误让我相信你有一个旧版本的PHP ..但是为了解决你的问题的一部分，我建议不要使用shell_exec来运行其他PHP代码..如果你需要捕获脚本的输出，你可以使用这样的输出缓冲区：

<?php

session_start();
$_SESSION['arg1'] = "whatever";
$_SESSION['arg2'] = "loginurl";

// Start capturing the output in a buffer
ob_start();
include 'otherevent.php';    
// Get the data and close the buffer
$output = ob_get_clean();
echo $output;

?>

如果您之后要立即回显输出，那么输出缓冲区没有意义。

检查您的PHP版本。该SDK需要5.4 +。

PHP - 作为命令行脚本执行Web脚本？

1 个答案: