如何在批处理脚本的变量中存储sql查询的输出? [英] how to store output of sql query in variable in batch script?

问题描述

我正在处理批处理脚本.

i am working on a batch script.

我想将行的计数存储在变量中.

i want to store the count of row's in variable.

喜欢

set var = mysql -uroot -proot -e从表中选择count(*)"；

set var = mysql -uroot -proot -e"select count(*) from table";

我也尝试过用其他方式做

i also tried to do it other way like

 set var= mysql -uroot -proot -e "select count(*) from table into outfile 'F:\count.txt'";

 for /f %%a in ("F:\count.txt") do ( 
set output = %%a
echo %output% 

pause 

)

在上面的代码中，变量输出"什么都没有显示(空).

In above code the variable "output" shows nothing(empty).

请帮帮我.

推荐答案

我可以在您的脚本中看到至少两个问题:

I can see at least two issues in your script:

1. 在IN( )中用双引号引起来的字符串被视为文字，而不是文件路径/名称，除非您指定usebackq选项，它强制执行不同的语义，从而将双引号字符串或未双引号的字符串都视为文件名.

1. A string in double quotes inside IN( ) is treated as a literal, not as a file path/name, unless you specify the usebackq option, which enforces different semantics, whereby either double-quoted string or non-quoted one is treated as a file name.

您要将<space>%%a存储到output<space>变量中，而不是将%%a存储到output中.

You are storing <space>%%a into the output<space> variable, not %%a into output.

在修复了这两个问题之后，将再有一个问题(可能只有一个).您正在为变量分配一个值，然后使用立即变量扩展(%var%)在同一括号中的块(这是您的循环体)中评估该变量.这不能按预期方式工作.事实是，带括号的块被完全解析为一个单元，即在执行第一个命令之前先解析其所有命令.您可能会猜到，在这种情况下，您的%output%表达式将没有任何结果，因为在解析时尚未为output赋值. (并且为 分配了一个值时，它不会更改任何内容，因为先前的(空)值已经替换了该表达式.)

After you've fixed those two, there will remain one (probably, just one) more issue. You are assigning a value to a variable and then evaluating the variable in the same bracketed block (which is your loop body) using immediate variable expansion (%var%). This cannot work as expected. The thing is, a bracketed block is parsed entirely as a single unit, i.e. all its commands are parsed before the first one executes. As you can guess, your %output% expression will in this case evaluate to nothing, because output is not yet assigned a value at the time of parsing. (And when it is assigned a value, it will change nothing, because the previous (empty) value will already have replaced the expression.)

您可以使用延迟变量扩展来解决此问题，可以猜测，延迟扩展使用了不同的评估时间.首先，您应该通过发出SETLOCAL EnableDelayedExpansion命令来启用延迟扩展，然后使用稍微不同的语法:!var!而不是%var%.

You can solve this using delayed variable expansion, which, as can be guessed, uses a different timing for evaluation. First, you should enable delayed expansion by issuing the SETLOCAL EnableDelayedExpansion command, then use a slightly different syntax: !var! instead of %var%.

因此，如果我们解决了上面提到的所有问题，则循环可能看起来像这样:

So, if we address all the issues mentioned above, the loop may look like this:

…
SETLOCAL EnableDelayedExpansion

FOR /F "usebackq" IN ("F:\count.txt") DO (
SET output=%%a
ECHO !output!
)

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