PHP :: Populating A Select Box From A Database

Reading Directories, Populating Select Box
Can anybody tell me how to read through a directory, get the file names and populate a select box(pulldown menu) with these names? I know how to loop through a directory using something like this:

while ($files = readdir($handle)) {

$ext=substr($file,-4);

if ($files != "." && $files != ".." && $ext == ".php") {

do some wild and crazy stuff here;

}

}

It's the populating the select box (pulldown menu) while doing this that I need the help on.

Populating SELECT Multiple Form Field And Inserting Into Db
I have an update form where I'm trying to populate a SELECT multiple field with a list of 48 categories, from tbl work_cat. And show, as SELECTED, the one or many choices that the user had previously selected from the 48 categories which are stored in tbl cat_relations as $relation_cat.

Then allow the user to update their selections and update the database. But I can't get my form to work. First problem, I can't get the SELECT field on the form to show the categories, and then I don't know where to go from there. Below are my form page and my processing page. Code:

Populating A Form Wth Database Info
I've coded a piece of code which populates a form with data read from the database:

$connection=mysql_connec ("localhost", "f2821842", "f2821842");
$result=mysql_select_db("QUERIES");
$query=mysql_query("Select * from Emp_Details where emp_num = '$employnum'");
while($row=mysql_fetch_array($query))
{$empname=$row['emp_name'];}

<form name="webregform" action="webadmin2.php" method="post">
<input name="requiredname" type="text" size="30" value="<?php echo "$empname"; ?>">
</form>

When I echo the $requiredname, I get spaces and no data, and I know that $empname is not a space-it does read a value in a database. 1. How can I get $requiredname to print a value?

Populating A Drop Box With Results From Database
I have a MySQL database with a table (category) with two fields, catId (int) and category (char(50)). What I want to do is to get all category names in this database and place all of them into a dropdown box on a web page so that the user can choose from the list of available categories.

Populating Form From Database, Then Passing Results To Next Page
I have a multiple select input in a form that's being populated by a row
from my database as such:
<input type="checkbox" name="subm[]" value="$row[ID]">

That part is working fine as I can check the displayed page using View
Source and see that the value is the correct row number from the
database. It is then being submitted on a form by $_POST method to
another page where I want to evaluate the checkboxes and display the
contents of the entire row that corresponds to each value="$row[ID]"
that have been checked. But I can't seem to get it to work. I'm having a
problem passing the selected value. Can someone point me in the right
direction?

$query = ("SELECT * FROM `table`");
$result = mysql_query($query);

print "<p>Data for Selections:";
print "<table border=2><tr><th>You chose:";

foreach ($_POST['subm'] as $value) {
print "<tr><td>";
print "$row[ID];
";
print mysql_field_name($result, 1) . ": " . $row[name]."<br>";
print mysql_field_name($result, 2) . ": " . $row[address]."<br>";
print mysql_field_name($result, 3) . ": " . $row[city]."<br>";
print "</td></tr>";
print "</table>
";
}

if (!isset($_POST['subm'])){
print "<p>No matching entry ";
}

mysql_close();

SELECT Error: No Database Selectedin Query: SELECT * FROM Schedule
I got the above error message after reading FAQ #10 which applied to my first error. The thing is, I thought I had selected a database. This script is copied from a tutorial, and I'm just too new at this to spot my error. PHP Code:

Populating A Second "select" Field
is there a way to populate a second select field in a form based on input for a first select without reloading and using only php? Code:

Database Value Select In Combo Box!
This function is intended to display the current database field value as selected.. along with other values as general options.

I can't figure out how to get it work. Help is appreciated!!

function retrieve_category()
{
$cat = mysql_query("SELECT category_id FROM category");
while ($current_row = mysql_fetch_row($cat))
{
$row = $current_row[0];
if ($row == $id)
{
printf("<option selected>%s</option>
",$current_row[0]);
}
else
{
printf("<option>%s</option>
",$current_row[0]);
}
}
}

Select One Record From A Database
I am trying to select one record from a database. This databse has like 10 tables. I do not know which table this record is in. What php and mysql syntax would I use to accomplish this?

