如何在同一个exe中启动进程 [英] How to Start a Process in a same exe
问题描述
大家好,
我正在开始一个流程 以下代码
ProcessStartInfo str = new processstartinfo(" abc.exe"," d:\\abc.xyz");
Process.start(str);
$
现在问题是甚至abc.exe已经打开,每次新文件新abc.exe都是open,但是exe支持在一个exe中打开多个文件
那么我怎样才能在同一个exe文件中使用processstartinfo打开多个文件?
Hello all,
I am starting a process with below code
ProcessStartInfo str = new processstartinfo("abc.exe", "d:\\abc.xyz");
Process.start(str);
Now the problem is even abc.exe is already open, every time for new file new abc.exe is open, however that exe is supported multiple file opening in one exe
So how can I do that for the multiple file open in same exe with processstartinfo??
谢谢。
推荐答案
无法回答,因为信息量太少。
Can not be answered, because there are too less infos.
- 你有abc.exe的代码吗?然后使用参数开关在文件上执行正确的作业。
- Do you have the code for abc.exe? Then use parameter switches to do the right jobs on the file.
- 该文件怎么办? (读/写,过滤,......)
- What is to do with the file? (read/write, filter, ...)
抱歉。
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