[PHP-DB] MYSQL Query problem...

[Ian Bert Tusil]

is there something wrong with my SQL Query? I didnt get any results neither errors..

?php
$link = mysql_connect('localhost', 'username') or die('Cannot connect to the 
database.');
mysql_select_db(profiles,$link);
$SQLQuery = INSERT INTO profyles (username, fname, mname, lname, address, country, 
gender, sexuality, status, bday, race, yahoo, msn, icq, email, webpage, hair, 
bodytype, eye, height, weight, facialhair, hobbies, interests, aboutme, profession, 
food, music, tvshows, authors, movies, rolemodel, place, visit, online, picture) 
VALUES ('$user', '$firstname', '$middlename', '$lastname ', '$address, '$country', 
'$gender', '$sexuality', '$status', '$birthday', '$race', '$yahoo', '$msn', '$icq', 
'$email'. '$webpage', 'haircolour', '$bodytype', '$eyecolour', '$height', '$weight', 
'$facialhair', '$hobbies', '$interest', '$aboutme', '$profession', '$food', '$music', 
'$tvshows', '$authors', '$movies', '$rolemodel', '$place', 0, 'y' , 'y');
mysql_query($SQLQuery,$link);
?


ian

Re: [PHP-DB] MYSQL Query problem...

[Fedde van Feggelen]

is there something wrong with my SQL Query? I didnt get any results 
neither errors..

?php
$link = mysql_connect('localhost', 'username') or die('Cannot connect to 
the database.');
mysql_select_db(profiles,$link);
$SQLQuery = INSERT INTO profyles (username, fname, mname, lname, address, 
country, gender, sexuality, status, bday, race, yahoo, msn, icq, email, 
webpage, hair, bodytype, eye, height, weight, facialhair, hobbies, 
interests, aboutme, profession, food, music, tvshows, authors, movies, 
rolemodel, place, visit, online, picture) VALUES ('$user', '$firstname', 
'$middlename', '$lastname ', '$address, '$country', '$gender', 
'$sexuality', '$status', '$birthday', '$race', '$yahoo', '$msn', '$icq', 
'$email'. '$webpage', 'haircolour', '$bodytype', '$eyecolour', '$height', 
'$weight', '$facialhair', '$hobbies', '$interest', '$aboutme', 
'$profession', '$food', '$music', '$tvshows', '$authors', '$movies', 
'$rolemodel', '$place', 0, 'y' , 'y');
mysql_query($SQLQuery,$link);
?
'$email'. '$webpage'  -- shouldn't this be:'$email', (comma 
;))'$webpage'



laters,

Fedde

~= Everybody lies, but it doesn't matter because nobody listens =~

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Re: [PHP-DB] MYSQL Query problem...

[John W. Holmes]

Ian Bert Tusil wrote:

is there something wrong with my SQL Query? I didnt get any results neither errors..
[snip]
mysql_query($SQLQuery,$link);
That's because you're not checking for errors...

mysql_query($SQLQuery,$link) or die(mysql_error());

--
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[PHP-DB] MySQL Query problem...

[NIPP, SCOTT V (SBCSI)]

I seem to have run into another problem with a query that I am doing
exactly like the example in the MySQL documentation, I think.  The following
query is not working on my PHP page, and it is not working from a MySQL
command line either.  This looks exactly the same as an example in the MySQL
documentation though.  Thanks in advance for the help again.

SELECT * FROM oncall WHERE TO_DAYS(NOW()) - TO_DAYS(ptime) = 3

This is the query that I am ultimately after:

SELECT * FROM oncall ORDER BY 'ptime' ASC WHERE TO_DAYS(NOW()) -
TO_DAYS(ptime) = 3

I am trying to display all of the records from the past three days.
This looks like the way to go about doing this, but if there is a better way
please let me know.  (Just as more info, the 'ptime' field is a datetime
type of field.)

Scott Nipp
Phone:  (214) 858-1289
E-mail:  [EMAIL PROTECTED]
Web:  http:\\ldsa.sbcld.sbc.com



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Re: [PHP-DB] mysql query problem

[Paul Burney]

on 7/26/01 11:11 AM, Steve Fitzgerald ([EMAIL PROTECTED]) wrote:

 ran the query in the MySQL client and got the ERROR 1054: Unknown column
 '$ContactID' in 'where clause'.  I know the column ContactID exists.

Of course it did.  It can't find the text $ContactID in the table.  Try
putting in a number (1,2,3, etc.) and see if it works.

If it does, try echo'ing out the $ContactID and the $sql to the screen.  See
if $ContactID really is set and then try the output of $sql in the MySQL
client.

?php

echo 'ContactID is: ' , $ContactID, 'br';
 
$sql=SELECT * FROM contacts WHERE ContactID='$ContactID';

echo 'SQL query is: ' , $sql, 'br';

$result = mysql_query($sql,$db);

printf(%s\n, mysql_result($result,FirstName));

printf(%sbr\n, mysql_result($result,LastName));

?

Hope that helps.

Sincerely,

Paul Burney

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