Randomly Select From Database?
this might seem like a stupid question, i should be able to do this. but on the front page of my site i want a thing to show a few users. but i dont want it to show the same people over and over. i want it to select different people each time. any ideas?

Selective Echoing And Database Select
I have written a search engine for business listings for the company I work for. Each listing is in a database, and that works fine. The problem though is two-fold.

The first part of the problem is that not all listings utilise every column of the database, and this leaves gaps when echoing the results. This is not what i had in mind, Any ideas/help would be appreciated.

The second part of the problem is related to the first in that there is a link for any listing that has a website, but not every listing has a website, so every listing having a "Visit our Site" link is impractical and shoddy. Also, i'm not sure how to echo the url from the database effectively in the first place.

Editing Mysql Database Row With Select Box
I have a small site, this small site has a few categories, now I can add items through my online admin area fine and delete them fine, but when it comes to editting them I am having some trouble, everything is working fine except the category column, because it is a drop down box I do not know how to have the category selected during the creation progress selected all the other info is loaded.

Multiple Select From Database Where Clause
I have a website with a database full of category names...I want to be able to chose lets say 5 of these categories and use a select from database clause where I could simply enter the category names I want to use but am not sure how I would go about doing it.

I can get the list to display just the Aprilia category using the code below:

$data1->q("SELECT * FROM categories WHERE cat_name = 'Aprilia' ORDER BY cat_name asc");

However...how could I get my site to display Aprilia, Ducati, Honda, Suzuki, Yamaha?

Using PHP And MYSQL To Select From Database Using Images
I'm trying to update a main page with a list of details matching a selection from a display of images that are displayed in a small window that have been selected from an image group using a select box in the main window, however I don't know how to do this and the main page is not receiving anything when clicking on an image so it's refreshing with empty data.

I've seen many many pages giving examples of how to do this with select boxes or how to do this if only one entry uses an image shown in the small window, but the image display window is displaying many different small images not a select box and the database itself has multiple entries that use the same image. Can anyone help? I have a few instant chat programs if someone wishes to discuss this with me in real-time to see what it is I'm trying to achieve.

Tick Box, Select And Update Database.
I am trying to change the value in table with a tick box. If the tick box is selected it changes the value in the field to 'y' and if it is not selected it changes the field to 'n'.

First of all I need to bring back the current state of the field and then I need to be able to update it from 'y' to 'n' as many times as I want. Code:

Population Select Menu Using Php And Database Values
I have no idea how to go about this. so i would apreciate some help i want to populate the values in a html select menu from database values in one table

so i want all the values from 1 particular field in the database to be displayed in the select menu as separate options.

Displaying A List Of Categories From A Database In A Select Box
Could someone please tell me why this outputs nothing

<select name=categories>
<?
$cat_array = get_categories();
foreach($cat_array as $this_cat)
{
echo "<option value="";
echo $this_cat["category_id"];
echo """;
echo ">";
echo $this_cat["category_name"];
echo "
";
}
?>
</select>

==========================================================

function get_categories()
{
//get the list of categories from the database

$conn = mysql_pconnect("localhost", "user", "pwd");
$query = "select * from categories";
$result = mysql_query($query);
if(!$result)
return false;

$num_cats = mysql_num_rows($result);
if($num_cats == 0)
return false;

$result = db_result_to_array($result);
return $result;
}

//A function that returns a query to the database as an array
function db_result_to_array($result)
{
$res_array = array();
for($count=0; $row=@mysql_fetch_array($result); $count++)
$res_array[$count] = $row;

return $res_array;
}

Connecting Two Database Populated Select Menus
I am able to grab the data from the database to populate the menus individually using.
PHP Code:

Select Specific Phrase From MYSQL Database
I am working on a World of Warcraft guild's website and I am trying to create a page where players can select different kinds of gear, i.e. the gear with the most Stamina.

