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Hi
I am trying to create a simple web comparision page using
php-mysql.
The web pages contains products. I would like to
provide viewers with the option to compare products. Atmost 4 products
should be compared at same time.
Therefore i propose to place a
small checkbox next to each product to see if 4 products are selected.
Once 4 products are checked and compare button is clicked the results
should display in pop-up window.
How do i write the query taking
into consideration the products selected ???
how do i display the
results from database onto website in a pop-up page ???
my
products table is as follows:
product-id
product-name
INSERT INTO
soldItems(itemId,qty,price,buyer,date,buyerStatus,sellerStatus,address,shipType,paymentType)
VALUES('$i',XXX,XXX,'$user','$time','0','0','$shipType','$paymentType'); />But qty,price must get so:
SELECT cart.qty,
cart.qty*items.price as price FROM cart,items WHERE cart.user='$user'
AND cart.itemId='$i' AND items.id='$i';
But with conditions />SELECT IF( EXISTS( SELECT * FROM delivery WHERE
deliveryId='$shipType' AND itemId='$i'), 1, 0)
AND
SELECT
IF( EXISTS( SELECT * FROM paymentType WHERE шd='$paymentType' AND
itemId='$i'), 1, 0)
it sould be in one query
I want to execute an UPDATE query to update a datetime field but I am
not getting the desired results!
Here is the query:
Syntax:
[ Download ] [ Hide ]Syntax: [ Download ] [ Show ]$result =
mysqli_query($con, "UPDATE `tasks` SET
`deadline`=".$_POST['dt'].", `status_id`=3 WHERE
`task_id`=".$_POST['id']."");
The
deadline is the datetime field and I am POSTING the dt from a text
input. Please help me. Thanks in advance!
Hello I made a request to display information from a base against I do
not get it to appear in column
For information in the database I
have dates with several slots
before(data from the base= />22/10/2013 HD:10h30 HF:11h30
22/10/2013 HD:18h30 HF:20h30 />22/10/2013 HD:21h30 HF:22h00
23/10/2013 HD:10h30 HF:11h30 />23/10/2013 HD:18h30 HF:20h30
25/10/2013 HD:10h30 HF:11h30 />25/10/2013 HD:18h30 HF:20h30
application to display the
information in column />22/10/2013---|-23/10/2013--|-23/10/2013--| />D10h30-F11h30|D18h30-F20h30|D18h30-F20h30| />D18h30-F20h30|D10h30-F11h30|
D21h30-F22h00|
Thank you />
Hello all,
I am somewhat new in writing codes in php and been
wondering about something:
Every book I read and tutorial I
follow, I never see any examples where relationships between tables
are established in mysql. everything seems to happen while coding in
php and joining when or where it is necessary. AM I better off not
establishing relationships in mysql and deal with relationship while I
write php codes or it better practice to reinforce relationship from
mysql first?
Ok I have a single table called opportunities, within that table I
have field called sales_stage. I need to get the percentage of rows
where sales_stage is "Closed Won" and also get percentage of rows
where sales_stage is "Closed Lost" And basically just display the two
headings Closed won and Closed Lost.
I'm a little bit lost with
how to do this one, here is an sql fiddle />http://sqlfiddle.com/#!2/ac28d/13
Syntax: [ Download ] [ Hide ]SXBB[id].writeCmd();Syntax: [ Download ]
[ Show ]$result = mysql_query ("SELECT DISTINCT
subscribed_upload.email FROM subscribed_upload LEFT JOIN unsubscribed
ON subscribed_upload.email = unsubscribed.email WHERE
unsubscribed.email IS NULL") or die
(mysql_error());
while($row =
mysql_fetch_array($result))
    { />   
    if
(filter_var($row[email],
FILTER_VALIDATE_EMAIL))
      { />      $csv_output .=
'"'.$row[email].'"';
     
$csv_output .= &q
0 down vote favorite

I am a rookie PHP and MongoDB
developer.
I have created a PHP web project with an HTML page
that contains an 'Add' button. The name of the page is awards.html.
The awards.html file contains its counterpart JavaScript file,
awards.js. A code is executed in this js file when the Add button is
clicked. This code sends an AJAX call to a PHP class elsewhere in the
project named, example.php which contains code to execute a function
called, connect() that is in an Awards.php file.
The source-code
of my files is given as follows:
Awards.html
<div class =
"divbottom">
<div id="divAddAward">
Hi everybody,
I've installed XAMPP on Linux Mint 14 successfully.
PHP works perfectly, but I need to test MySQL. How can I do that? I've
found a lot of tutorials to test MySQL, but to do that first I need to
access the MySQL shell using the following command:
Syntax: [
Download ] [ Hide ]Syntax: [ Download ] [ Show ]mysql -u root -p />Problem is that, when I'm issuing that command from Terminal, it's
showing the following message:
Syntax: [ Download ] [ Hide
]Syntax: [ Download ] [ Show ]The program 'mysql' is currently not
installed. You can install it by typing:
sudo apt-get install
mysql-client-core-5.5
 
How can I overcome t
Hello ;
Please refer to the attached script. The information that
I entered through a PHP form does not go int the mysql table. I have
tried so many things but it does not work. It always gives me the
"ERROR" message.

Syntax: [ Download ] [ Hide
]SXBB[id].writeCmd();Syntax: [ Download ] [ Show ]<?php />$connect = mysql_connect
("localhost","root","abc1234")
or die ("error");
mysql_select_db
("rameca") or die ("Error Connecting With
The Database");
$queryget1 = mysql_query
("SELECT id FROM quatation " ) or die ("
error with ta

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