title: bash: find and grep with regex in shellscript
body: <p>I have several versions of a certain software (lets call it <em>MySoftware</em>)installed and I like to find the path to a specific version with a combination of <code>find</code> and <code>grep</code>.
  Assume I have the following versions:
  1.12.0<br />
  1.12.2<br />
  1.42.2</p>
  <p>It is stored in the following way:</p>
  <pre><code>~/src/MySoftware/1.12.0/...
  ~/src/MySoftware/1.12.2/...
  ~/src/MySoftware/1.42.2/...
  </code></pre>
  <p>In a shell I could do something like <code>find . -name MySoftware | grep 1.12.0</code>. This is working since the command is giving me the <code> ~/src/MySoftware/1.12.0/</code> path.</p>
  <p>However, when switching to a shell script, I try to do this:</p>
  <pre><code>find . -name &quot;MySoftware&quot;  -exec grep &quot;1\.12\.0&quot; {} ';'
  </code></pre>
  <p>The example above is, however, not returning anything and I have no idea why. Other tries with <code>grep -HF &quot;1.12.0&quot;</code> are also not working. I am grateful for any advice</p>