All the player's items are stored in the database and every item have a field in the Items-table containing the item's tooltip. This could look something like:

Boots of the Nexus Warden
Soulbound
Feet Cloth
97 Armor
+27 Stamina
+17 Intellect
Durability 35 / 35
Equip: Improves spell hit rating by 18.
Equip: Increases damage and healing done by magical spells and effects by up to 21.

I then want to select the item with i.e. the most Stamina for all the different slots; Head, Chest, Legs etc.

If I write this:
$query = mysql_query("SELECT substring(item_tooltip, LOCATE('Stamina', item_tooltip)) AS maxor FROM roster_items WHERE member_id = ེ'") or die(mysql_error());

It returns the following:
Stamina +17 Intellect Durability 35 / 35 Equip: Improves spell hit rating by 18. Equip: Increases damage and healing done by magical spells and effects by up to 21.

But I want for it to show the leading characters also (in this case "+27 " before "Stamina") and then not show anything else after the phrase "Stamina". So it shows "+27 Stamina" and nothing else for the above mentioned item. How do I accomplish this?

And then, secondly how do I then get it to show only the item with the most Stamina after it selected the above?

Hope this makes any sense.

I've tried to search these forums and found something a little along the lines of my problem but haven't been able to adjust them to my needs and I therefore need your help which will be greatly appreciated.

Reusing An Object For Common Select To Database
I've noticed that after I finished this app, I had repeated a similar task over and over and would like to modularize it. The repeated task was connecting to the database, executing a select, putting all the results of the columns into different variables of an object. PHP Code:

Simple Filter Of Database With $_GET And Select SQL
I need help with my logic statement with a simple search engine on a MySQL table.
I have three form text input fields named f_name, l_name and dept. I am using the GET method to gather data the user enters in each input field.

So far I have been able to build a check to see that the submit has been clicked but I am having trouble with assembling the SQL statement because I need for the search engine to be able to filter the results based on the information provided in 1-3 of the fields.

I could use if statements with some and clauses but that would mean that I would have to account for each combination. Something like

If(($_GET[‘f_name’] != “”) || ($_GET[‘l_name’] !=) || ($_GET[‘dept] !=)){

$sql = "SELECT * FROM facform WHERE f_name LIKE '$_GET[f_name]',
l_name LIKE '$_GET[‘l_name]', dept LIKE '$_GET[dept]' ”;}

The problem with this is that I would have to get every combination and it’s not very scalable. How can I create this SQL statement dynamically? Code:

Error In Generating Table From Database Select Statment
I am trying to dynamically create a table with the results from a select statement. Below is the code that is not working. Any help would be oh so greatly appreciated. This is for a final at school which is due in an hour. This is the only thing I can't seem to make work. Code:

Select Specific Snippet Of Text, Replace With A Database Result
I'm having some problems trying to do something; I just can't think of a way to do it. Basically, I want to search for a specific snippet of text in a body of text, then replace it with the requested database result with some thrown in HTML.

Users will be able to type the following (or something along these lines, depending on what works best): [image=12345]Insert caption here[/image]

Then I need it to search for the requested image in the appropriate table and return the image's url (for an example, let's say the table is called "images", and the columns required are "imageid" and "imageurl". Code:

Multiple Select From Listbox And Execute A Sql Query Select Statement
Supposing i have a html form with 2 listbox and i name it status and resolution.

So it goes <select name='status[]'>... <select name='resolution[]'>

Then i have a another php page.
$status = $_POST['status'];
$resolution = $_POST['resolution'];

Now my question is, how do i do the sql query statement?

mysql_query("SELECT * from bugs where status=$status and resolution=$resolution");

The problem is, i select 3 options form the status listbox eg. new, closed, duplicate. And from the resolution, i chose fixed, invalid.

PHP + MYSQL, Populating Drop Down Box
I am looking to generate a dropdown box from MYSQL data:

db name = h2, table = Working, Column = Home.

Populating Two Combo Boxes
I have one combo box (Category) which is populated by a MySQL table using php. I am trying to use the onchange javascript to submit the value selected so that I can populate the other combo box (sub-category) based on what was previously choosen. So can anybody give me a clue on how to submit the first boxes value so that the second can see what was selected (javascript function).

