Add docx get keys operation

README.md

@@ -111,6 +111,15 @@ Run Char Index:  0   1   2   3   4   5   6      7  8  9 0  1  2   3   4   5   6
 
 All done, now you Word document is fully replaced keeping all the format.
 
+## Get document keys - docx_get_keys
+
+You can get all the keys present in the Word document by calling the function `docx_get_keys`:
+
+```python
+keys = docx_get_keys(doc) # Let's suppose the Word document has the keys: ${name} and ${phone}
+print(keys)  # ['name', 'phone']
+```
+
 ## Replace blocks - docx_blocks
 
 You can define a block in your Word document and set if it is going to be removed or not. The format required for key blocks are exactly like tags `HTML`, as following:

setup.cfg

@@ -1,6 +1,6 @@
 [metadata]
 name = python-docx-replace
-version = 0.4.3
+version = 0.4.4
 author = Ivan Bicalho
 author_email = ivanribeirob@gmail.com
 description = Replace words and remove blocks inside a Word document without losing format

src/python_docx_replace/__init__.py

@@ -1,6 +1,8 @@
+import re
 from typing import Any
 
-from python_docx_replace.exceptions import EndTagNotFound, InitialTagNotFound, TableIndexNotFound
+from python_docx_replace.exceptions import (EndTagNotFound, InitialTagNotFound,
+                                            TableIndexNotFound)
 from python_docx_replace.paragraph import Paragraph
 
 __all__ = ["docx_replace", "docx_blocks", "docx_remove_table"]
@@ -89,6 +91,24 @@ def docx_remove_table(doc: Any, index: int) -> None:
         raise TableIndexNotFound(index, len(doc.tables))
 
 
+def docx_get_keys(doc: Any) -> set:
+    """
+    Search for all keys in the Word document and return a list of unique elements
+
+    ATTENTION: The required format for the keys inside the Word document is: ${key}
+
+    For a document with the following content: "Hello ${name}, is your phone ${phone}?"
+    Result example: ["name", "phone"]
+    """
+    result = set()  # unique items
+    for p in Paragraph.get_all(doc):
+        paragraph = Paragraph(p)
+        matches = re.finditer(r"\$\{([^{}]+)\}", paragraph.get_text())
+        for match in matches:
+            result.add(match.groups()[0])
+    return list(result)
+
+
 def _handle_blocks(doc: Any, initial: str, end: str, keep_block: bool) -> bool:
     # The below process is a little bit complex, so I decided to comment each step
     look_for_initial = True

src/python_docx_replace/paragraph.py

@@ -62,6 +62,9 @@ def clear_tag_and_after(self, key, keep_block) -> None:
         block_handler = BlockHandler(self.p)
         block_handler.clear_key_and_after(key, keep_block)
 
+    def get_text(self) -> str:
+        return self.p.text
+
     def _simple_replace_key(self, key, value) -> None:
         # try to replace a key in the paragraph runs, simpler alternative
         for run in self.p.runs: