tweaks to overdrive, added text2int

src/calibre/ebooks/metadata/sources/overdrive.py

@@ -17,7 +17,7 @@ from lxml import html
 from lxml.html import soupparser
 
 from calibre.ebooks.metadata import check_isbn
-from calibre.ebooks.metadata.sources.base import Source
+from calibre.ebooks.metadata.sources.base import Source, Option
 from calibre.ebooks.metadata.book.base import Metadata
 from calibre.ebooks.chardet import xml_to_unicode
 from calibre.library.comments import sanitize_comments_html
@@ -58,11 +58,13 @@ class OverDrive(Source):
             self.parse_search_results(ovrdrv_data, mi)
             if ovrdrv_id is None:
                 ovrdrv_id = ovrdrv_data[7]
+
+            if get_full_metadata():
+                self.get_book_detail(br, ovrdrv_data[1], mi, ovrdrv_id, log)
+
             if isbn is not None:
                 self.cache_isbn_to_identifier(isbn, ovrdrv_id)
 
-            self.get_book_detail(br, ovrdrv_data[1], mi, ovrdrv_id, log)
-
             result_queue.put(mi)
 
         return None

src/calibre/utils/text2int.py

@@ -0,0 +1,63 @@
+#!/usr/bin/env  python
+__author__ = "stackoverflow community"
+__docformat__ = 'restructuredtext en'
+"""
+Takes english numeric words and converts them to integers.
+Returns False if the word isn't a number.
+
+implementation courtesy of the stackoverflow community:
+http://stackoverflow.com/questions/493174/is-there-a-way-to-convert-number-words-to-integers-python
+"""
+
+import re
+
+numwords = {}
+
+def text2int(textnum):
+    if not numwords:
+
+        units = [ "zero", "one", "two", "three", "four", "five", "six",
+                "seven", "eight", "nine", "ten", "eleven", "twelve",
+                "thirteen", "fourteen", "fifteen", "sixteen", "seventeen",
+                "eighteen", "nineteen"]
+
+        tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", 
+                "seventy", "eighty", "ninety"]
+
+        scales = ["hundred", "thousand", "million", "billion", "trillion", 
+                'quadrillion', 'quintillion', 'sexillion', 'septillion', 
+                'octillion', 'nonillion', 'decillion' ]
+
+        numwords["and"] = (1, 0)
+        for idx, word in enumerate(units): numwords[word] = (1, idx)
+        for idx, word in enumerate(tens): numwords[word] = (1, idx * 10)
+        for idx, word in enumerate(scales): numwords[word] = (10 ** (idx * 3 or 2), 0)
+
+    ordinal_words = {'first':1, 'second':2, 'third':3, 'fifth':5, 
+            'eighth':8, 'ninth':9, 'twelfth':12}
+    ordinal_endings = [('ieth', 'y'), ('th', '')]
+    current = result = 0
+    tokens = re.split(r"[\s-]+", textnum)
+    for word in tokens:
+        if word in ordinal_words:
+            scale, increment = (1, ordinal_words[word])
+        else:
+            for ending, replacement in ordinal_endings:
+                if word.endswith(ending):
+                    word = "%s%s" % (word[:-len(ending)], replacement)
+
+            if word not in numwords:
+                #raise Exception("Illegal word: " + word)
+                return False
+
+            scale, increment = numwords[word]
+
+        if scale > 1:
+            current = max(1, current)
+
+        current = current * scale + increment
+        if scale > 100:
+            result += current
+            current = 0
+
+    return result + current