如何将字符串扩展到一定长度 [英] How to extend string to certain length

问题描述

基本上现在我的程序给了我这个输出:

Hey basically right now my program gives me this output:

BLABLABLA
TEXTEXOUAIGJIOAJGOAJFKJAFKLAJKLFJKL
TEXT
随机字符超出限制的更多文本

BLABLABLA
TEXTEXOUAIGJIOAJGOAJFKJAFKLAJKLFJKL
TEXT
MORE TEXT OF RANDOM CHARACTER OVER LIMIT

是for循环的结果.现在，这就是我想要的:

which is a result of for loop. Now here's what i want:

• 如果字符串超过10个字符，则剪掉其余部分并添加两个点& ;；以冒号结尾"..:"
• 否则(如果字符串少于10个字符)用空格填充空格，以便使用

所以在我提供的示例中，我想要这样的东西作为输出:

so on the example i provided i'd want something like this as output:

BLABLABLA  :  
TEXTEXOUA..:  
TEXT       :
MORE TEXT..:

我也解决了问题的第一部分(当字符超过10个字符时)，只有第二部分给我带来了麻烦.

I also solved the first part of the problem (when its over 10 characters), only the second one gives me trouble.

AMOUNT=definition here, just simplyfying so not including it
for (( i=1; i<="$AMOUNT"; i++ )); do
STRING=definition here, just simplyfying so not including it
DOTS="..:"
STRING_LENGTH=`echo -n "$STRING" | wc -c`
if [ "$STRING_LENGTH" -gt 10 ]
    then
          #Takes 
          STRING=`echo -n "${STRING:0:10}"$DOTS` 
    else
          #now i dont know what to do here, how can i take my current $STRING 
          #and add spaces " " until we reach 10 characters. Any ideas?
    fi

推荐答案

Bash提供了一种简单的方法来获取存储在变量中的字符串的长度: $ {#STRING}

Bash provides a simple way to get the length of a string stored in a variable: ${#STRING}

STRING="definition here, just simplyfying so not including it"
if [ ${#STRING} -gt 10 ]; then
    STR12="${STRING:0:10}.."
else
    STR12="$STRING            "         # 12 spaces here    
    STR12="${STR12:0:12}"
fi
echo "$STR12:"

您发布的预期输出与问题中的要求不符.我试图遵循这些要求，而忽略了示例预期输出和您发布的代码.

The expected output you posted doesn't match the requirements in the question. I tried to follow the requirements and ignored the sample expected output and the code you posted.

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