Using sqlite3 DB-API multiple parameter substitution in SELECT statements Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) The Ask Question Wizard is Live! Data science time! April 2019 and salary with experience Should we burninate the [wrap] tag?“SELECT … WHERE … IN” with unknown number of parametersMultiple variables in a 'with' statement?Substitute multiple whitespace with single whitespace in Pythonsqlite3 square brackets in a select statementSelecting multiple columns in a pandas dataframesqlite3 command line tools don't work in UbuntuPython threading, why can i launch thread only once?sqlite3 select statement fails with "parameters are of unsupported typeWhy can parameter substitution not be used with tables in sqlite3?Parameter substitution for a SQLite with multiple “IN” clausePython and SQLite3 SELECT statement

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Using sqlite3 DB-API multiple parameter substitution in SELECT statements



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?“SELECT … WHERE … IN” with unknown number of parametersMultiple variables in a 'with' statement?Substitute multiple whitespace with single whitespace in Pythonsqlite3 square brackets in a select statementSelecting multiple columns in a pandas dataframesqlite3 command line tools don't work in UbuntuPython threading, why can i launch thread only once?sqlite3 select statement fails with "parameters are of unsupported typeWhy can parameter substitution not be used with tables in sqlite3?Parameter substitution for a SQLite with multiple “IN” clausePython and SQLite3 SELECT statement



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0















Given:



letters = list("abc")


I'd like to get all rows in characters that contain any of the letters in letters in their c column. I can do this, but only with python's string operations, which isn't suitable given it's vulnerabilities.



Ideally (my example is simplified) this would be using the GLOB clause.



E.g.



>>> cur.execute(**the statement here**)
>>> print(cur.fetchall())
>>> [('a',), ('b',), ('c',)]


Creation of the db:



import sqlite3
import string

def char_generator():
for c in string.ascii_lowercase:
yield (c,)

con = sqlite3.connect(":memory:")
cur = con.cursor()
cur.execute("create table characters(c)")

cur.executemany("insert into characters(c) values (?)", char_generator())









share|improve this question






















  • I don't think it's quite a dupe, but this question might help.

    – glibdud
    Mar 8 at 16:28

















0















Given:



letters = list("abc")


I'd like to get all rows in characters that contain any of the letters in letters in their c column. I can do this, but only with python's string operations, which isn't suitable given it's vulnerabilities.



Ideally (my example is simplified) this would be using the GLOB clause.



E.g.



>>> cur.execute(**the statement here**)
>>> print(cur.fetchall())
>>> [('a',), ('b',), ('c',)]


Creation of the db:



import sqlite3
import string

def char_generator():
for c in string.ascii_lowercase:
yield (c,)

con = sqlite3.connect(":memory:")
cur = con.cursor()
cur.execute("create table characters(c)")

cur.executemany("insert into characters(c) values (?)", char_generator())









share|improve this question






















  • I don't think it's quite a dupe, but this question might help.

    – glibdud
    Mar 8 at 16:28













0












0








0








Given:



letters = list("abc")


I'd like to get all rows in characters that contain any of the letters in letters in their c column. I can do this, but only with python's string operations, which isn't suitable given it's vulnerabilities.



Ideally (my example is simplified) this would be using the GLOB clause.



E.g.



>>> cur.execute(**the statement here**)
>>> print(cur.fetchall())
>>> [('a',), ('b',), ('c',)]


Creation of the db:



import sqlite3
import string

def char_generator():
for c in string.ascii_lowercase:
yield (c,)

con = sqlite3.connect(":memory:")
cur = con.cursor()
cur.execute("create table characters(c)")

cur.executemany("insert into characters(c) values (?)", char_generator())









share|improve this question














Given:



letters = list("abc")


I'd like to get all rows in characters that contain any of the letters in letters in their c column. I can do this, but only with python's string operations, which isn't suitable given it's vulnerabilities.



Ideally (my example is simplified) this would be using the GLOB clause.



E.g.



>>> cur.execute(**the statement here**)
>>> print(cur.fetchall())
>>> [('a',), ('b',), ('c',)]


Creation of the db:



import sqlite3
import string

def char_generator():
for c in string.ascii_lowercase:
yield (c,)

con = sqlite3.connect(":memory:")
cur = con.cursor()
cur.execute("create table characters(c)")

cur.executemany("insert into characters(c) values (?)", char_generator())






python sqlite3






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 8 at 16:14









DaveDave

1158




1158












  • I don't think it's quite a dupe, but this question might help.

