Question

我想在tar中使用exec命令和Php备份我的网站，我写了一个小脚本来做到这一点但没有任何反应......哪里有错？我有php 5.6.5并托管具有exec启用和tar命令的linux。这是我想要做的Php示例。

<?php
  $root = $_SERVER['DOCUMENT_ROOT'];
  # root is /web/htdocs/www.example.com/home/
  $name = "backup_" . date("[d-m-Y][H-i]") . ".tar.gz";
  # name is backup_[25-02-2015][18-57].tar.gz
  $skip = "*.gz";
  # skip is the file I want to exclude (example: skip backup_[25-02-2015][18-57].tar.gz)
  if ((substr($_SERVER['DOCUMENT_ROOT'],-1,1) == "/") && (substr($_SERVER['PHP_SELF'],0,1) =="/")) {
    $sdir = $_SERVER['DOCUMENT_ROOT'] . substr(dirname($_SERVER['PHP_SELF']),1);
  } else {
    $sdir = $_SERVER['DOCUMENT_ROOT'] . dirname($_SERVER['PHP_SELF']);
  }
  # sdir is /web/htdocs/www.example.com/home/bak/ and is the path where the script lives
  # out is the output
  # oky is the success o failed exec command
  function backup()  {
    exec("tar -cvf $sdir/$name $root/* --exclude='$sdir/$skip' ", $out, $oky);
  }
  backup();
  if (!$oky) {
    echo "$out: Backup Completed!";
  } else {
    echo "$out: Backup Not Completed!";
  }
?>

感谢任何帮助！

Answer 1

$oky和$out是局部变量。它们不在功能之外。 $sdir，$name和$root未在函数中定义。

方法1 - 参数：

function backup($sdir,$name,$root,$salt)  {
    exec("tar -cvf $sdir/$name $root/* --exclude='$sdir/$salt' ", $out, $oky);
    return array("oky"=>$oky, "out"=>$out);
}

$result = backup($sdir, $name, $root, $salt);

if (!$result["oky"]) {
    echo $result["out"].": Backup Completed!";
} else {
    echo $result["out"].": Backup Not Completed!";
}

方法2 - 不要使用功能：

exec("tar -cvf $sdir/$name $root/* --exclude='$sdir/$salt' ", $out, $oky);

if (!$oky) {
    echo "$out: Backup Completed!";
} else {
    echo "$out: Backup Not Completed!";
}

方法3 - 全局变量：

function backup()  {
    global $sdir,$name,$root,$salt,$oky,$out;
    exec("tar -cvf $sdir/$name $root/* --exclude='$sdir/$salt' ", $out, $oky);
}

backup();

if (!$oky) {
    echo "$out: Backup Completed!";
} else {
    echo "$out: Backup Not Completed!";
}

exec php tar命令动态备份网站

1 个答案: