Hi All,
Could there be some missing field in my configuration that is causing this ?
Or could it be a code issue with the shibboleth authentication part ?
It’s really strange that the shib auth throws a SQL error.
When I try to log in to the new VCL installation, it does take me to my
authentication page, where I enter my credentials.
That part works fine. It’s only when I get redirected back that I get an
error message on the browser, then an email with the message below:
The error message doesn’t seem to include the entire MySQL line, just a
piece of it.
Is there a piece of the shibboleth configuration that I am missing ?
What are the “WebSecrectKeys” that the backtrace is mentioning ?
Could it be the shibboleth authentication token ?
You have an error in your SQL syntax; check the manual that corresponds to your
MariaDB server version for the right syntax to use near ') AS s LEFT JOIN
cryptsecret cs ON (ck.id = cs.cryptkeyid AND cs.secretid = s.id' at line 1
SELECT ck.id as cryptkeyid, ck.pubkey as cryptkey, s.id as
secretid, s.cryptsecret AS mycryptsecret FROM cryptkey ck JOIN (SELECT secretid
as id, cryptsecret FROM cryptsecret WHERE cryptkeyid = ) AS s LEFT JOIN
cryptsecret cs ON (ck.id = cs.cryptkeyid AND cs.secretid = s.id) WHERE
ck.hosttype = 'web' AND cs.secretid IS NULL AND ck.id !=
-- Al Quiros
Enterprise Systems
On 10/11/18, 1:21 PM, "Evelio Quiros" <[email protected]> wrote:
Hi All,
I am working on a new VCL 2.5 installation using Shibboleth.
The test script in the documentation seems to work correctly.
But when I try to log into the new VCL using Shibboleth, I get a MySQL
error:
You have an error in your SQL syntax; check the manual that
corresponds to your MariaDB server version for the right syntax to use near ')
AS s LEFT JOIN cryptsecret cs ON (ck.id = cs.cryptkeyid AND cs.secretid = s.id'
at line 1
SELECT ck.id as cryptkeyid, ck.pubkey as cryptkey, s.id as
secretid, s.cryptsecret AS mycryptsecret FROM cryptkey ck JOIN (SELECT secretid
as id, cryptsecret FROM cryptsecret WHERE cryptkeyid = ) AS s LEFT JOIN
cryptsecret cs ON (ck.id = cs.cryptkeyid AND cs.secretid = s.id) WHERE
ck.hosttype = 'web' AND cs.secretid IS NULL AND ck.id !=
ERROR(101): General MySQL error
Mode was
Backtrace:
=-=-=-=-=-=-=-=-=-=-=-=
Call#:1 => index.php:addLoginLog() (line#:187)
Call#:2 => authentication.php:checkMissingWebSecretKeys()
(line#:580)
Call#:3 => utils.php:doQuery() (line#:3075)
Backtrace with Arguments:
=-=-=-=-=-=-=-=-=-=-=-=
Call#:1 => index.php:addLoginLog() (line#:187)
Arguments(4)
Argument#: 1 => evquir@FIU
Argument#: 2 => shibboleth
Argument#: 3 => 3
Argument#: 4 => 1
-----------------------
Call#:2 => authentication.php:checkMissingWebSecretKeys()
(line#:580)
Arguments(none):
-----------------------
Call#:3 => utils.php:doQuery() (line#:3075)
Arguments(1)
Argument#: 1 => SELECT ck.id as cryptkeyid, ck.pubkey as cryptkey,
s.id as secretid, s.cryptsecret AS mycryptsecret FROM cryptkey ck JOIN (SELECT
secretid as id, cryptsecret FROM cryptsecret WHERE cryptkeyid = ) AS s LEFT
JOIN cryptsecret cs ON (ck.id = cs.cryptkeyid AND cs.secretid = s.id) WHERE
ck.hosttype = 'web' AND cs.secretid IS NULL AND ck.id !=
-----------------------
Any ideas on what could be causing this issue ?
Thanks,
-- Al Quiros
Enterprise Systems
