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Commit 217afe2

Browse files
simplify: use direct node.children check instead of has_children() method
Even simpler approach: - Replace node.has_children() with direct if node.children: - Remove unnecessary has_children() methods from all classes - Pythonic and direct - empty lists are falsy, non-empty are truthy - Less code, same functionality This is the most straightforward way to check for children in Python.
1 parent 5efe631 commit 217afe2

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2 files changed

+3
-7
lines changed

2 files changed

+3
-7
lines changed

‎src/gitingest/output_formatter.py

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -156,7 +156,7 @@ def _create_tree_structure(
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display_name = node.get_display_name()
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tree_str += f"{prefix}{current_prefix}{display_name}\n"
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159-
if node.has_children():
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if node.children:
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prefix += " " if is_last else "│ "
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for i, child in enumerate(node.children):
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tree_str += _create_tree_structure(query, node=child, prefix=prefix, is_last=i == len(node.children) - 1)

‎src/gitingest/schemas/filesystem.py

Lines changed: 2 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -112,9 +112,7 @@ def get_display_name(self) -> str:
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"""Get display name for tree view. Override in subclasses."""
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return self.name
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115-
def has_children(self) -> bool:
116-
"""Return whether this node has children to display."""
117-
return False
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@property
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def content(self) -> str:
@@ -167,9 +165,7 @@ def get_display_name(self) -> str:
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"""Directories get a trailing slash."""
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return self.name + "/"
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170-
def has_children(self) -> bool:
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"""Directories have children if the list is not empty."""
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return bool(self.children)
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def render_tree(self, prefix: str = "", *, is_last: bool = True) -> list[str]:
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"""Render the tree representation of this directory."""

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