如何测试窗口（按标题）是否已从命令提示符打开？

Question

在批处理文件中，我打开一个具有如下特定名称的窗口：

@echo off    
start "my log" /D \logs\ /I powershell -nologo -noexit -command "$host.ui.RawUI.WindowTitle = 'My log'"; get-content logfile.log -wait

所以窗口标题是“我的日志”。

仅当窗口尚未打开时，如何运行此命令。是否有bat文件命令来测试这个？我宁愿不使用程序或powershell命令，只是一个简单的bat文件cmd如果可能的话。

像这样的东西：

@echo off    
if EXISTS window("My log") goto skip

start "my log" /D \logs\ /I powershell -nologo -noexit -command "$host.ui.RawUI.WindowTitle = 'My log'"; get-content logfile.log -wait

:skip

Answer 1

@For /f "Delims=:" %A in ('tasklist /v /fi "WINDOWTITLE eq New Folder"') do @if %A==INFO echo Prog not running

有关批次的更多信息。另见for /?。

&    seperates commands on a line.

&&    executes this command only if previous command's errorlevel is 0.

||    (not used above) executes this command only if previous command's errorlevel is NOT 0

>    output to a file

>>    append output to a file

<    input from a file

|    output of one command into the input of another command

^    escapes any of the above, including itself, if needed to be passed to a program

"    parameters with spaces must be enclosed in quotes

+ used with copy to concatinate files. E.G. copy file1+file2 newfile

, used with copy to indicate missing parameters. This updates the files modified date. E.G. copy /b file1,,

%variablename% a inbuilt or user set environmental variable

!variablename! a user set environmental variable expanded at execution time, turned with SelLocal EnableDelayedExpansion command

%<number> (%1) the nth command line parameter passed to a batch file. %0 is the batchfile's name.

%* (%*) the entire command line.

%<a letter> or %%<a letter> (%A or %%A) the variable in a for loop. Single % sign at command prompt and double % sign in a batch file.


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Answer 2

总是使用任务列表，这是一个例子：

@echo off
for /f "tokens=*" %%a in ('tasklist /fi "WINDOWTITLE eq My log"') do if "%%a" == "INFO: No tasks are running which match the specified criteria." start "my log" /D \logs\ /I powershell -nologo -noexit -command "$host.ui.RawUI.WindowTitle = 'My log'"; get-content logfile.log -wait

没错，你只需2行即可。