将 PHP 变量传递给 Bash 脚本，然后启动它
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Pass PHP variables to a Bash script and then launch it
提问by Axiol
So, basically, what I want to do is execute this Bash script :
所以，基本上，我想做的是执行这个 Bash 脚本：
#!/bin/bash
path="./"
filename="Ellly.blend"
file=${path}${filename}
description="Test of the api with a simple model"
token_api="ff00ff"
title="Uber Glasses"
tags="test collada glasses"
private=1
password="Tr0b4dor&3"
curl -k -X POST -F "fileModel=@${file}" -F "filenameModel=${filename}" -F "title=${title}" -F "description=${description}" -F "tags=${tags}" -F "private=${private}" -F "password=${password}" -F "token=${token_api}" https://api.sketchfab.com/v1/models
Inside a PHP function. Problem is, I don't really see how I can pass to it some variables and then wait for it to end before resuming the PHP function.
在 PHP 函数中。问题是，我真的不知道如何将一些变量传递给它，然后在恢复 PHP 函数之前等待它结束。
Any help ?
有什么帮助吗？
回答by Levi Morrison
Note: I didn't do all the settings, just enough I hope you get the idea.
注意：我没有做所有的设置，就足够了，我希望你能明白。
Option 1: Passing parameters
选项 1：传递参数
PHP:
PHP：
$file = escapeshellarg($file);
$filename = escapeshellarg($filename);
// escape the others
$output = exec("./bashscript $file $filename $tags $private $password");
Bash:
重击：
#!/bin/bash
filename=
file=
description="Test of the api with a simple model"
token_api="ff00ff"
title="Uber Glasses"
tags=
private=
password=
...
Option 2: Using environment variables
选项 2：使用环境变量
PHP:
PHP：
putenv("FILENAME=$filename");
putenv("FILE=$file");
putenv("TAGS=$tags");
putenv("PRIVATE=$private");
putenv("PASSWORD=$PASSWORD");
$output = exec('./bash_script');
Bash:
重击：
filename=$FILENAME
file=$FILE
description="Test of the api with a simple model"
token_api="ff00ff"
title="Uber Glasses"
tags=$TAGS
private=$PRIVATE
password=$PASSWORD
...