Populating Dropdown List
I am wanting to know how to populate the <SELECT> dropdown menu with all existing values from the DB and also have the value associated with that id selected. This is an updating area of my admin. Any ideas? Here's the code so far. At the moment it only retrieves the value assigned to that id. PHP Code:

Populating A Drop Down Menu
How would I go about populating a Drop Down menu from data in MySQL? If you can show me wither a URL to learn this, or if you feel like telling me here, that would be great.

Dymanic Populating Checkboxes
i like to retrieve a value from DB,
accordingly i like to display the checkbox, either checked or not.

Now im able to retrieve the value from DB, i can show even show the
checkbox, but i cant make it checked????

the Code i used to display as follows,

#Require the database class
require_once('../dbinfoinc.php');

#Get an array containing the resulting
record
$query = "SELECT * FROM event";
$result = mysql_query($query);

$num=mysql_numrows($result);
mysql_close();

$i=0;
while ($i < $num) {
if(($i % 2) == 0) { $c =
'"TableDetail"' }
else { $c = '"TableDetail2"' }

$event_id=mysql_result($result,$i,"event_id");

$event_name=mysql_result($result,$i,"event_name");

$event_publish=mysql_result($result,$i,"publish");
// if(($event_publish) == 0) {
$event_publish = 'No' }
// else { $event_publish = 'YES'
}
$take_action = 'Delete'
echo "<tr>
<td Class='navText'
align='center' class=$c>$event_id</td>
<td class='navText'
class=$c>$event_name</td>
<td class='navText'
align='center' class=$c><input type='checkbox'
name='publish' if($event_publish==1){'CHECKED'}?></td>
<td class='navText'
align='center' class=$c><a
href='delete_event.php?event_id=$event_id'>$take_action</a></td></tr>";

$i++;
}
?>

Re-populating A Form If Errors
I am learning php. I have created a simple Web page with a form.
After receing the POST request, I do some error checking. In case of
errors, I would like to re-post the page, but with the current values
for each field.

I have organized my files like this:
1) Webpage with the form (all in html, form action points to a php
file)
2) php file that receives the request.

Populating Form Text Box.
I have a PHP form for entering data on-line into a mySQL
table..
Because of the intended search facility, I require one field in the form
to be completed with exact and precise item names and spelling.
In testing, I had achieved this with look-up tables.
Ideally I would like the selected item from one of five drop-down lists
to autopopulate the form text field when selected.
Alternatively at least the option of copy and paste.
Copy and paste - for some reason will not work.
Autofill worked fine on a test form with no database connection
and in standard HTML and javascript, however, once the PHP
is added to the page neither method works.
Could anyone please point me in the direction of how to get the
autofill to work?

Populating Drop Down And Table
I have a database with the following fields.
Name | Company | Date

values in each column could be repeated, or not. as in there could be several same names with the same company with different dates, or different names with same company.

How do I populate a table with this info and have drop down boxes, so that I can narrow down the search? For example:
Name | Company | Date
12 | 1 | 1929
13 | 1 | 1929
14 | 1 | 1929
12 | 2 | 1929
12 | 4 | 1929
13 | 1 | 1941
12 | 6 | 1929

So if in the drop down I select '12' under name, only those entries with 12 are shown, and then I can further sort it by selecting only '1' under Company.
Hope I'm clear, I manage to complicate things when I post them.

Populating An Auto_increament Attribute
I have just designed a BD. there are companies with auto increament which are integer. When I try to populate it after i tool value from another form and try to insert it into a table i recieve an error! I don't ask the companyID value from user as it is auto increament. Don't tell me to put NULL for that entry as it didn't work as well! Code:

Populating Table Information
I am trying to dynamically generate bgcolors with to help populate some information for a site. Here is the code:

Auto Populating A Drop Down Box
Basically I'm setting up a website which needs an populated drop down box made up from all fields in a specific column of a table in a mysql db....

