bash - 如果块在明显应该运行时没有运行，则 Bash

Question

编码：

first=true
mountpoint=" "
partlist=`df -h | grep "^/dev"` # get partition info
for i in $partlist              # loop through info
do
  if [[ ${i:0:1} = "/" ]]       # items starting with / are what I'm interested in
  then
    if [[ $first ]]             # The first instance in each pair is the device name
    then
      mountdev=$i
      mountpoint=" "
      first=false
    else                        # The second instance is the device mount point
      mountpoint=$i
      first=true
    fi
    if [[ $mountpoint != " " ]] # If the mountpoint was just set (last to be set
                                #   before printing config)
    then
      # Print config
      echo "${mountpoint} \${fs_size ${mountpoint}"
      echo "USED \${fs_used ${mountpoint}}\${alignr}\${fs_free ${mountpoint}} FREE"
      echo "\${fs_bar 3,300 ${mountpoint}}"
      echo "READ \${diskio_read  ${mountdev}}\${alignr}\${diskio_write  ${mountdev}} WRITE"
      echo ""
    fi
  fi
done

问题：

目标是创建一个脚本，该脚本将为我的每台计算机生成我喜欢的 conky 配置，而无需为每台计算机编辑文件。以上是一个片段，它生成报告我的磁盘使用信息的部分。不幸的是，我无法让它输出任何东西。我向其中抛出了各种调试回声，除了实际输出配置的 if 语句外，它似乎都正常工作。我无法弄清楚它有什么问题。有什么建议或帮助吗？

提前谢谢你:)

Answer 1

这是错误的行：

if [[ $first ]]             # The first instance in each pair is the device name

那总是评估为真。[[ true ]]or是or[[ false ]]的同义词，它总是正确的。[[ -n true ]][[ -n false ]]

你的意思可能是

if "$first"

或者

if [[ $first == true ]]