Bash 命令嵌套条件语句失败，但为什么

Question

我有一个 bash 脚本，其中运行有很多条件语句，但是当我运行脚本时，它只会命中以下行：if [ $? -eq 0 ]

touch /tmp/converted.json
tool test --json --all-projects | tee >> /tmp/converted.json
ret=$?
set -o pipefail

#no vulns
if ((ret == 0)); then
  echo "Hooray!"

#found vulns
elif ((ret == 1)); then
then                    
  #check single project
  if [[ $firstchar == "{" ]]
  then         
    echo "SINGLE PROJECT"
  elif [[ $firstchar == "[" ]]
  then
    echo "MULTIPROJECT"
  else
    echo "didnt get a single or multiproject"
  fi
#error with your command
elif ((ret == 2)); then
then
  echo "An error occurred"

#no manifest
elif ((ret == 3)); then
then
  echo "Ensure your repo has a proper required manifest file"

else
  echo "Something else occurred"
fi

那么，为什么会一直满足if((ret==0))条件呢？

谢谢

Answer 1

$? 具有执行的 last 命令的退出状态。 elif [ $? -eq 1 ] 的最后一个命令是 [ 在 if [ $? -eq 0 ] 执行的。

if [ $? -eq 0 ]      # executes `[`
then 
    : dont care
elif [ $? -eq 1 ]    # $? here has a nonzero the exit status of `[`

在执行你感兴趣的命令后调用任何命令之前，只需将退出状态保存到变量中。在bash中你可以使用算术展开((...))。

some_command
ret=$?
if ((ret == 0)); then
    : dont care
elif ((ret == 1)); then

Bash 命令嵌套条件语句失败，但为什么

Bash Command Nested conditional statements failing, but why

bash

logic