title: How to replace double quotes in findstr batch file
body: <p>I'm not sure what happened to my old account so had to create this new one. I am having trouble with a batch file I wrote. I have it working where the user will input certain information and then the input will replace text within a file. However, the issue I am having that I couldn't seem to find an answer to exactly on here or help me out is...how do you replace data within quotes? I tried escaping the quotes and i am not sure if it is finding it and just not replacing it or what. Here is the part</p>
  
  <pre><code>setlocal DisableDelayedExpansion
  (
    for /F "usebackq delims=" %%a in ("%drive%:\bdi\%bdi1%\importer.config") do (
      set "str=%%a"
      setlocal EnableDelayedExpansion
      set "str=!str:""saveTcpMessage"" value=""True""="saveTcpMessage" value="False"!"
  
      echo(!str!
      endlocal
    )
  ) &gt; %drive%:\bdi\%bdi1%\newfile.txt
  %drive%:
  cd "%drive%:\bdi\%bdi1%"
  del importer.config
  rename newfile.txt importer.config
  pause
  </code></pre>
  
  <p>In the above example I want the batch file to find the following string in a file called importer.config</p>
  
  <pre><code>&lt;add key="saveTcpMessage" value="True" /&gt;
  </code></pre>
  
  <p>If it find the above value I want it to replace it with:</p>
  
  <pre><code>&lt;add key="saveTcpMessage" value="False" /&gt;
  </code></pre>
  
  <p>I am pretty sure the find part works OK as I see you have to escape the dbl quotes..but the replace part isnt working. I have tried using double quotes in the replace too but with no avail. </p>
  
  <p>Any help would be appreciated.</p>