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hello. I have a table populated by name. The names are links when clicked will pop up a form for editing. each input (for editing) is populated by the equal value from database but im having trouble with one particular text box. this text box is for date of birth …

#include <iostream> #include <cstdlib> using namespace std; // function prototype for add int add (int x, int y); int sub (int q, int r); int mul (int s, int t); int divi (int u, int v); int main () { int a,b; char ch; loop:cout<<"ENTER CHOICE OF OPERATION: + - …

I need to make one statement to select either of the items in listbox and show the same label, regardless of which item is selected. I've tried someting like If Me.ListBox1.SelectedItem = "R115" Or "Bay, Charles" Then Label1.Text = "Address Information Avaliable" End If But it doesn't work. The code …

Hi! Can anyone know the result of this query: SELECT COALESCE(MAX(ID),0) + 1 FROM Table123

Hello, so I was trying few relations between tables and I noted an unexpected behaviour when using `group_concat()`. Let say I have three tables: fruits, fruitstock and fruitprices. These are the structures: create table fruits ( id tinyint unsigned not null auto_increment primary key, name varchar(50) not null, quality varchar(50) …

can someone help me to find out why my label display is listing the result of the textboxes instead of adding up the values of the text boxes? here are my codes: Dim strstudentName As String = (" ") Dim dblAssignment1 As Double Dim dblAssignment2 As Double Dim dblAssignment3 As …

<?php $username = $_SESSION['username']; $result = 'sample'; $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } $sql = "Select FirstName from Person Where LoginID = (Select LoginID from UserLogin Where Username = '$username')"; $result = mysql_query($sql,$con); echo $result; ?> I'm kinda new to this server-side scripting …

I am making a mysql query that gathers comments for a user, it works properly, I connect to the database and recieve the comments for a user, but it is not showing the first comment for the user. Here is the code function comments(){ //connect to db connect_db_members(); //start query …

>Hi guys,..i've two arrray data1 and data2 then i do intersect to get same value between data1 and data2, and >use count to get total same value,..how to ARSORT count result from intersect array multidimensional? <?php $data1 = array( array( '7' => 'chelsea everton', '8' => 'everton villa', '9' => …

Hi every one!! I need help to match user data from database inorder to log in to admin panel. Here is what I am trying. it works ok but it is giving dialog box containing text "Sorry! You have no access to Database" here is the code. Please help String …

i have writeen following query in my code. $tquery="SELECT questions.id,questions.date,questions.title,users.user_name,subjects.subject FROM questions INNER JOIN users ON questions.student_id=users.id INNER JOIN topics ON questions.topic_id=topics.id INNER JOIN subjects ON topics.subject_id=subjects.id WHERE subjects.id=".$_SESSION['sub1']." OR subjects.id=".$_SESSION['sub2'].";"; but when i am trying to echoing its result it is not displaying anything.

Hello, I want to combine all my database table in to one line so i can put it all in the url e.g. Name | Times __________________________ test1 | 1 __________________________ test2 | 2 __________________________ This is how i want it to look in the url (Result.php?loggedin=true&links=test1-1_newline_test2-2)(_newline_ means that will …

HI. I've tried other codes, but they don't seem to work. What I'm trying to do, is grab a result from google's shopping page, and display it on my page. I only want the title, which is in the tag:[CODE]<h3 class="result-title">TITLE</h3>[/CODE] How do I do that in PHP? Thanks!

Hi All Im trying to get this to work, but can only get it to work without putting the form tags in, and wont work with more than one set of the inputs. the below is the result when i use ajax to search for item names, then want to …

Hi all, I am new to php and I have a problem that I am looking into it for 1 day and can't understand it please help. here is my html code : <html> <head> <title> Calculation Form </title> </head> <body> <form action=calculate.php method="POST"> value one: <input type="text" name"val1" size=10 …

Hi Everyone, I have been trying to understand mysql a little further - with some small steps in the right direction. but I have come across a problem I am unable to fix / get working. I have a form, that has a variable $country - I am trying to …

I'm stumped as to why this might be happening. A website I had completed for a client over a year ago, was working fine until a couple of days ago when the web host went and moved the site onto a new server without warning any of their customers (myself …

Hi Everyone- We are building a database that needs table(purchaser) to check table(paypal) for a matching telephone number. Table(paypal) has 'purchaser telephone' as PK and table(purchaser) has 'purchaser telephone' as FK. I can insert and see the data in both tables, but running a select query returns an empty result. …

HELP!! I try to bind a textbox to a datatable but i cant.. I checked if the datatable is filled (it includes the results of a sql query)and it s ok.. i can depict the result in a combo box but not in textbox. I wait for only one result …

Okay, So I want an array to get userinformation by the username that is provided. But in the other tables, it only has a ID field to link with... [CODE]$sql = "mos_users.id as userid, block, username, email, lastvisitDate, firstname, lastname, avatar, cb_age, userip FROM mos_users, mos_comprofiler, mos_comprofiler_plug_iplog WHERE mos_users.username = …

hi guys having a problem getting a single value from a mysql query getting all sorts of problems - heres my latest version below [code=php]mysql_connect($host,$user,$password) or die( "Unable to select database"); mysql_select_db("mydatabase"); // select database to use. $user = 84; $sql1="SELECT `order_id` FROM `table1` WHERE `user_id`= '$user'"; $array = mysql_fetch_array('$sql1'); …