Re: [PHP-DB] About mysql_connect()
Use PDO and try/catch. I think it's much easier. On Feb 6, 2008 7:07 PM, Prabath Kumarasinghe [EMAIL PROTECTED] wrote: Hi All When I stop mysql database whole application will break in mysql_connect() function even without giving any error message.Is there are any way to put exception handling with mysql_connect() function. Cheers Prabath Looking for last minute shopping deals? Find them fast with Yahoo! Search. http://tools.search.yahoo.com/newsearch/category.php?category=shopping -- 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] Re: Query Object in PHP?
Ansering myself. DB_DataObject http://pear.php.net/package/DB_DataObject/ On 2/25/07, js [EMAIL PROTECTED] wrote: Hi list, Is there any implementation of Query Object [1] written in PHP? I spent some time searching on the net but no clues're found. Thank you in advance. [1] http://www.martinfowler.com/eaaCatalog/queryObject.html -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Query Object in PHP?
Hi list, Is there any implementation of Query Object [1] written in PHP? I spent some time searching on the net but no clues're found. Thank you in advance. [1] http://www.martinfowler.com/eaaCatalog/queryObject.html -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] MySQL query question
i want to search the table called comments, and i want to count the number of times artid appears. then i want to group them all together, but then i want to order them by most appearances to fewest. so if artid 1 appeared 40 times and it was the most, i want that to be retrieved first. if artid 3 appeared 38 times and it was second most, i want that retrieved second... etc and so on. this code ive given is wrong, but i know that it has to be something like this or that im close. if any of you could help me id really appreciate it. if theres a shortcut with this code that would be great, otherwise i have to run my php around in a ton of different mysql queries trying to find out which has the most and have it ordered from highest to lowest. thank you for your help. ive tried mysql.com too and i cant find a thing there. plus the mailing lists for mysql, no one ever responds! thank you php list, you are my only hope. ;P -james $mostcomquery = SELECT artid, COUNT(*) FROM comments GROUP BY artid ORDER BY (COUNT(*)) DESC LIMIT 5;
[PHP-DB] mktime question
i have a page that displays an essay someone has written. and people who come to the site can submit comments. the comments are taken quickly in a form and posted. everything is posting perfect except for the time to get the time i dont know what to use? time() or date() or mktime() ??? im really stuck. i dont know how to submit the time to the MySQL database so it takes the correct format for the time area. i have a substr() script that formats it when it is retrived and displayed, but i do not know how to make the INSERT INTO with the time. if you can help me out i thank you! -james
[PHP-DB] page hit tracking
i am using php MySQL and i want to know how i can make a hidden value for certain entries in a table and how many times they are called. for instance, say i have a table of vegetables people can look at. the page displays all of the vegetables, and then it generates links to each one and if the user clicks on one then the page just calls itself and displays more information about the vegetables. now, how do i make a counter where theres a field in the table where it increases by 1 everytime someone looks at a particular vegetable. i really am new to this so thank you for your help. thanks -james
[PHP-DB] mysql date/time question
ok, 2 questions. first, how do i do a mysql query that orders by date? like, say i want to have 5 newest members who have signed up? i want it to display only 5 also. how do i do that? like this? $newestquery = SELECT artid,title,artdate,artauthor FROM article WHERE artdate = $today ORDER BY artdate DESC LIMIT 5; and also, my next question is how do i format the display of date and time from how MySQL stores it to where i want it displayed differently in PHP? like, mysql will store it as -DD-MM and i want it displayed in php as date(n.j.y.) and time displayed as $time = date(g:ia); my time and date columns are 2 different columns in my TABLE, and i want to know how to format the date to enter into mysql and how to format it when i retrive it from mysql. thanks for your help. -james (im a newbie programmer)
[PHP-DB] counting a value
i want to know what i can use to count the number of times a value appears in a column listing where different values are listed. for example, say in the column of Favorite Number we have 100 students, and for each student his or her own row. and each student can pick a number from 1-10. So, we have a column of 100 rows and each field can contain either 1-10.Now, i want to know how can i count the number of times lets say the #2 appears in this list and give that count a number. so if #2 appears 43 times in that column, then i want it to say 2 is listed 43 times. how do i do this? thank you for your help with my beginner newbie questions. -james
[PHP-DB] counting a value REVISED pt2
p.s.- im using PHP with MySQL... if that helps any. thanks i want to know what i can use to count the number of times a value appears in a column listing where different values are listed. for example, say in the column of Favorite Number we have 100 students, and for each student his or her own row. and each student can pick a number from 1-10. So, we have a column of 100 rows and each field can contain either 1-10.Now, i want to know how can i count the number of times lets say the #2 appears in this list and give that count a number. so if #2 appears 43 times in that column, then i want it to say 2 is listed 43 times. how do i do this? thank you for your help with my beginner newbie questions. -james
[PHP-DB] CREATE TABLE problem
ok im making this page and every single time i try to execute it, it tells me it was not successful. the database name, user,pw,and local host are all correct. i have no idea what do do, and i have 2 more tables besides this one i wanted to create but this was kind of my template for it and i cant even get it to work. im still really bad with this kind of stuff so any help i could get from you is really appreciated. thanks. here is the code: html head titleUntitled Document/title /head body ?php $host = localhost; $user = user; $password = ; $dbname = swwdb; $link = mysql_connect($host,$user,$password); $query = CREATE TABLE staff ( staffid INT(3) NOT NULL AUTO_INCREMENT UNSIGNED, name VARCHAR(255) NOT NULL, login VARCHAR(10) NOT NULL, password VARCHAR(8) NOT NULL, picaddy VARCHAR(255) NOT NULL, email VARCHAR(255) NOT NULL, staffbio TEXT NOT NULL, createdate DATE NOT NULL, tagline VARCHAR(255) NOT NULL, PRIMARY KEY(staffid,login) ); if (mysql_db_query ($dbname,$query,$link)) { print (the query was successfull); } else { print (the query was NOT success); } mysql_close($link); ? /body /html
[PHP-DB] CREATE TABLE problem redefined
ok i did the mysql_error DIE thing, and now it tells me this: You have an error in your SQL syntax near 'UNSIGNED, name VARCHAR(255) NOT NULL, login VARCHAR(10) NOT NULL, password' at line 2 i have no idea what it means. i dont think any of those are taken or reserved by PHP or mySQL. any further help would be great. thank you. -james
[PHP-DB] get a total comment listing
im making a website where visitors can comment on essays that they read. i have a table that contains every comment written and each comment has its own ID for primary key and articleid column for which essay it is written for. now, i want to be able to tally up how many comments there are for each particular article on a page that sums up all of the essays for a user, so that you can tell which ones have had new comments and which havent. its quite basic, but im stuck as far as what to do. any help would be great, thanks. -james