[PHP-DB] Re: Error is SQL(works in phpMyAdmin)
Here is one line of my new query: INSERT INTO article_keyword (aid,keyword,weight) VALUES (3, 'gases', 13.42282); I get the same error: You have an error in your SQL syntax near '; INSERT INTO article_keyword (aid,keyword,weight) VALUES (3, 'gases', 13.42282)' at line 1 -- JJ Harrison [EMAIL PROTECTED] www.tececo.com - Original Message - From: Joni Järvinen [EMAIL PROTECTED] To: JJ Harrison [EMAIL PROTECTED] Sent: Saturday, July 27, 2002 3:26 PM Subject: Re: [PHP-DB] Error is SQL(works in phpMyAdmin) Hi. The correct way is: INSERT INTO table VALUES (to what colums) 'data'; Try reading PHP manual from mysql_query and from mySQL manual the INSERT statements format. HTH. -Joni - Original Message - From: JJ Harrison [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, July 27, 2002 6:07 AM Subject: [PHP-DB] Error is SQL(works in phpMyAdmin) I do my SQL in phpMyAdmin or at the command line and it works fine. When I try and do my query through PHP I get this(Source to show there are no br etc in it): Query failed: INSERT INTO article_keyword VALUES (3, 'earth', 8.94855); INSERT INTO article_keyword VALUES (3, 'gases', 13.42282); INSERT INTO article_keyword VALUES (3, 'called', 8.94855); INSERT INTO article_keyword VALUES (3, 'atmosphere', 17.89709); INSERT INTO article_keyword VALUES (3, 'allows', 4.47427); INSERT INTO article_keyword VALUES (3, 'light', 6.71141); INSERT INTO article_keyword VALUES (3, 'pass', 4.47427); INSERT INTO article_keyword VALUES (3, 'absorbed', 4.47427); INSERT INTO article_keyword VALUES (3, 'earth\'s', 8.94855); INSERT INTO article_keyword VALUES (3, 'and', 26.84564); INSERT INTO article_keyword VALUES (3, 'converted', 4.47427); INSERT INTO article_keyword VALUES (3, 'heat', 4.47427); INSERT INTO article_keyword VALUES (3, 'energy.', 4.47427); INSERT INTO article_keyword VALUES (3, 'energy', 6.71141); INSERT INTO article_keyword VALUES (3, 'natural', 4.47427); INSERT INTO article_keyword VALUES (3, 'greenhouse', 29.08277); INSERT INTO article_keyword VALUES (3, 'effect', 6.71141); INSERT INTO article_keyword VALUES (3, 'car', 6.71141); INSERT INTO article_keyword VALUES (3, 'temperature', 6.71141); INSERT INTO article_keyword VALUES (3, 'inside', 4.47427); INSERT INTO article_keyword VALUES (3, '200', 4.47427); INSERT INTO article_keyword VALUES (3, 'years', 4.47427); INSERT INTO article_keyword VALUES (3, 'population', 4.47427); INSERT INTO article_keyword VALUES (3, 'six', 4.47427); INSERT INTO article_keyword VALUES (3, 'increase', 8.94855); INSERT INTO article_keyword VALUES (3, 'gas', 4.47427); INSERT INTO article_keyword VALUES (3, 'carbon', 4.47427); INSERT INTO article_keyword VALUES (3, 'concentrations', 4.47427); INSERT INTO article_keyword VALUES (3, 'temperatures', 4.47427); INSERT INTO article_keyword VALUES (3, 'lower', 6.71141); INSERT INTO article_keyword VALUES (3, 'causing', 4.47427); INSERT INTO article_keyword VALUES (3, 'warming', 8.94855); INSERT INTO article_keyword VALUES (3, 'our', 4.47427); INSERT INTO article_keyword VALUES (3, '(the', 4.47427); INSERT INTO article_keyword VALUES (3, 'troposphere)', 4.47427); INSERT INTO article_keyword VALUES (3, 'climate', 4.47427); INSERT INTO article_keyword VALUES (3, 'change', 4.47427); INSERT INTO article_keyword VALUES (3, 'ozone', 4.47427); brYou have an error in your SQL syntax near '; INSERT INTO article_keyword VALUES (3, 'gases', 13.42282); INSERT INTO article' at line 1 why is it so? -- JJ Harrison [EMAIL PROTECTED] www.tececo.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] select from two tables
I have the following code:
$query = SELECT * FROM xoops_album,xoops_artist WHERE
xoops_album.artist_id = xoops_artist.artist_id;
$albumby = mysql_query($query) or die(Select Failed!);
echotdcenter;
if (mysql_num_rows($albumr)) {
while ($album = mysql_fetch_array($albumr))
{
echo ($album[album]);
echoi by ; echo ($albumby[xoops_artist.artist]); echoibr;
}
echo /center/td;
}
What I'm trying to do is select the album title from one table ( this
works ok ) and get the artist name from another table ( this is the part
I can't get to work ). I believe it has something to do with the code
above as the rest works ok. The result I end up with is below:
Waterloo by ( this is where I want to put the artist name)
Thanks for in help in advance.