    – glibdud
    Mar 8 at 16:28

















  • I don't think it's quite a dupe, but this question might help.

    – glibdud
    Mar 8 at 16:28
















I don't think it's quite a dupe, but this question might help.

– glibdud
Mar 8 at 16:28





I don't think it's quite a dupe, but this question might help.

– glibdud
Mar 8 at 16:28












1 Answer
1






active

oldest

votes


















1














Maybe this sample can help you.



import sqlite3
import string

def char_generator():
for c in string.ascii_lowercase:
yield (c,)


con = sqlite3.connect(":memory:")

def initdb():
cur = con.cursor()
cur.execute("create table characters(c)")

cur.executemany("insert into characters(c) values (?)", char_generator())

def search(value):
values = [c for c in value]
cur = con.cursor()
cur.execute('SELECT * FROM characters WHERE c IN (0)'.format(','.join(['?' for c in values])), values)
return cur.fetchall()


if __name__ == '__main__':
initdb()
print(search("abcde"))


This code uses parameters. So you do not need to worry about SQL Injection.






share|improve this answer























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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Maybe this sample can help you.



    import sqlite3
    import string

    def char_generator():
    for c in string.ascii_lowercase:
    yield (c,)


    con = sqlite3.connect(":memory:")

    def initdb():
    cur = con.cursor()
    cur.execute("create table characters(c)")

    cur.executemany("insert into characters(c) values (?)", char_generator())

    def search(value):
    values = [c for c in value]
    cur = con.cursor()
    cur.execute('SELECT * FROM characters WHERE c IN (0)'.format(','.join(['?' for c in values])), values)
    return cur.fetchall()


    if __name__ == '__main__':
    initdb()
    print(search("abcde"))


    This code uses parameters. So you do not need to worry about SQL Injection.






    share|improve this answer



























      1














      Maybe this sample can help you.



      import sqlite3
      import string

      def char_generator():
      for c in string.ascii_lowercase:
      yield (c,)


      con = sqlite3.connect(":memory:")

      def initdb():
      cur = con.cursor()
      cur.execute("create table characters(c)")

      cur.executemany("insert into characters(c) values (?)", char_generator())

      def search(value):
      values = [c for c in value]
      cur = con.cursor()
      cur.execute('SELECT * FROM characters WHERE c IN (0)'.format(','.join(['?' for c in values])), values)
      return cur.fetchall()


      if __name__ == '__main__':
      initdb()
      print(search("abcde"))


      This code uses parameters. So you do not need to worry about SQL Injection.






      share|improve this answer

























        1












        1








        1







        Maybe this sample can help you.



        import sqlite3
        import string

        def char_generator():
        for c in string.ascii_lowercase:
        yield (c,)


        con = sqlite3.connect(":memory:")

        def initdb():
        cur = con.cursor()
        cur.execute("create table characters(c)")

        cur.executemany("insert into characters(c) values (?)", char_generator())

        def search(value):
        values = [c for c in value]
        cur = con.cursor()
        cur.execute('SELECT * FROM characters WHERE c IN (0)'.format(','.join(['?' for c in values])), values)
        return cur.fetchall()


        if __name__ == '__main__':
        initdb()
        print(search("abcde"))


        This code uses parameters. So you do not need to worry about SQL Injection.






        share|improve this answer













        Maybe this sample can help you.



        import sqlite3
        import string

        def char_generator():
        for c in string.ascii_lowercase:
        yield (c,)


        con = sqlite3.connect(":memory:")

        def initdb():
        cur = con.cursor()
        cur.execute("create table characters(c)")

        cur.executemany("insert into characters(c) values (?)", char_generator())

        def search(value):
        values = [c for c in value]
        cur = con.cursor()
        cur.execute('SELECT * FROM characters WHERE c IN (0)'.format(','.join(['?' for c in values])), values)
        return cur.fetchall()


        if __name__ == '__main__':
        initdb()
        print(search("abcde"))


        This code uses parameters. So you do not need to worry about SQL Injection.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 8 at 17:03









        andercruzbrandercruzbr

        33619




        33619





























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