bash 执行mysql查询后捕获退出代码
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Catching exit code after executing mysql query
提问by andresg3
I'm working on a bash script that will query mysql. I would like to add some error checking. Let say if the following query for some reason fails I want to catch the exit error and exit the script. For example this is part of my script.
我正在编写一个将查询 mysql 的 bash 脚本。我想添加一些错误检查。假设以下查询由于某种原因失败，我想捕获退出错误并退出脚本。例如，这是我脚本的一部分。
QUERY="SELECT DISTINCT `TABLE_SCHEMA`, `TABLE_NAME`
FROM `information_schema`.`TABLES`
WHERE table_schema NOT IN ( 'mysql', 'information_schema', 'performance_schema' )"
mysql -u user -pPASSWD --batch -N -e "$QUERY" | while read DATABASE TABLE;
do
...
...
...
done
How could i catch the exit code after the scripts run the "$QUERY". I was thinking something like this. But it doesn't seem to work.
在脚本运行“$QUERY”后，我如何捕获退出代码。我在想这样的事情。但它似乎不起作用。
mysql -u user -pPASSWD --batch -N -e "$QUERY" echo $? | while read DATABASE TABLE;
Any ideas
有任何想法吗
回答by fedorqui 'SO stop harming'
You are in the good way: $?is the flag to check:
你是个好方法：$?是要检查的标志：
$ mysql -h mydb <<< "SELECT * FROM MyDB.some_table_that_exists;"
$ echo $?
0
$ mysql -h mydb <<< "SELECT * FROM MyDB.asdfasdfasdf;"
ERROR 1146 (42S02) at line 1: Table 'MyDB.asdfasdfasdf;' doesn't exist
$ echo $?
1
So what you can do is to execute the query and then:
所以你可以做的是执行查询，然后：
if [ $? ... ]; then ...
