Boskor

Hello guys... Iam developing a stock maintenance system for a hotel (PHP-MySQL) in which the client will be updating the consumption everyday. Suppose the items are like Pepsi Buns Sauce Then everyday consumption he needs to keep track. He will login and enter the consumption details for 23rd March then …

Good day) Advice please - is it permissible to use this syntax - [CODE] insert into EXAM_MARKS ( EXAM_DATE) values (str_to_date('26-05-2000', '%d,%m,%y')); [/CODE] This line does not work. I can not understand on what went wrong. Thank you for your answers)

I know there are many other threads about this same error, but none have helpped me, here is my code: [CODE] <?php $mysql_id = mysql_connect('mysql3.000webhost.com', 'a2778852_556875', 'pendolino390'); mysql_select_db('a2778852_555676', $mysql_id); $result = mysql_query("SELECT personalexperience, sex, age, sexuality, FROM Personal_experience"); while($row = mysql_fetch_array($result)) { echo $row['personalexperience'], $row['sex'], $row['age'], $row['sexuality']; } mysql_close($mysql_id); ?>[/CODE] …

Hi All, I am making website in php. I want to add funds to my account using paypal api. Please help me how i can do that? Regards

[CODE]<?php require_once('upper.php'); // Connects to your Database require_once('database.php'); //This checks to see if there is a page number. If not, it will set it to page 1 if (!(isset($pagenum))) { $pagenum = 1; } //Here we count the number of results //Edit $data to be your query $query = "SELECT …

How can I use explode function in php I already implode some data in mysql [CODE] // customer table cust_id, impid, 1, 1,2 // sports table sport id, sport name 1, football 2, basketball [/CODE] How can I explode and make a php/mysql query

I am just trying to link a reference numbers using php/mysql 1) Each reference number will select (see code) 2) Post that number to next page and query the info. How can I post reference ID and query in the page 2. [CODE] <? // Page 1 refno.php $cid= $_SESSION["c_id"]; …

I have this code: [CODE]$sql=mysql_query("select distinct one.ime_kategorije AS ime_kat,count, two.kategorija_g AS kategorija_g FROM (SELECT *, COUNT(DISTINCT kategorija_g) AS count FROM galerija_slik LEFT JOIN kategorija ON galerija_slik.id_kat=kategorija.id_kategorija GROUP BY id_kat) AS one JOIN ( select * from galerija_slik group by kategorija_g) AS two WHERE one.id_kat = two.id_kat;"); while ($row=mysql_fetch_array($sql)) { $kategorija=$row['ime_kat']; …

I have another problem with mysql. I have 3 tables (Users, User_status and Analysis). Users and User_status are in relation (Users.User_status_name=User_status.user_status) and Users and Analysis are in relation (Users.Username=Analysis.Tipster). I want to have table in php: [CODE]Tipster Form boskor 1W 1V 0L slopart 0W 1W 2L [/CODE] In table Analysis …

Hello I have another problem. Here is previous thread: [url]http://www.daniweb.com/forums/post1237196.html#post1237196[/url] and my code: [CODE]$tipster = mysql_query("SELECT SUM(Analysis.Profit), Users.id, Users.Country_name, Users.Username, User_status.id_status FROM Users INNER JOIN User_status ON Users.User_status_name=User_status.user_status left outer JOIN Analysis ON Users.Username=Analysis.Tipster WHERE User_status.id_status>=3 GROUP BY Tipster ORDER BY SUM(Analysis.Profit) DESC"); while ($row = mysql_fetch_array($tipster)) { $sel_tipster=$row['Username']; echo …

I have one problem with php and mysql. I have 3 tables (Users, User_status and Analysis) and want to display users and profit in a php table order by profit. So I wrote this code: [CODE]$tipster = mysql_query("SELECT Analysis.Profit, Users.Username, User_status.id FROM Analysis, Users INNER JOIN User.status ON Users.User_status_name=User_status.user_status WHERE …