在PHP中替代exec？

时间：2011-04-16 05:18:02

标签： php

我有一个在后台运行的PHP脚本，可以将图片发布到不同的网站。上传图片后，它会调用脚本运行如下：

$user_id = "77";
exec("/usr/bin/php5 upload_auto.php?user_id=$user_id");

在调用之后，它会运行如下代码：

<?php
$the_picture_id_to_check = $_GET['user_id'];
mysql_connect("localhost","********", "********") or die(mysql_error());
mysql_select_db(myflashp_pictures) or die (mysql_error());
require_once('connect/twitter_files/twitteroauth/twitteroauth.php');
require_once('connect/twitter_files/config.php');

?>

<!--- Upload Tumblr --->

<?php
$result_tumblr = mysql_query("SELECT * FROM upload_tumblr WHERE user_id=$the_picture_id_to_check"); 
$num_rows_find = mysql_num_rows($result_tumblr);
while($row_tumblr=mysql_fetch_array($result_tumblr)){

// Get Username
$the_id = "".$row_tumblr['id']."";
$picture_id = "".$row_tumblr['picture_id']."";
$picture_url = "".$row_tumblr['picture_url']."";
$user_id = "".$row_tumblr['user_id']."";
$caption = "".$row_tumblr['caption']."";
$album_id = "".$row_tumblr['album_id']."";

// Authorization info
$monday_result = mysql_query("SELECT * FROM photo_albums WHERE id='$album_id'");
$monday_query_row=mysql_fetch_array($monday_result);
$tumblr_id = $monday_query_row['tumblr_id']; 

$monday_result2 = mysql_query("SELECT * FROM connections_tumblr WHERE id='$tumblr_id'");
$monday_query_row2 = mysql_fetch_array($monday_result2);
$tumblr_email = $monday_query_row2['email']; 
$tumblr_password = $monday_query_row2['password']; 

// Prepare POST request
$request_data = http_build_query(
array(
    'email'             => $tumblr_email,
    'password'          => $tumblr_password,
    'type'              => 'photo',
    'source'            => $picture_url,
    'caption'           => $caption,
    'generator'         => 'FlashPics iPhone App'
)
);

// Send the POST request (with cURL)
$c = curl_init('http://www.tumblr.com/api/write');
curl_setopt($c, CURLOPT_POST, true);
curl_setopt($c, CURLOPT_POSTFIELDS, $request_data);
curl_setopt($c, CURLOPT_RETURNTRANSFER, true);
$result = curl_exec($c);
$status = curl_getinfo($c, CURLINFO_HTTP_CODE);
curl_close($c);

// Check for success
if ($status == 201) {
mysql_query("UPDATE pictures SET tumblr_id='$result' WHERE id='$picture_id'");
echo "Success! The new Tumblr post ID is $result.<br>\n";
mysql_query("DELETE FROM upload_tumblr WHERE id='$the_id'");
} else if ($status == 403) {
echo 'Bad email or password for Tumblr.';
} else if ($status == 400) {
echo "Error ($status): $result\n";
mysql_query("DELETE FROM upload_tumblr WHERE id='$the_id'");
} else {
echo "Error ($status): $result\n";
}

}

if ($num_rows_find == 0) {
echo "There's nothing to post to Tumblr.";
}

?>

<br /><br />



<!--- Upload Email --->

<?php

include_once('class.phpmailer.php');

$result_email = mysql_query("SELECT * FROM upload_email WHERE status='processing' AND user_id='$the_picture_id_to_check'"); 
$num_rows_find = mysql_num_rows($result_email);
mysql_query("UPDATE upload_email SET status='$num_rows_find' WHERE id='1'");
while($row_email=mysql_fetch_array($result_email)){

// Get Username
$to = "".$row_email['email']."";


$get_info_result = mysql_query("SELECT * FROM users WHERE id='".$row_email['user_id']."'");
$get_info_query_row = mysql_fetch_array($get_info_result);
$get_username = $get_info_query_row['username'];
$get_first_name = $get_info_query_row['first_name'];
$get_last_name = $get_info_query_row['last_name']; 

