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:

Is there a piece of the shibboleth configuration that I am missing ?

-- 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
     
     
    
    
    

Reply via email to