Question

我正在编写一个脚本来合并漫画1110的xkcd webcomic磁贴。到目前为止，我已经编写了大约100行，而不是分别对图像的每个象限进行编写，我决定使用函数来执行此操作，为每个象限重复并使用每个象限的变量使每次函数的行为不同。我想能够做到这样的事情，除了每次都有不同的价值。

a='((n=$north; n>=1; n--))';
for ${a};
    do "echo Hello World!"
done

我的问题是，我在for循环表达式中使用变量时遇到了麻烦，它变为“第8行：eval”$ a“'：不是有效的标识符”，如果我尝试将它放在括号中，它就会出现“第30行：语法错误：需要算术表达式”，然后“第30行：语法错误：`（（$ {a}））”。这就是我到目前为止所拥有的。在有人回答之前，是的，我知道它有错误。

#!/bin/bash
# Date Created: September 20th, 2012
# Created by Jonathan Bondhus - https://github.com/jbondhus
# Credit to Andreas Reichinger for analysis of image tile placement, xkcd.com for comic. Used ImageMagick code from Antonio Frascarelli.
#
#  This script will merge the files from xkcd comic number 1110, "Click and Drag".

function function_join() {
    echo "#";
    if [[ quadrant -eq 1 ]]; then
    echo "# First Quadrant:"; # Join together the first quadrant
    function_quadrant
    else
        if [[ quadrant -eq 2 ]]; then
        echo "# Second Quadrant:"; # Join together the second quadrant
        function_quadrant
        else
            if [[ quadrant -eq 3 ]]; then
            echo "# Third Quadrant:"; # Join together the third quadrant
            function_quadrant
            else
                if [[ quadrant -eq 4 ]]; then
                echo "# Fourth Quadrant:"; # Join together the fourth quadrant
                function_quadrant
                fi
            fi
        fi
    fi
    echo "#";
    for ${a};
    do
        echo "# row number "$n" ...";
        convert $b;
        for `eval "$c"`;
        do
            if [ $d ];      # If the tile file doesn't exist, an empty square will be used.
                then convert $e;
                else convert $f;
            fi;
        done;
        convert $g;    # Append the image onto the current quadrant
        rm -f $h;
        echo "# ... OK";
        echo "#";
    done;
}

function function_quadrant() {
    if [[ quadrant -eq 1 ]]; then # Set the variables for function_join
    quadrantName="first"
    a='((n=$north; n>=1)); n--'
    b='-size 0x1 xc:white resultn$n''w.png'
    c='((w=$west; w>=1; w--))'
    d='-e $n''n$w''w.png'
    e='+append resultn$n''w.png $n''n$w''w.png resultn$n''w.png'
    f='+append resultn$n''w.png _blank.png resultn$n''w.png'
    g='-append ${quadrantName}Quadrant.png resultn$n''w.png ${quadrantName}Quadrant.png'
    h='resultn$n''w.png'
    else
        if [[ quadrant -eq 2 ]]; then
        quadrantName="second"
        a='n=$north; n>=1; n--'
        b='-size 0x1 xc:white resultn$n''e.png'
        c='e=$east; e<=20; e++'
        d='-e $n''n$e''e.png'
        e='+append resultn$n''e.png $n''n$e''e.png resultn$n''e.png'
        f='+append resultn$n''e.png _blank.png resultn$n''e.png'
        g='-append ${quadrantName}Quadrant.png resultn$n''e.png ${quadrantName}Quadrant.png'
        h='resultn$n''e.png'
        else
            if [[ quadrant -eq 3 ]]; then
            quadrantName="third"
            a='s=1; s<=$south; s++'
            b='-size 0x1 xc:black results$s''w.png'
            c='w=$west; w>=1; w--'
            d='-e $s''s$w''w.png'
            e='+append results$s''w.png $s''s$w''w.png results$s''w.png'
            f='+append results$s''w.png _black.png results$s''w.png'
            g='-append ${quadrantName}Quadrant.png results$s''w.png ${quadrantName}Quadrant.png'
            h='results$s''w.png'
            else
                if [[ quadrant -eq 4 ]]; then
                quadrantName="fourth"
                a='s=1; s<=$south; s++'
                b='-size 0x1 xc:black results$s''e.png'
                c='e=1; e<=20; e++'
                d='-e $s''s$e''e.png'
                e='+append results$s''e.png $s''s$e''e.png results$s''e.png'
                f='+append results$s''e.png _black.png results$s''e.png'
                g='-append ${quadrantName}Quadrant.png results$s''e.png ${quadrantName}Quadrant.png'
                h='results$s''e1.png'
                else # If more than 4 quadrants, just to be safe, exit
                    exit 1
                fi
            fi
        fi
    fi
}

for (( quadrant = 0; quadrant < 4; quadrant++ )); do
    function_join
done

echo "#################################"
echo "Joining completed!"
echo "#################################"

exit 0

Answer 1

./north.sh 
Hello World!  4 5
Hello World!  3 4
Hello World!  2 3
Hello World!  1 2
Hello World!  0 1

cat north.sh

#!/bin/bash
north=5;
for ((n=$north; n>=1; --n))
do
        ((north--))
     echo "Hello World!  $north $n"
done

Answer 2

不敢。语法解析发生在参数扩展之前（毕竟需要首先识别参数）。我建议重构代码以传递数据，而不是可执行代码。

是否可以将变量作为for循环表达式？

2 个答案: