如何检查外部外壳脚本退出代码

Question

我有一个简单的"主"shell脚本，它做一些准备工作，然后调用另一个将文件上传到ftp站点的shell脚本。我想知道如何等待并检查被调用的 shell 脚本的退出代码，以及如何轻松检查 FTP 文件是否真的成功上传并提供正确的退出代码（0 或 1）

谢谢

主脚本：

#!/bin/sh
# check for build tools first
FTP_UPLOAD_SCRIPT=~/Desktop/ftp_upload.sh
if [ -f "$FTP_UPLOAD_SCRIPT" ]; then
    echo "OK 3/5 ftp_upload.sh found. Execution may continue"
else
    echo "ERROR ftp_upload.sh not found at $FTP_UPLOAD_SCRIPT. Execution cannot continue."
exit 1
fi
  # upload the packaged installer to an ftp site
  sh $FTP_UPLOAD_SCRIPT
  # check the ftp upload for its exit status
  ftp_exit_code=$?
  if [[ $ftp_exit_code != 0 ]] ; then
    echo "FTP ERRORED"
    exit $ftp_exit_code
  else
    echo $ftp_exit_code
    echo "FTP WENT FINE"
  fi
  echo "n"
  exit 0

ftp_upload_script：

#!/bin/sh
FTP_HOST='myhost'
FTP_USER='myun'
FTP_PASS='mypass'
FTPLOGFILE=logs/ftplog.log
LOCAL_FILE='local_file'
REMOTE_FILE='remote_file'
ftp -n -v $FTP_HOST <<SCRIPT >> ${FTPLOGFILE} 2>&1
quote USER $FTP_USER
quote PASS $FTP_PASS
binary
prompt off
put $LOCAL_FILE $REMOTE_FILE
bye
SCRIPT
echo $!

Answer 1

我认为您要查找的是exit $?而不是echo $!在FTP脚本的底部。

使用

echo 将简单地打印到stdout但不会返回退出代码（因此应使用exit）。 特殊$?是前一个进程的返回码，而不是$!，是进程ID。

Answer 2

我想出了一个解决方案，既要仔细检查ftp上传，又要提供适当的退出代码。

... ftp upload first ... then ...
# this is FTP download double-check test
ftp -n -v $FTP_HOST <<SCRIPT >> ${FTPLOGFILE} 2>&1
quote USER $FTP_USER
quote PASS $FTP_PASS
binary
prompt off
get $REMOTE_FILE $TEST_FILE
bye
SCRIPT
#check to see if the FTP download test succeeded and return appropriate exit code
if [ -f "$TEST_FILE" ]; then
  echo "... OK FTP download test went fine. Execution may continue"
  exit 0
else
  echo "... ERROR FTP download test failed. Execution cannot continue"
  exit 1
fi