测试字符串在Bash中是否包含非空白字符

Question

我的脚本正在阅读并显示id3标签.我试图让它回应未知如果该字段是空白但我尝试的每个if语句将无法工作.id3标签是固定大小的,因此它们永远不会为空,但如果没有值,它们将填充空白区域.IE标题标签的长度为30个字符.到目前为止我已经尝试过

echo :$string: #outputs spaces between the 2 ::

if [ -z "$string" ] #because of white space will always evaluate to true

x=echo $string | tr -d ' '; if [ -z "$string" ]; #still计算结果为true但回声:$ x:it echos ::

剧本

#!bin/bash
echo "$# files";
while [ "$i" != "" ];
do
   TAG=`tail -c 128 "$i" | head -c 3`;
   if [ $TAG="TAG" ]
   then
      ID3[0]=`tail -c 125 "$1" | head -c 30`;
      ID3[1]=`tail -c 95 "$1" | head -c 30`;
      ID3[2]=`tail -c 65 "$1" | head -c 30`;
      ID3[3]=`tail -c 35 "$1" | head 4`;
      ID3[4]=`tail -c 31 "$i" | head -c 28`;
      for i in "${ID3[@]}"
      do
         if [ "$(echo $i)" ] #the if statement mentioned
         then
            echo "N/A";
         else
            echo ":$i:";
         fi
      done
   else
      echo "$i does not have a proper id3 tag";
   fi
   shift;
done

Answer 1

这些答案中的许多答案都比它们应该更复杂,或者更不易读.

[[ $string = *[[:space:]]* ]]  && echo "String contains whitespace"
[[ $string = *[![:space:]]* ]] && echo "String contains non-whitespace"

Answer 2

您可以使用bash的正则表达式语法.

它要求你使用双方括号[[ ... ]](一般来说更通用).
该变量不需要引用.不得引用 正则表达式本身

for str in "         "  "abc      " "" ;do
    if [[ $str =~ ^\ +$ ]] ;then 
      echo -e "Has length, and contain only whitespace  \"$str\"" 
    else 
      echo -e "Is either null or contain non-whitespace \"$str\" "
    fi
done

产量

Has length, and contain only whitespace  "         "
Is either null or contain non-whitespace "abc      " 
Is either null or contain non-whitespace ""