implement ls command in python

Author: Pezhman-Azizi

implement-shell-tools/ls/ls.py

@@ -0,0 +1,49 @@
+#!/usr/bin/env python3
+import sys
+import os
+
+def list_dir(path=".", show_all=False):
+    try:
+        entries = os.listdir(path)
+        if not show_all:
+            entries = [e for e in entries if not e.startswith(".")]
+        return sorted(entries)
+    except FileNotFoundError:
+        print(f"ls: cannot access '{path}': No such file or directory", file=sys.stderr)
+        return []
+    except NotADirectoryError:
+        # If it's a file, just return it
+        return [path]
+
+def main():
+    args = sys.argv[1:]
+
+    # Flags
+    show_one_per_line = "-1" in args
+    show_all = "-a" in args
+
+    # Remove flags from args
+    paths = [arg for arg in args if arg not in ("-1", "-a")]
+
+    if not paths:
+        paths = ["."]  # default to current directory
+
+    for i, path in enumerate(paths):
+        entries = list_dir(path, show_all=show_all)
+
+        if len(paths) > 1:
+            print(f"{path}:")
+
+        for entry in entries:
+            if show_one_per_line:
+                print(entry)
+            else:
+                print(entry, end="  ")
+        if not show_one_per_line:
+            print()  # newline after the row
+
+        if i < len(paths) - 1:
+            print()
+
+if __name__ == "__main__":
+    main()