RE: [PHP-DB] select from two tables
Well, I'm just a beginner myself, but I would say that: echoi by ; echo ($albumby[xoops_artist.artist]); echoibr; should at least be: echoi by ; echo ($album[xoops_artist.artist]); echoibr; or even: echoi by ; echo ($album[artist]); echoibr; Hope this helps, Herman -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] select from two tables
Ok I tired this but it did not help, but thank you. -Original Message- From: Herman Verkade [mailto:[EMAIL PROTECTED]] Sent: Sunday, 28 July 2002 1:18 a.m. To: 'Barry Rumsey' Cc: [EMAIL PROTECTED] Subject: RE: [PHP-DB] select from two tables Well, I'm just a beginner myself, but I would say that: echoi by ; echo ($albumby[xoops_artist.artist]); echoibr; should at least be: echoi by ; echo ($album[xoops_artist.artist]); echoibr; or even: echoi by ; echo ($album[artist]); echoibr; Hope this helps, Herman -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Error is SQL(works in phpMyAdmin)
Hi You have an error in your SQL syntax near '; ^ As far as I know MySQL doesn't require a trailing *;* at the end of the query. Regards Conni -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] select from two tables
Hi Barry
$query = SELECT * FROM xoops_album,xoops_artist WHERE
xoops_album.artist_id = xoops_artist.artist_id;
If you only need two fields, it is not necessary to retrieve all
fields
$query = SELECT xoops_album.album, xoops_artist.artist
FROM xoops_album, xoops_artist
WHERE xoops_album.artist_id = xoops_artist.artist_id;
$albumby = mysql_query($query) or die(Select Failed!);
Your resultset contains then the only two required fields.
echotdcenter;
if (mysql_num_rows($albumr)) {
while ($album = mysql_fetch_array($albumr)) {
echo $album[album] iby $album[artist] /ibr; // new
line
}
echo /center/td;
}
Hope it helps
Conni
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Re: [PHP-DB] Error is SQL(works in phpMyAdmin)
It turns out phpMyAdmin actually splits up the queries and does them singaly. -- JJ Harrison [EMAIL PROTECTED] www.tececo.com Cornelia Boenigk [EMAIL PROTECTED] wrote in message 00ef01c23580$2ed210c0$7f54fea9@zwerg98">news:00ef01c23580$2ed210c0$7f54fea9@zwerg98... Hi You have an error in your SQL syntax near '; ^ As far as I know MySQL doesn't require a trailing *;* at the end of the query. Regards Conni -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: SQL order by queries
In article [EMAIL PROTECTED], [EMAIL PROTECTED] says... Can you d arithmetic division on two fields in an ORDER BY?? eg SELECT * FROM table_name ORDER BY field1 / field2 DESC Can it be done? Probably would work - give it a try :-) If you want to use the value you are ordering on, try ELECT * FROM table_name ORDER BY field1 / field2 AS something DESC then you can retrieve the reuslt of the division as $something. Cheers -- David Robley Temporary Kiwi! Quod subigo farinam -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Pulling a data from array of checkboxes
I am passing a value of checkboxes in a form to a new page like this (see example here: http://www.gibsonusa.com/test/check/) In a new page how can I pull the data from a MySQL database according to each value I passed from a previous page? In other words: Assume that the value of a checkbox equal to the ID of the record in a database. The value of checkbox 1 (for example) is 12345 and the value of checkbox2 is 67890. After passing those values to a new page how can I pull the rest of the information from a database for those values? _ Chat with friends online, try MSN Messenger: http://messenger.msn.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Pulling a data from array of checkboxes
Hi
In a new page how can I pull the data from a MySQL database
according to
each value I passed from a previous page?
Not tested, just an idea;-)
For example
$chk[0] contains 12345, the value of checkbox1
$chk[1] contains 67890, the value of checkbox2 and so on.
?php
foreach($chk as $key = $val) {
$sql = SELECT field1, field2 ... from yourtable WHERE ID = $val;
$res = mysql_query($sql);
if ($res) {
$row = mysql_fetch_row($res);
// do something with the data in the retrieved row
}
}
Greetings
Conni
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