It was great information for

dzoapps — Tue, 21 Jul 2015 07:07:24 +0000

It was great information for everyone to inrease the knowledge for MySQL

i am developing chat

anilkumar123 — Fri, 17 Apr 2015 12:57:39 +0000

i am developing chat application..
what is my problem here by.

i am unable to passing messages to particular user.if am passing any message that store default on db and seeing same message in all what ever users logined on chat.

please give me solutions for it.

james21 — Thu, 22 Jan 2015 03:08:10 +0000

`<?php //create connection into a database then include 'conn.php'; $sql="INSERT INTO tablename (first,second) VALUES ('$_POST['first']','$_POST['second']')'; $result=mysql_query($sql); // if successfully insert data into database, displays message "Successful". if($result){ echo "Successful"; echo "<BR>"; } else { echo "ERROR"; } ?>`

Ricky nkonya wrote:mysql>

latif pala — Sun, 08 Sep 2013 07:52:43 +0000

Ricky nkonya wrote:

mysql> select * from link;
+--------+----------+----
| userID | memberID |
+--------+----------+----
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
| 0 | 0 |
+--------+----------+-----
5 rows in set (0.00 sec)

mysql> select * from list;
+----------+--------+-------+--------+--------+------
|memberID| userID| title | fnem | midnem |
+----------+--------+-------+--------+--------+------
| 1 | 0 | Dr | madee| j |
| 2 | 0 | Ms | janeth | R |
| 3 | 0 | Ms | jane | m |
+----------+--------+-------+--------+-----

i have three tables CUSTOMERS,LIST,LINK... my problem is that i have a table called LINK that has the userID(which is the primary key for the CUSTOMERS table) and memberID(which is the primary key for the LIST table) when i insert information in the customers table am expecting the userID and the memberID in the LINK table to be filled with the values that appear in the CUSTOMERS table.But when i add the information there is no any connection between the three tables... WHAT is the appropriate query to accomplish the joint.

Thank you guys. Special

selvamraja — Sun, 08 Sep 2013 05:35:22 +0000

Thank you guys. Special thanks to pr0gr4mm3r. This was very helpful.

mysql> select * from

Ricky nkonya — Fri, 20 Apr 2012 15:05:01 +0000

This is great , thanks to you

David26 — Thu, 15 Mar 2012 12:12:38 +0000

This is great , thanks to you I got this information. I appreciate your work, thanks for taking this opportunity to discuss this, the thread is really helpful.

You might be able to setup a

iigwk — Fri, 27 Jan 2012 14:20:58 +0000

You might be able to setup a view with the appropriate joins, and insert into the view. I've tried it a few times, and it seems to work ok.

Here is my php code i want

Guru — Thu, 12 Jan 2012 06:42:11 +0000

Here is my php code i want that whenever i run my HTML page new enter will get stored in my database, please help me in this code ,here below is my PHP code.

<?php
$con= mysql_connect("localhost","root","ozone2412");

if(!$con){
   die("could not connect: ". mysql_error());
}
mysql_select_db("gurneet",$con);
$sql="INSERT INTO login(Username , Password, Id, Role) VALUES ('$_POST['username']','$_POST['password']','$_POST['id']','$_POST['role']')";
or die("data not inserted");

$exec=mysql_query($sql,$con);
 
$sql_1="INSERT INTO login(Username , Password , Id ,Role) VALUES ('$_POST['username']','$_POST['password']','$_POST['id']','$_POST['role']')";
 
$exec=mysql_query($sql_1,$con);

if(!mysql_query($sql,$con)){
  die("ERROR: ".mysql_error());
}
echo "1 record added";

mysql_close($con);
?>

Here is my php code i want

Guru — Thu, 12 Jan 2012 06:42:00 +0000

Here is my php code i want that whenever i run my HTML page new enter will get stored in my database, please help me in this code ,here below is my PHP code.

<?php
$con= mysql_connect("localhost","root","ozone2412");

if(!$con){
   die("could not connect: ". mysql_error());
}
mysql_select_db("gurneet",$con);
$sql="INSERT INTO login(Username , Password, Id, Role) VALUES ('$_POST['username']','$_POST['password']','$_POST['id']','$_POST['role']')";
or die("data not inserted");

$exec=mysql_query($sql,$con);
 
$sql_1="INSERT INTO login(Username , Password , Id ,Role) VALUES ('$_POST['username']','$_POST['password']','$_POST['id']','$_POST['role']')";
 
$exec=mysql_query($sql_1,$con);

if(!mysql_query($sql,$con)){
  die("ERROR: ".mysql_error());
}
echo "1 record added";

mysql_close($con);
?>