sorinbuda
Posts: 27
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| Posted: 06/13/2010, 1:17 PM |
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What is the right syntax to do something like that in PHP?
Res = CCExecSQL ("INSERT INTO bla_track VALUES ('" & Request.ServerVariables("REMOTE_HOST") & "',1,1,DATEADD( hh , +3, GETDATE()) );", Conn, true)
I found the code above at
http://forums.yessoftware.com/posts.php?post_id=73107
Thank you in advance,
Sorin
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magus
Posts: 98
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| Posted: 06/14/2010, 4:22 AM |
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Hi,
{
$db = new clsDBdatabasename();
$SQL = "INSERT INTO bla_track (column_name1,column_name2,column_name3,column_name4) ".
"VALUES (". $db->ToSQL($_SERVER['REMOTE_HOST'],ccsText) .", 1, 1,". time() .")";
$db->query($SQL);
$db->close();
}
Will probably work when you name your columns. It inserts the current unix timestamp. For other date formats look up the date() function in the php manual.
Regards,
Don A
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sorinbuda
Posts: 27
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| Posted: 06/14/2010, 6:06 AM |
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Something does not work:
{
$db = new clsDBdbname();
$SQL = "INSERT INTO tablename (ip) ".
"VALUES (". $db->ToSQL($HTTP_SERVER_VARS["REMOTE_HOST"],ccsText) .")";
$db->query($SQL);
$db->close();
}
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mamboBROWN
Posts: 1713
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| Posted: 06/15/2010, 5:05 PM |
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sorinbuda,
Is (ip) suppose to be a variable?? If it is then it should be $ip.
Also, you could echo out the generated query to make sure that it is correctly being generated (just an suggestion).
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magus
Posts: 98
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| Posted: 06/16/2010, 9:36 AM |
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Hi,
What doesn't work is that you are using double quotes for two separate purposes in a single statement.
Use single quotes around REMOTE HOST and you should see an improvement.
Regards,
Don A
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sorinbuda
Posts: 27
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| Posted: 06/19/2010, 4:12 AM |
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Now the code looks as follows:
{
$db = new clsDBdbname();
$SQL = "INSERT INTO tablename (ip) ".
"VALUES (". $db->ToSQL($HTTP_SERVER_VARS['REMOTE_HOST'],ccsText) .")";
$db->query($SQL);
$db->close();
}
yet no inproovements. No IP address registered in the database.
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magus
Posts: 98
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| Posted: 06/19/2010, 5:18 AM |
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Hi sorinbuda,
Have you tested that
$HTTP_SERVER_VARS['REMOTE_HOST']
works as opposed to:
$_SERVER['REMOTE_HOST']
that I recommended.
Also,
Do you have the correct new clsDBdbname();
Is the tablename correct?
Is the column named 'ip'
Regards,
Don
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sorinbuda
Posts: 27
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| Posted: 06/19/2010, 5:37 AM |
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Yes, I've tryed that in the very beginning.
The other settings are correct because all other information is inserted in the database. The full scipt is:
{
$db = new clsDBfsmanagement();
$SQL = "INSERT INTO utilizatori_statistici_acces (login,parola,ip,data_ora_acces) ".
"VALUES (". $db->ToSQL(CCGetParam("login",0),ccsText) .",". $db->ToSQL(CCGetParam("password",0),ccsText) .",". $db->ToSQL($HTTP_SERVER_VARS['REMOTE_HOST'],ccsText) .",". $db->ToSQL(CCGetParam("data_ora_acces",0),ccsText) .")";
$db->query($SQL);
$db->close();
}
the only thing which is not inserted in the database is the IP address.
Thank you in advance,
Sorin
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magus
Posts: 98
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| Posted: 06/19/2010, 6:24 AM |
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and if you echo the $SQL what is the statement being sent to the server?
echo $SQL;
Is there anything obviously wrong with the generated syntax?
$db->query($SQL);
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sorinbuda
Posts: 27
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| Posted: 06/21/2010, 5:17 AM |
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I have solved the problem with a small piece of javascript. However I am looking forward to find a solution for making the script above to work because the javascript I have implemented now is affecting the menu of the page.
Quote :
and if you echo the $SQL what is the statement being sent to the server?
echo $SQL;
Is there anything obviously wrong with the generated syntax?
$db->query($SQL);
Sorry I am a begginer I do not anderstand your piece of advise.
I get no errors in the page. And after executing the script above I can find in the database all data but the IP address. (user, pasword, data)
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magus
Posts: 98
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| Posted: 06/26/2010, 5:33 AM |
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Hi Sorinbuda,
Sorry I missed your previous message.
If you're not getting anything back it generally means you have a syntax error in your php. Look for unclosed (), {}, "", '' pairs, and make sure that each statement ends with a semicolon. ; A comma instead of a fullstop and vice versa will also cause problems.
The advice above is to echo $SQL statement to screen so you can see what is being sent to the database server.
Include
echo $SQL;
exit();
in the code previously recommended before your
$db->query($SQL);
statement.
This will print the sql to your screen and you can then try it in a tool like phpmyadmin to see if the sql works.
Regards,
Don A
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