function repeatString()
{
    local -r string="${1}"
    local -r numberToRepeat="${2}"

    if [[ "${string}" != '' && "${numberToRepeat}" =~ ^[1-9][0-9]*$ ]]
    then
        local -r result="$(printf "%${numberToRepeat}s")"
        echo -e "${result// /${string}}"
    fi
}

样本运行

$ repeatString 'a1' 10 
a1a1a1a1a1a1a1a1a1a1

$ repeatString 'a1' 0 

$ repeatString '' 10 

参考库:https://github.com/gdbtek/linux-cookbooks/blob/master/libraries/util.bash

如果你想重复一个字符n次，n是一个变量的次数，这取决于，比如说，字符串的长度，你可以这样做:

#!/bin/bash
vari='AB'
n=$(expr 10 - length $vari)
echo 'vari equals.............................: '$vari
echo 'Up to 10 positions I must fill with.....: '$n' equal signs'
echo $vari$(perl -E 'say "=" x '$n)

它显示:

vari equals.............................: AB  
Up to 10 positions I must fill with.....: 8 equal signs  
AB========  

printf -- '=%.0s' {1..100}

双破折号——表示“命令行标志的结束”，所以不要试图解析命令行选项后面的内容。

如果你想打印破折号字符，而不是=字符，多次，不包括双破折号-这是你会得到的:

$ printf '-%.0s' {1..100}
bash: printf: -%: invalid option
printf: usage: printf [-v var] format [arguments]

为什么不创建这样的一行函数呢:

function repeat() { num="${2:-100}"; printf -- "$1%.0s" $(seq 1 $num); }

然后，你可以这样调用它:

$ repeat -
----------------------------------------------------------------------------------------------------

或者像这样:

$ repeat =
====================================================================================================

或者像这样:

$ repeat '*' 8
********

没有简单的方法。但是举个例子:

seq -s= 100|tr -d '[:digit:]'
# Editor's note: This requires BSD seq, and breaks with GNU seq (see comments)

或者是一种符合标准的方式:

printf %100s |tr " " "="

还有一个tput代表，但对于我手头的终端(xterm和linux)，它们似乎不支持它:)

如果你想在echo和printf的不同实现之间遵循posix并保持一致性，和/或shell而不仅仅是bash:

seq(){ n=$1; while [ $n -le $2 ]; do echo $n; n=$((n+1)); done ;} # If you don't have it.

echo $(for each in $(seq 1 100); do printf "="; done)

.．.将在所有地方产生与perl -E 'say "=" x 100'相同的输出。

我的答案有点复杂，可能并不完美，但对于那些希望输出大数字的人来说，我能够在3秒内完成大约1000万。

repeatString(){
    # argument 1: The string to print
    # argument 2: The number of times to print
    stringToPrint=$1
    length=$2

    # Find the largest integer value of x in 2^x=(number of times to repeat) using logarithms
    power=`echo "l(${length})/l(2)" | bc -l`
    power=`echo "scale=0; ${power}/1" | bc`

    # Get the difference between the length and 2^x
    diff=`echo "${length} - 2^${power}" | bc`

    # Double the string length to the power of x
    for i in `seq "${power}"`; do 
        stringToPrint="${stringToPrint}${stringToPrint}"
    done

    #Since we know that the string is now at least bigger than half the total, grab however many more we need and add it to the string.
    stringToPrint="${stringToPrint}${stringToPrint:0:${diff}}"
    echo ${stringToPrint}
}