Question

我有一个像这样的XML文件：

  <fruit>
      <desc>xxx / yyy / zzz</desc>
  </fruit>
  <fruit>
      <desc></desc>
  </fruit>
  <fruit>
      <desc>abc / def / ghi</desc>
  </fruit>

我使用批处理脚本来修复节点＆＃34; desc＆＃34;。这是修复后的结果：

  <fruit>    
      <desc>xxx/yyy/zzz</desc>    
  </fruit>    
  <fruit>    
      <desc>N/A</desc>    
  </fruit>    
  <fruit>    
      <desc>abc/def/ghi</desc>    
  </fruit>

如何在不扫描所有文件两次的情况下更换2个不同的字符串？这是脚本：

@echo off
setlocal enabledelayedexpansion

set "input_xml=xmlfile.xml"

set "search=^<desc^>^</desc^>"
set "search2= / "
set "replace=^<desc^>N/A^</desc^>"
set "replace2=/"
for /f "delims=" %%i in ('type "!input_xml!" ^& break ^> "!input_xml!" ') do (
  set FixNullNode=%%i
  echo !FixNullNode:%search%=%replace%! >>"!input_xml!" 
)
for /f "delims=" %%v in ('type "!input_xml!" ^& break ^> "!input_xml!" ') do (
  set FixSpaceSlash=%%v
  echo !FixSpaceSlash:%search2%=%replace2%! >>"!input_xml!" 
)
pause

由于

Answer 1

将第二个替换移动到第一个命令。

@echo off
setlocal enabledelayedexpansion

set "input_xml=xmlfile.xml"

set "search=^<desc^>^</desc^>"
set "search2= / "
set "replace=^<desc^>N/A^</desc^>"
set "replace2=/"
for /f "delims=" %%i in ('type "!input_xml!" ^& break ^> "!input_xml!" ') do (
  set "Fix=%%i"
  set "Fix=!Fix:%search%=%replace%!"
  echo !Fix:%search2%=%replace2%! >>"!input_xml!" 
)

批处理脚本一次查找和替换2个不同的字符串

1 个答案: