Bash + MySQL -d反引号问题 [英] Bash + MySQL -d backtick issue

问题描述

我正在按照社区精神为社区项目创建解决方案，以免费监测河流水位.这项工作的最终结果将是一个系统，该系统将从河流水位探测器中获取数据，并为在线社区生成图形.

I am helping to create a solution for a community project to monitor river levels for free, in the community spirit. The eventual product of this effort will be a system that takes data from a river level probe, and generates a graph for an online community.

我很早就开始从事该项目，并且据我所知，Bash正在使用它从包含数据的探针中获取一个文本文件，并将其推送到MySQL数据库中.直到我遇到绊脚石之前，一切都进行得非常顺利.我的许多数据库表列都是数字，因此必须使用MySQL命令行工具反引号与之交互.不幸的是，我想从MySQL语句中获取输出并将其放入变量中，但是我认为由于两组反向标记，MySQL无法获得正确的命令.

I am early on with the project, and as I am familiar with Bash am using this to take a text file from the probe containing the data and push it into a MySQL database. All has been going really nicely until I hit a stumbling block. A number of my database table columns are numbers and so to interact with these using the MySQL command line tool backticks have to be used. Unfortunately I want to take the output from a MySQL statement and put it into a variable, but I think due to the two sets of back ticks MySQL is not getting the right command.

我用一种讨厌的方法来捏造它，就像这样使它工作效率低下:

I have fudged it with a nasty work around like so which works all be it inefficiently:

mysql -N -D $targetDatabase -e "select \`"$timeSample"\` from RiverDataDays where date="$dateOfFile";" >tmpValue

dbEntry=`cat tmpValue`

echo $dbEntry

但是实际上，我要执行的操作是将其直接推入这样的变量中:

But actually, what I want to do is push it straight into an variable like this:

dbEntry=`mysql -N -D $targetDatabase -e "select \`"$timeSample"\` from RiverDataDays where date="$dateOfFile";"`

echo $dbEntry

推荐答案

尝试执行此操作:

dbEntry="$(printf "SELECT \140%s\140 FROM 'RiverDataDays' WHERE date = '%s';\n" "$timeSample" "$(<tmpValue )" | mysql -N -D "$targetDatabase")"
echo "$dbEntry"

或

dbEntry="$(printf "SELECT \`%s\` FROM 'RiverDataDays' WHERE date = '%s';\n" "$timeSample" "$(<tmpValue )" | mysql -N -D "$targetDatabase")"
echo "$dbEntry"

反引号(`)用于旧式命令替换，例如

The backquote (`) is used in the old-style command substitution, e.g.

foo=`command`

The

foo=$(command)

建议使用

语法. $()中的反斜杠处理不足为奇，并且$()更易于嵌套.参见 http://mywiki.wooledge.org/BashFAQ/082

\140

是反引号的八进制表示形式，请参见

is the octal representation of a backtick, see

man ascii

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