$get_info_result = mysql_query("SELECT * FROM pictures WHERE id='".$row_email['picture_id']."'");
$get_info_query_row = mysql_fetch_array($get_info_result);
$get_caption = $get_info_query_row['caption'];
$get_picture_url = $get_info_query_row['long_url'];
$get_short_string = $get_info_query_row['short_string'];

if ($get_caption == "") {
$get_caption = "I have shared a FlashPic with you.";
}


$mail             = new PHPMailer();
$mail->From       = "share@myflashpics.com";
$mail->FromName   = "FlashPics";
$mail->Subject    = "$get_first_name $get_last_name has shared a FlashPic with you.";

$mail->IsHTML(true);
$mail->Body = "
<html><body style='margin: 0px; padding: 0px;'>
<div style='padding: 15px; width: 480px; font-family: Helvetica; margin: 10px; border: 1px solid #d7d7d7;'>
<div style='font-weight: bolder; font-size: 32px; margin: 8px 0px 20px 0px;'><font color='#353535'>Flash</font><font color='#f8bc49'>Pics</font></div>

<div style='font-size: 20px; color: #353535; font-weight: normal; text-transform: lowercase;'>$get_username</div>

<div style='margin-top: 15px; font-size: 14px; line-height: 18px;'>$get_caption<br>
<br>
<img src=$get_picture_url width='480'>
<br><br>
<a href='http://myflashpics.com/picture/$get_short_string' style='color: #000000; text-decoration: underline;'>view full size image</a><br>
<br>
Sincerely,<br>
The FlashPics Team</div>

</div>
</body></html>";

$mail->AltBody    = "$get_username
$get_caption

http://myflashpics.com/picture/".$row_email['picture_id']."

Yours Truly,
The FlashPics Team";


$the_id = "".$row_email['id']."";

$pieces = explode(", ", "$to");

$one = $pieces['0'];
$two = $pieces['1']; 
$three = $pieces['2']; 
$four = $pieces['3']; 
$five = $pieces['4']; 

if ($one != "") {
$mail->AddBCC("$one");
}
if ($two != "") {
$mail->AddBCC("$two");
}
if ($three != "") {
$mail->AddBCC("$three");
}
if ($four != "") {
$mail->AddBCC("$four");
}
if ($five != "") {
$mail->AddBCC("$five");
}

if(!$mail->Send()) {
echo "Failed sending email to <b> $one, $two, $three, $four, $five</b>." . $mail->ErrorInfo;
mysql_query("UPDATE upload_email SET status='error' WHERE id='$the_id'");
} else {
echo "Sent email to <b> ".$row_email['email']."</b> successfully.<br>";
mysql_query("UPDATE upload_email SET status='sent' WHERE id='$the_id'");
}

}

if ($num_rows_find == 0) {
echo "There are no pictures to email out.";
}

?>

所以基本上只有某些部分可以使用（我没有发布我的所有代码）。我刚发布了什么不起作用。 但是如果我从浏览器手动运行它，所有部分都可以运行。如果你想看到我的php_info，我有here。

提前致谢，
库尔顿

PS：“不工作”，我的意思是没有任何反应，我看到没有错误

2 个答案:

答案 0 :(得分：3)

从评论主题，我认为exec无效。

尝试：

  1. 将路径文件更改为绝对路径：

  2. 此外，因为它意味着在shell上运行，所以将参数更改为shell脚本的工作原理：

    exec("/usr/bin/php5 /path/to/whatever/your/file/is/upload_auto.php $user_id");

    3. 注意

    请参阅我的更改参数：

    upload_auto.php?user_id=$user_id

    到

    upload_auto.php $user_id

    shell脚本如何工作。

    要获取该参数，您应该使用$argv：

    $user_id = $argv[1];

    或者，为了能够在浏览器和shell上运行，请将其更改为：

    $user_id = (php_sapi_name() == 'cli' && empty($_SERVER['REMOTE_ADDR'])) ? $argv[1] : $_GET['user_id'];

答案 1 :(得分：1)

您正在从命令行调用php脚本，但您正在尝试从$ _GET变量中获取值。它不会起作用。 尝试从argv [1]中获取值。 $ the_picture_id_to_check = argv [1];

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