使用来自另一个shell脚本的if语句的返回码函数

我有 shell 脚本已调用 script_1.sh 如下图所示

# query
batch_count_query="SELECT COUNT(*) AS COUNT FROM ${db_name}.${table_name} WHERE BATCH_STATUS = "RUNNING";"

echo "************** $batch_count_query ******************************"

# Invoke the query
resp1=$(hive -e "$batch_count_query")


# return status check
if [ $? -eq 0 ]; then
    echo "******************* Command Ran Successfully ******************** "
    echo "Return message is ***************** ${resp1} ****************** "
else
    echo "******************* Error during the command execution ******************** "
    echo "Return message is ***************** ${resp1} ****************** "
    exit 1
fi

现在,当查询失败时,脚本将成功退出并显示错误消息。

现在我想对脚本进行一些更改。我想用 return status check 许多脚本中的一部分代码。 所以我试图在中创建一个函数 session_helper.sh 文件如下

# find return status of the command
command_exec_status ()
{
message=$1
if [ $? -eq 0 ]; then
    echo "******************* Command Ran Successfully ******************** "
    echo "Return message is ***************** ${message} ****************** "
else
    echo "******************* Error during the command execution ******************** "
    echo "Return message is ***************** ${message} ****************** "
    exit 1
fi
}

新shell脚本 script_2.sh 如下:

source /home/$USER/session_helper.sh

# query
batch_count_query="SELECT COUNT(*) AS COUNT FROM ${db_name}.${table_name} WHERE BATCH_STATUS = "RUNNING";"

echo "************** $batch_count_query ******************************"

# Invoke the query
resp1=$(hive -e "$batch_count_query")   

# find status based on return code
command_exec_status $resp1

当我使用上述新脚本时,即使查询失败。这项工作没有失败。

我在这里做错了什么。什么是正确的方法。