我试图将MySQL结果存储到全局bash数组变量中,但我不知道该怎么做.
我应该将MySQL命令结果保存在一个文件中并在我的for循环中逐行读取文件以进行其他处理吗?
例:
#Database, table uses
user password
Pierre aaa
Paul bbb
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命令:
$results = $( mysql –uroot –ppwd –se « SELECT * from users );
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我想要results包含两行.
你可以试试这个:
mapfile result < <(mysql –uroot –ppwd –se "SELECT * from users;")
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比
echo ${result[0]%$'\t'*}
echo ${result[0]#*$'\t'}
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要么
for row in "${result[@]}";do
echo Name: ${row%$'\t'*} pass: ${row#*$'\t'}
done
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Nota这将正常工作,而行只有2个字段.更多是可能的,但变得棘手
while IFS=$'\t' read name pass ;do
echo name:$name pass:$pass
done < <(mysql -uroot –ppwd –se "SELECT * from users;")
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i=0
while IFS=$'\t' read name[i] pass[i++];do
:;done < <(mysql -uroot –ppwd –se "SELECT * from users;")
echo ${name[0]} ${pass[0]}
echo ${name[1]} ${pass[1]}
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有一个小工具(在github上)或在我自己的网站上:( shellConnector.sh你可以使用:
cd /tmp/
wget -q http://f-hauri.ch/vrac/shell_connector.sh
. shell_connector.sh
newSqlConnector /usr/bin/mysql '–uroot –ppwd'
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就这样.现在,为demo创建临时表:
echo $SQLIN
3
cat >&3 <<eof
CREATE TEMPORARY TABLE users (
id bigint(20) unsigned NOT NULL PRIMARY KEY AUTO_INCREMENT,
name VARCHAR(30), date DATE)
eof
myMysql myarray ';'
declare -p myarray
bash: declare: myarray: not found
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该命令myMysql myarray ';'将发送;然后执行内联命令,但由于mysql不会解决任何问题,变量$myarray不会存在.
cat >&3 <<eof
INSERT INTO users VALUES (1,'alice','2015-06-09 22:15:01'),
(2,'bob','2016-08-10 04:13:21'),(3,'charlie','2017-10-21 16:12:11')
eof
myMysql myarray ';'
declare -p myarray
bash: declare: myarray: not found
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好的,那么现在:
myMysql myarray "SELECT * from users;"
printf "%s\n" "${myarray[@]}"
1 alice 2015-06-09
2 bob 2016-08-10
3 charlie 2017-10-21
declare -p myarray
declare -a myarray=([0]=$'1\talice\t2015-06-09' [1]=$'2\tbob\t2016-08-10' [2]=$'3\tcharlie\t2017-10-21')
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此工具处于构建的早期阶段...您必须在重新使用之前手动清除变量:
unset myarray
myMysql myarray "SELECT name from users where id=2;"
echo $myarray
bob
declare -p myarray
declare -a myarray=([0]="bob")
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