使用Bash比较PHP版本号? [英] Comparing PHP version numbers using Bash?

问题描述

我有这个脚本，应该确保用户当前的PHP版本在一定范围内，尽管它应该工作，但是某个地方存在一个错误，使其认为该版本超出范围，有人可以看看吗?并告诉我该如何解决?

I have this script that should make sure that the users current PHP version is between a certain range, though it SHOULD work, there is a bug somewhere that makes it think that the version is out of range, could someone take a look and tell me what I can do to fix it?

function version { echo "$@" | gawk -F. '{ printf("%d.%d.%d\n", $1,$2,$3); }'; }

phpver=`php -v |grep -Eow '^PHP [^ ]+' |gawk '{ print $2 }'`

if [ $(version $phpver) > $(version 5.2.13) ] || [ $(version $phpver) < $(version 5.2.13) ]; then
  echo "PHP Version $phpver must be between 5.2.13 - 5.3.15"
  exit
fi

推荐答案

此处是比较版本的方法.

Here's how to compare versions.

使用sort -V:

using sort -V:

function version_gt() { test "$(printf '%s\n' "$@" | sort -V | head -n 1)" != "$1"; }

示例用法:

first_version=5.100.2
second_version=5.1.2
if version_gt $first_version $second_version; then
     echo "$first_version is greater than $second_version !"
fi

pro:

• 比较精美版本字符串的可靠方法:
  • 支持任意长度的子部分(例如:1.3alpha.2.dev2> 1.1吗?)
  • 支持字母顺序排列(即:1.alpha< 1.beta2)
  • 支持大尺寸版本(即1.10003939209329320932> 1.2039209378273789273吗?)
  • solid way to compare fancy version strings:
    • support any length of sub-parts (ie: 1.3alpha.2.dev2 > 1.1 ?)
    • support alpha-betical sort (ie: 1.alpha < 1.beta2)
    • support big size version (ie: 1.10003939209329320932 > 1.2039209378273789273 ?)
    • 通常非常有用，带有3个参数:(即:1.2< my_version< 2.7)

    缺点:

    • 使用许多不同的程序来调用不同的程序.因此效率不高.
    • 使用的是sort的较新版本，可能无法在您的计算机上使用 系统. (与man sort一起检查)
    • uses a lot of various calls to different programs. So it's not that efficient.
    • uses a pretty recent version of sort and it might not be available on your system. (check with man sort)

    没有sort -V:

    without sort -V:

    ## each separate version number must be less than 3 digit wide !
    function version { echo "$@" | gawk -F. '{ printf("%03d%03d%03d\n", $1,$2,$3); }'; }
    

    示例用法:

    first_version=5.100.2
    second_version=5.1.2
    if [ "$(version "$first_version")" -gt "$(version "$second_version")" ]; then
         echo "$first_version is greater than $second_version !"
    fi
    

    pro:

    • 快速解决方案，因为它仅调用1个子进程
    • 兼容得多的解决方案.

    缺点:

    • 非常特定的版本字符串必须:
      • 仅具有1、2或3个部分的版本. (不包括"2.1.3.1")
      • 每个部分都只能是数字(不包括"3.1a")
      • 每个部分不能大于999(不包括"1.20140417")
      • quite specific, version string must:
        • have version with 1, 2 or 3 parts only. (excludes '2.1.3.1')
        • each parts must be numerical only (excludes '3.1a')
        • each part can't be greater than 999 (excludes '1.20140417')

        关于脚本的评论:

        我看不到它如何工作:

        如注释中所述
        • >和<是非常特殊的shell字符，您应将它们替换为-gt和-lt
        • 即使替换了字符，也无法将版本号视为整数或浮点数.例如，在我的系统上，php版本是5.5.9-1ubuntu4.
        • as stated in a comment > and < are very special shell character and you should replace them by -gt and -lt
        • even if you replaced the characters, you can't compare version numbers as if they where integers or float. For instance, on my system, php version is 5.5.9-1ubuntu4.

        但是您的函数version()已经非常聪明地编写了，并且可以绕过经典问题，即按字母数字排序不会对数字进行数字排序(按字母顺序1 ＜11 ＜2，这在数字上是错误的)可以为您提供帮助.但是要小心:bash不支持任意大的数字(如果要与32位系统兼容，请尝试将数字保持在32位以下，因此该数字应为9位长).因此，我已经修改了您的代码(在第二种方法中，不使用sort -V)仅对版本字符串的每个部分强制使用3位数字.

        But your function version() is quite cleverly written already and may help you by circumventing the classical issue that sorting alphabetically numbers won't sort numbers numerically ( alphabetically 1 < 11 < 2, which is wrong numerically). But be carefull: arbitrarily large numbers aren't supported by bash (try to keep under 32bits if you aim at compatibility with 32bits systems, so that would be 9 digit long numbers). So I've modified your code (in the second method NOT using sort -V) to force only 3 digits for each part of the version string.

        编辑:应用了@phk改进，因为它非常聪明，并使用sort删除了第一个版本中的子流程调用.谢谢.

        EDIT: applied @phk amelioration, as it is noticeably cleverer and remove a subprocess call in the first version using sort. Thanks.

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