python3 test.py
will return a JSON string that contains all 6 letter words with the second letter as s and the third letter as a.
{
"words":
["asadha", "asanga", "asarum", "bsarch", "isaiah", "isatis", "osasco", "psalms", "usable", "usacil", "usance"],
"meanings": [
["the fourth month of the Hindu calendar"],
["Indian religious leader and founder of the Yogacara school of Buddhism in India (4th century)"],
["wild ginger"],
["a bachelor's degree in architecture"],
["(Old Testament) the first of the major Hebrew prophets (8th century BC)", "an Old Testament book consisting of Isaiah's prophecies"],
["Old World genus of annual to perennial herbs: woad"],
["a city in southeastern Brazil; suburb of Sao Paulo"],
["an Old Testament book consisting of a collection of 150 Psalms"],
["capable of being put to use", "fit or ready for use or service", "convenient for use or disposal"],
["a defense laboratory of the Criminal Investigation Command; the United States Army's primary forensic laboratory in support of criminal intelligence"],
["the period of time permitted by commercial usage for the payment of a bill of exchange (especially a foreign bill of exchange)", "(economics) the utilization of economic goods to satisfy needs or in manufacturing", "accepted or habitual practice"]
],
"total": 11,
"count": 11,
"time": "0.03261208534240723",
"status": "success",
"message": "",
"start": 0
}Change wordPattern = "*sa***" in test.py line 7 to change the pattern
Set wordFinder.variable = "l" in test.py after line 14
variable v can be either of : f | l | r | b
f = fixed
l = left
r = right
b = both
f : LENGTH(w.lemma) = 6 AND SUBSTR(lemma, 2, 1) = 's' AND SUBSTR(lemma, 3, 1) = 'a'
l : w.lemma LIKE '%sa___'
r : w.lemma LIKE '_sa%'
b : w.lemma LIKE '%sa%'
underscore specifies a single character instance in SQL