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Exec启动php作为后台进程(Exec to launch php as background process)

 我正在努力获取一个php文件，以使用PHP的exec（）在后台运行。 作为第一个测试，我尝试过：  
exec("ls -l > logfile.txt 2> errfile.txt &");
 
 这工作正常。 logfile.txt获得一个目录列表。  
 按照php文档中的说明，由于exec启动了在后台运行的进程，标准输出（和标准错误）被重定向到一个文件。  
 现在，我试试  
exec("/usr/bin/php -f /path/to/my.php > logfile.txt 2> errorfile.txt &");
 
 它似乎没有任何反应。  
 这里是我正在尝试的测试文件：  
 alpha.php  
<?php 
    $version="a"; 
    // Go do something we do not need to wait for. 
    exec("/usr/bin/php -f /path/to/beta.php > logfile.txt 2> errorfile.txt &"); 
?>

<html> 
    <head><title>Test</title></head> 
    <body> 
    <p>This is Alpha version <?php echo $version; ?></p> 
    </body> 
</html>
 
 beta.php  
<?php 
    if (!($fp = fopen('/home/johnst12/public_html/workshops/admin/betadata.txt', 'w'))) { exit;} 
    fprintf($fp, "Proof that Beta executed."); 
    fclose($fp); 
?>
 
 如果我直接运行beta.php，它工作正常。 Betadata.txt获取消息。  
 如果我运行alpha.php来启动beta.php，则不会创建betadata.txt。 logfile.txt和errorfile.txt保持为空（预期）。  
 我确信php的路径和我的php文件的路径是正确的。  
 谷歌搜索澄清没有收获。 一些共同的主题似乎是（a）资源耗尽？ （b）目标php文件没有权限？ 资源似乎不太可能。 脚本的权限是全局读取644（rw-r - r--）。 我尝试添加执行（755）以防万一它会帮助。 它没有任何区别。  
 PHP版本5.3.21 Linux / Apache系统。 safe_mode关闭  
 我错过了什么？ 谢谢。 

I am struggling with getting a php file to run in the background with PHP's exec(). As a first test, I tried :  
exec("ls -l > logfile.txt 2> errfile.txt &");
 
That works fine. logfile.txt gets filled with a directory listing. 
Per instructions in the php documentation, since the exec kicks off a process that runs in the background, standard out (and standard error) are redirected to a file.  
Now, I try 
exec("/usr/bin/php -f /path/to/my.php > logfile.txt 2> errorfile.txt &");
 
It appears nothing happens.  
Here are test files that I'm trying: 
alpha.php 
<?php 
    $version="a"; 
    // Go do something we do not need to wait for. 
    exec("/usr/bin/php -f /path/to/beta.php > logfile.txt 2> errorfile.txt &"); 
?>

<html> 
    <head><title>Test</title></head> 
    <body> 
    <p>This is Alpha version <?php echo $version; ?></p> 
    </body> 
</html>
 
beta.php 
<?php 
    if (!($fp = fopen('/home/johnst12/public_html/workshops/admin/betadata.txt', 'w'))) { exit;} 
    fprintf($fp, "Proof that Beta executed."); 
    fclose($fp); 
?>
 
If I run beta.php directly, it works fine. Betadata.txt gets the message.  
If I run alpha.php to launch beta.php, betadata.txt is not created. logfile.txt and errorfile.txt remain empty (expected). 
I am sure that the path to php, and the path to my php file are correct.  
Googling for clarification has not been fruitful. A couple of common themes seem to be (a) running out of resources? (b) lack of permission on the target php file? Out of resources seems unlikely. The permission on the script is global read 644 (rw-r--r--). I tried adding execute (755) just in case it would help. It made no difference.  
PHP version 5.3.21 Linux/Apache system. safe_mode Off 
What am I missing? Thanks.

原文：https://stackoverflow.com/questions/16290784

更新时间：2022-08-16 07:08

最满意答案

 以下不是语法  
<ArgType>genericMethod()
 
 看起来类型参数必须在一个点之后  
SomeClass.<ArgType>genericMethod()
this.<ArgType>genericMethod()
p.<ArgType>genericMethod()
super.<ArgType>genericMethod()
SomeClass.super.<ArgType>genericMethod()
SomeClass.this.<ArgType>genericMethod()

The following is not the syntax  
<ArgType>genericMethod()
 
It seems the type arguments must come after a dot as in 
SomeClass.<ArgType>genericMethod()
this.<ArgType>genericMethod()
p.<ArgType>genericMethod()
super.<ArgType>genericMethod()
SomeClass.super.<ArgType>genericMethod()
SomeClass.this.<ArgType>genericMethod()

Exec启动php作为后台进程(Exec to launch php as background process)

最满意答案

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