Boskor

Try with such sql table: example: id item date quantity 1 pepsi 23rd Mar 30 ... With php you will insert items in this table and he can add what he want (no limits)

[QUOTE=vedro-compota;1394833]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)[/QUOTE] I am using this date function in php forms: …

[QUOTE=Cstonehouse;1399588]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] …

[QUOTE=ravian48;1399056]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[/QUOTE] First you check their site, they have documentation for this: [url]https://www.x.com/index.jspa[/url]

Pagination looks good. Try this: Change $PHP_SELF with $_SERVER['PHP_SELF']...

[QUOTE=nukabolhi;1393671]Any one figured out the problem, pls let me know[/QUOTE] First your sql query is not good. You have to have relation between those 2 tables. There are simple example: [CODE]$sql1=mysql_query("SELECT * FROM customer,sports WHERE customer.impid=sports.id"); while ($row=mysql_fetch_array($sql)) { $sport_name=$row['name']; ... } [/CODE] Than you write your explode function.

[QUOTE=nukabolhi;1393336]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 …