windows - 如何仅从结果列表中获取第一个结果并设置为 var？

Question

编写批处理脚本来查找正确的文件。

dir /o:-d /a:-d F:\ | findstr /i ".target."

有上面的脚本，我的目标是只有第一个文件，并想存储文件名以供以后脚本使用。

有谁知道如何获取第一个文件并设置为 var？

Answer 1

Here's one example for retrieving the first line of your slightly modified command line, and save the line into a variable line1. The /b parameter is added onto dir command, since we only want the file name. The | character in the command line also needs to be escaped by a ^ character.

@echo off
setlocal
set line1=
for /f "tokens=* delims=" %%A in ('dir /o:-d /a:-d /b F:\ ^| findstr /i ".target."') do (
  set line1=%%A
  rem exit loop immediately once we got the line
  goto next
)
:next
if "%line1%" != "" (
  echo First Line:
  echo %line1%
) else (
  echo No file is found
)