PHP - How Do I Populate A Database From A Number Grid In Mysql?

Hey Guys

I have the following table: (login)


I need that table to be echoed he


But it needs to filter it for the school they are logged in as.

Help: When they login they have to select what school they are at so you can use that code.
example: PHS = Poole High School

Hi Good day Can someone pls help me regarding with this matter.
I have a page for inserting records. And I have a combobox that is fetched through the specific field in database. I also have a table of records, for each row i have an href edit which throws the values to its respective texbox and combobox. 
THE PROBLEM is that when i click the href edit, the value from a selected row was not displayed from the combobox

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=351349.0

original sql
Code: [Select]
-- --------------------------------------------------------

--
-- Table structure for table `countries`
--

CREATE TABLE `countries` (
  `id` int(6) NOT NULL auto_increment,
  `value` varchar(250) NOT NULL default '',
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=243 ;

--
-- Dumping data for table `countries`
--

INSERT INTO `countries` VALUES (1, 'Vancouver');

New Sql
Code: [Select]
-- --------------------------------------------------------

--
-- Table structure for table `countries`
--

CREATE TABLE `countries` (
  `id` int(6) NOT NULL auto_increment,
  `value` varchar(250) NOT NULL default '',
  `code` varchar(12) NOT NULL default '',
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=243 ;

--
-- Dumping data for table `countries`
--

INSERT INTO `countries` VALUES (1, 'Vancouver', 'BC-50');

PHP code for the job
Code: [Select]
<?php

// PHP5 Implementation - uses MySQLi.
// mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase');
$db = new mysqli('localhost', 'root' ,'password', 'weather');

if(!$db) {
// Show error if we cannot connect.
echo 'ERROR: Could not connect to the database.';
} else {
// Is there a posted query string?
if(isset($_POST['queryString'])) {
$queryString = $db->real_escape_string($_POST['queryString']);

// Is the string length greater than 0?

if(strlen($queryString) >0) {
// Run the query: We use LIKE '$queryString%'
// The percentage sign is a wild-card, in my example of countries it works like this...
// $queryString = 'Uni';
// Returned data = 'United States, United Kindom';

// YOU NEED TO ALTER THE QUERY TO MATCH YOUR DATABASE.
// eg: SELECT yourColumnName FROM yourTable WHERE yourColumnName LIKE '$queryString%' LIMIT 10

$query = $db->query("SELECT your_column FROM your_db_table WHERE your_column LIKE '$queryString%' LIMIT 10");
if($query) {
// While there are results loop through them - fetching an Object (i like PHP5 btw!).
while ($result = $query ->fetch_object()) {
// Format the results, im using <li> for the list, you can change it.
// The onClick function fills the textbox with the result.

// YOU MUST CHANGE: $result->value to $result->your_colum
          echo '<li onClick="fill(\''.$result->value.'\');">'.$result->value.'</li>';
          }
} else {
echo 'ERROR: There was a problem with the query.';
}
} else {
// Dont do anything.
} // There is a queryString.
} else {
echo 'There should be no direct access to this script!';
}
}
?>

What the original code does
1) you start typing your city (eg. Van)
2) when you type the first letter (eg. V), it looks into mysql and auto fills a dropdown menu with all possible cities

What i need it to do
1) you start typing your city (eg. Van)
2) when you type the first letter (eg. V), it looks into mysql and auto fills a dropdown menu with all possible cities
3) when you find your city you click it or press enter, and it POST's the city code as well

now how do i munipulate the script to do that...

another thing, when i put the extra sql entry in "code", the auto fill stopped working, why?

Thanks

Hi everyone.

I have a combo box which lists usernames and onchange, the username value is passed to a textbox. however, I have 3 textboxes i need to populate based on the selection of the combobox: username. department and email. i have the username going into a textbox, but i'm not sure how to pass department and email into two other textboxes.

I'd appreciate any help you could provide. Thanks.

Code: [Select]
<script>
function CBtoTB()
{document.getElementById("username").value=document.getElementById("usernameselect").value}
 </script>

<?php  
$result=mysql_query("select Username, EMail, Department from users"); 
$options=""; 
while ($row=mysql_fetch_array($result)) { 
$username=$row["Username"]; 
$options.="<OPTION VALUE=\"$username\">".$username.'</option>'; 
} 
?>
<select name="usernameselect" id="usernameselect" onchange="CBtoTB()">
<option value="">&lt; select user &gt;<?php echo $options ?></option>
</select>
 <input name="username" type="text" id="username" value="<?php echo $username ?>" size="25" readonly="readonly" />

I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database.

Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too.

Thanks in advance!

Like the title says. I have 4 of the same number in my database, but I only want my website to display one of the 4.

How can I do something like this?

How do I call the last ID number from a table in MySql and echo it? This is what I have so far:

Code: [Select]
$query = mysql_query("SELECT idnum FROM ready_aclassft");

Maybe im comeing at it all wrong.

I am designing a mmorpg guild site, and decided to make a coordinate database. A user will enter their own coords, x and y, and the 10 closest coords to them will be displayed.

I need help doing this. My table is:

id
coordx
coordy
alliance
region

I am using $_GET to retrieve the user's input. so
$_GET['mex']
$_GET['mey']

Thanks in advance for your help.

How to text a phone number in database
as well as record recieved textmessages into mysql database.

Any help here? i've been googling for ages.

My current project requires me to save the number of views an image (banner ad) has appeared on users website. The problem here is that the image will be appearing on several users website and i want to be able to save the number of times it has appeared on my database. I know how to do this if the image was on the same server as the database, but not when the image and database are on completely different servers.

This is somewhat like analytic where it tracks the visitors to a page. Does anyone how i can accomplish this?

i figured i would need to create a JavaScript code to provide to the user so they can place it on the site but how do i get that code to connect to my database?

Hey guys,
I m not yet an experienced coder. SO faced alot of problems. Here, I'm trying to generate a UNIQUE RANDOM NUMBER set between two numbers. Repetative occurance must be avoided by all means as I want every number so generated bears a unique value. In other words, every values that made entry into the field of my database should be differant from each others.


<?php
$conN=mysql_connect("localhost","root","");
if(!$conN)
{
die('error'.mysql_error());
}
mysql_select_db("freebie_allusers",$conN);
$UIN=mt_rand(1,5);
$locateUIN="SELECT UIN FROM user_info WHERE UIN='".$UIN."'";
$fetchUIN=mysql_query($locateUIN);
$resultUIN=mysql_num_rows($fetchUIN);
if($resultUIN>0)
{



WHAT CODE IS REQUIRED HERE SO AS TO GENERATE UNIQUE RANDOM NUMBER?






}

else 
echo $UIN;
?>

Thanx in advance

hi!, is it possible to use rand() to generate unique number that will be saved in a database as a primary key? Since i dont want to have numbers like 00001 incrementing on the table. They dont look like real account numbers... i need something like 45642  or 95452 and the like.

thanks in advance.

Hello,

I am using the following code to display images managed by a MySQL database.  Basically another program manages a bunch of images, but this script displays certain ones (ones with INCLUDE = 1 in the database) on my main page.  My question is, is there an easy way to limit the number of images it displays, say to 5?  I'm not too concerned which images actually display (ascending or descending)... or better yet, random!  Most importantly, I only want five to display.  Each image will be linked to the full page, which displays all the images.

Any ideas?

Thanks!


Code: [Select]
<?php
$username="XXXXXXX";
$password="XXXXXXX";
$database="XXXXXXX";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM ft_form_12 WHERE col_24='1'";
// $query="SELECT * FROM ft_form_12";
// SELECT * FROM ft_form_12 WHERE col_24='1'
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>

<?php
$i=0;
while ($i < $num) {

$f20=mysql_result($result,$i,"col_23"); //Photo file name
$f21=mysql_result($result,$i,"col_24"); //INCLUDE
?>

<a href="http://www.domain.com/display_whole_page.shtml"><img src="http://www.domain.com/the_file/pictures/<?php echo $f20; ?>" height="50" border="0"></a>&nbsp;&nbsp;&nbsp;
               
<?php
$i++;
}
?>

Hi,

I need to write some code for this but unsure where to start.

Basically i have a database of say 10 rows. I need to be able to input a number and some sort of fair algorithm will choose one of the rows.

The number will be from 6 to 10 numbers long, and will need to pick the same result each time based on that number.

Its for a lotto game