SQL SELECT Statement problem
Bob Sneidar
bobsneidar at iotecdigital.com
Tue Sep 15 12:40:22 EDT 2015
I will have to go back and check. I seem to remember having a problem with sqLite using just '%:1%' and reading that the solution for sqLite was LIKE ':1%' or LIKE '%:1%' or LIKE '%:1' or LIKE ':1' whereas I did not have the problem with mySQL.
Bob S
> On Sep 15, 2015, at 09:31 , Peter Haworth <pete at lcsql.com> wrote:
>
> I found that LIKE :1 where the :1 variable contains %whatever% with no
> quotes around it works, at least ion sqlite.
>
> I'm not sure I understand your comment about this not finding rows that
> begin with, end with, or contain the string in sqlite. It's always worked
> for me.
>
>
> On Tue, Sep 15, 2015 at 9:15 AM Bob Sneidar <bobsneidar at iotecdigital.com>
> wrote:
>
>> Not sure if it matters, but I use LIKE '%:1%'. Also remember that with
>> sqLite, and unlike mySQL, this form will not find records beginning with,
>> ending with or explicitly is whatever replaces :1. Why the implementations
>> of the different SQL engines behave differently escapes me, other than to
>> think that the people who develop these engines don't have a lot of respect
>> for standards and why we need them.
>>
>> Bob S
>>
>>
>>> On Sep 2, 2015, at 12:52 , Peter Haworth <pete at lcsql.com> wrote:
>>>
>>> Having an issue with the following statement in SQLite.
>>>
>>> SELECT col1,col2 FROM TableA WHERE colid IN (:1)
>>>
>>> This is executed with
>>>
>>> put revQueryDatabase(gdbid,tsql,"tArray") into tCursor
>>>
>>> If tArray[1] contains a single integer, the SELECT works, if tArray[1]
>>> contains a comma separated list of integers, no records are returned,
>> even
>>> though I know there are qualifying entries.
>>>
>>> If I replace ":1" in the SELECT with 1,2 the rows are correctly returned,
>>> but if tArray[1] contains 1,2 then no rows are returned.
>>>
>>> On another related issue, I remember a discussion about the correct
>> syntax
>>> for using a parameter list variable with a LIKE statement but can't find
>>> it. I have tried:
>>>
>>> LIKE ':1' with :1 containing %abc%
>>> LIKE '%:1%' with :1 containing abc
>>> LIKE :1 with :1 containing '%abc%'
>>>
>>> None of these return the correct data. Anyone remember how to get this
>> to
>>> work?
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