Question

我想在一段时间后从shell脚本执行MySQL查询。我写了脚本，但输出不是预期的。它从数据库再次打印相同的值。有人可以帮忙。

SCRIPT - demo.sh

#!/bin/bash
echo "Shell Script to Query a Database!"
echo "--Querying from the Database starts--"
cd "C:\Program Files\MySQL\MySQL Server 5.7\bin"
for i in {1..5};
do 
./mysql -u root -proot@123 << EOF
use catalyst_latest;
select id from ci_master limit 1;
EOF
echo "Wait for 2 seconds for the next ID." 
sleep 2; 
done
echo "--Query Stopped!--"

输出

$ ./demo.sh
Shell Script to Query a Database!
--Querying from the Database starts--
mysql: [Warning] Using a password on the command line interface can be   insecure.
id
282
Wait for 2 seconds for the next ID.
mysql: [Warning] Using a password on the command line interface can be insecure.
id
282
Wait for 2 seconds for the next ID.
mysql: [Warning] Using a password on the command line interface can be insecure.
id
282
Wait for 2 seconds for the next ID.
mysql: [Warning] Using a password on the command line interface can be insecure.
id
282
Wait for 2 seconds for the next ID.
mysql: [Warning] Using a password on the command line interface can be insecure.
id
282
Wait for 2 seconds for the next ID.
--Query Stopped!--

你能看到每次都有282人返回吗？所以我想在2秒后从数据库中获取下一个ID。请告诉我怎么做。提前谢谢。

Answer 1

您需要将返回的id值记住到bash变量中，然后在查询的WHERE子句中重用该变量，如下所示：

select id from ci_master where id > $id limit 1

要将命令的输出写入变量，请使用以下内容：

id=$(mysql <options> -Nsqe "<query>")

选项N用于跳过列名（如id），选项s用于更多静默输出，选项q用于不缓存结果，选项e用于命令执行。

最初，$id变量应设置为某个值，如1或表中现有值的最小值。

Answer 2

#!/bin/bash

echo "Shell Script to Query a Database!"
echo "--Querying from the Database starts--"
cd "C:\Program Files\MySQL\MySQL Server 5.7\bin"
for i in {1..10};
do 
temp=`./mysql -u root -proot@123 << EOF
use catalyst_latest;
select id from ci_master limit $i;
EOF`
result=''
for j in $temp 
do 
result=$j
done
#date;
echo "$result"
echo "Wait for 1 seconds for the next ID." 
sleep 1; 
done
echo "--Query Stopped!--"

在一段时间间隔之后从shell脚本调用MySQL查询

2 个答案: