在 bash 中从 mysql 获取 count val 的最简单方法
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Easiest way to get count val from mysql in bash
提问by ghostdog74
Maybe i should use python or perl but i dont know any.
也许我应该使用 python 或 perl，但我不知道。
I have 4 statements and i would like to check if there are any errors longer then an hour. My user is setup so i dont need to enter a mysql user/pass. This statement is in mysql_webapp_error_check.sh
我有 4 条语句，我想检查是否有任何错误超过一个小时。我的用户已设置，所以我不需要输入 mysql 用户/密码。该语句在 mysql_webapp_error_check.sh 中
#!/bin/bash
mysql testdb -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);"
How do i make it give me the return value (count(*)) instead of printing to screen?
我如何让它给我返回值（计数（*））而不是打印到屏幕？
Then i'll write an if statement and output to stdout/err for cron to use to email me (otherwise i want the script to be silent so nothing is emailed unless theres a problem)
然后我会写一个 if 语句并输出到 stdout/err 以便 cron 用来给我发电子邮件（否则我希望脚本保持静音，所以除非出现问题，否则不会通过电子邮件发送任何内容）
采纳答案by DigitalRoss
#!/bin/bash
echo show databases\; | mysql -u root | (while read x; do
echo "$x"
y="$x"
done
echo "$y"
)
回答by MarcoHager
Searched the same, -s for silent works exactly for me.
搜索相同的， -s 对我来说完全适用。
#!/bin/bash
result=`mysql testdb -s -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);"`
echo result = .$result.
PS.: There is also a --batch parameter in my mysql Ver 14.14 Distrib 5.1.49 which "Write fields without conversion. Used with --batch" so its a little off-topic here, but should be mentioned here.
PS.：在我的 mysql Ver 14.14 Distrib 5.1.49 中也有一个 --batch 参数，它“无需转换即可写入字段。与 --batch 一起使用”，因此这里有点题外话，但应在此处提及。
回答by ghostdog74
in bash, you use $()syntax.
在 bash 中，您使用$()语法。
#!/bin/bash
ret=$(mysql testdb -e "select count(*) from tbl where last_error_date < DATE_SUB(NOW(), INTERVAL 1 HOUR);")
if [[ "$ret" > 0 ]];then
echo "there is count"
else
echo "no count"
fi
回答by Wrikken
I usually do this:
我通常这样做：
var=`mysql -e "SELECT COUNT(*) FROM ...\G" | awk '/COUNT/{print }/'`
回答by XDjuj
For my part I simply use grep -vto exclude the line printing count(*)from the return of MySQL.
就我而言，我只是用来从 MySQL 的返回中grep -v排除行打印count(*)。
So I get the counter like that:
所以我得到这样的计数器：
db_name="NAME_DB";
db_user="USER_DB";
db_pwd="PWD_DB";
counter=`mysql -u${db_user} -p${db_pwd} ${db_name} -e "SELECT count(*) FROM my_table WHERE something = '1';" | grep -v "count"`;
echo "Count for request: $counter";
I use it for some Wordpress stuff this way, reading databases infos from the wp-config.phpfile:
我以这种方式将它用于一些 Wordpress 的东西，从wp-config.php文件中读取数据库信息：
wp_db_infos="wp-config.php";
wp_db=`cat ${wp_db_infos} | grep "DB_NAME" | awk -F ', ' '{print }' | awk -F "'" '{print }'`;
wp_user=`cat ${wp_db_infos} | grep "DB_USER" | awk -F ', ' '{print }' | awk -F "'" '{print }'`;
wp_pwd=`cat ${wp_db_infos} | grep "DB_PASSWORD" | awk -F ', ' '{print }' | awk -F "'" '{print }'`;
img_to_update=`mysql -u${wp_user} -p${wp_pwd} ${wp_db} -e "SELECT count(*) FROM wp_offres WHERE maj_img = '1';" | grep -v "count"`;
回答by Vladimir Perepechenko
local count=$(mysql -u root --disable-column-names --batch --execute "SELECT COUNT(*) FROM mysql.user WHERE user = '$DstDbName'")
if [[ "$count" > 0 ]]
then
fi
--batch - do clear output w/o borders --disable-column-names - prints only row with value
--batch - 不带边框清除输出 --disable-column-names - 仅打印带值的行
no creasy AWK used :)
没有使用任何皱巴巴的 AWK :)
