我正在尝试将string替换为id：

import sqlite3

TABLES = ['table1', 'table2', ...]
PAR_TYPES_STRINGS = ['foo', 'bar', ...]
PAR_TYPES_ID = [1, 2, ...]

def upateTable(table, parTypeId, parTypeString):
  return f"UPDATE {table} SET par_type = {parTypeId} WHERE par_type = {parTypeString};"

conn = sqlite3.connect('existing_db.db')
cursor = conn.cursor()

for table in TABLES:
  for index, parTypeString in enumerate(PAR_TYPES_STRINGS):
    parTypeId = PAR_TYPES_ID[index]
    cursor.execute(upateTable(table, parTypeId, parTypeString))
conn.commit()
cursor.close()

退货：
sqlite3.OperationalError:无此列：浮
我静态地使用par_type作为列名，并将其与parTypeString（动态地包含foo）进行比较。

WHERE par_type = {parTypeString} < dynamically passed variable
      ^
      hardcoded column name

然而，它说“no such column foo”，这是for loop从PAR_TYPES_STRINGS中取出的第一个元素，为什么呢？