Here's the code I've made up using various tutorials....

<?php
$user = "";
$host = ""
$password = ""
$dbName =  ""

/* make connection to database */
mysql_connect($host, $user, $password) OR DIE( "Unable to connect
to database");
mysql_select_db($dbName); //did you forget this line?

$sql = "SELECT model FROM usedVehicles";
$query = mysql_query($sql) or die(mysql_error() . "<br>$sql"); //use the or die(...) part ONLY IN DEVELOPMENT
?>

<form action="action" method="post">
<select name="option">

<?php
while ($row = mysql_fetch_array($result)) {
    echo "<option value="" . $row['model'] . "">" . $row['model'] . "</option>
";
}
?>
</select>
<input type="submit">
</form>

I've left out the connection details for obvious reasons... When I upload and try to test this, jus a blank drop down appears... there are definately fields in the column as I have tried the query on phpMyAdmin.

Populating Fresh Data
Im building a live web chat using php, mysql, javascript, ajax I am fine with sending and retreiving data in real time using ajax as the transfer protocol However, my problem, which is becoming a nightmare, is how to I display it !! Lets say the below is the chat screen,

<div id="messages">
Jamie - Hello
Peter = Hi
Jamie = Nice Day
Peter = I know
</div>

Lets say Jamie's next message is "What are you doing today". I can send that off to the database no problem and I can get peters live chat to pull that message in, however, how do I get it to display underneath Peter = I know

As a practice I was using

document.getelementbyid('messages').innerHTML = document.getelementbyid('messages').innerHTML + "<BR>What are you doing today"

That works, but, lets just say the chat gets to 1000 lines, for each new message those 1000 lines have to be pulled out via javascript, a new line written to it and then printed back in the div

Im not sure but I'd take a good guess at this not being optimal and will cause CPU load on the users machine

What is the best way to do this! Chat messages are deleted after 1 minute, but that doesnt matter because by then they have already been sent the the users chat. Just incase you are thinking of completely repopulating there chat with every message they have sent since the chat started..

Populating Form Item From SQL Enum Set Rev#2
$sql = "SHOW COLUMNS FROM table LIKE 'subject' ";
$qry = mysql_query($sql) or die("Query not valid: " . mysql_error());
$res = mysql_fetch_object($qry);
// This returns a row with a field 'Type' containing 'enum(...)'

$res->Type = str_replace('enum('', '', $res->Type);
$res->Type = str_replace('')', '', $res->Type);
//now the string is something like this: one','two','three

$temp = explode('','',$res->Type);
//temp is an array with as much elements as the enum

while ($temp)
print array_pop($temp);

This way is much more easier using the elements since they are elements
of an array. Cleaver isn't it?

Populating Array With Mysql Results
I'm returning a result set of one field in a table, and want to populate an array with the results. I can't for the life of me figure out a simple way to do this without using mysql_fetch_array() to cycle through the results, append that to a var, then explode that into a var and pass that. PHP Code:

Populating Dynamically Generated TextFields
If one assigns something to the value iattribute off an input /text field, that value will be displayed when the form is loaded. My problem is, when building a form dynamically, the only way I can get the text to show is by doing the above, however, as this is an update form, when the form POSTS to processUpdate.php,
both the data entered by the user AND the data in the value attribute are inserted/updated.

So, how does one dynamically build a form where the data is displayed in the text fields but value atttribute is not set. PHP Code:

Populating A Drop Down List With The Year
i'm trying to loop the years from 1960 - 2010 with the function get_year() below:
PHP Code:

Populating Form Fields With Variables
The script itself works fine: mail gets sent. If there are any missing fields the form displays again with a list of the missing fields, but I can't get the form fields of the reloaded form to be populated with teh existing values. PHP Code:

Populating DHTML Menu From MySQL
I have a client that would like to have drop down menus added to a nav
bar that is generated from MySQL. Is it possible to have a dynamically
driven DHTML menu from MySQL?

Populating Arrays From MySQL Query
I have the following code: http://pastebin.com/746601

The field 'material' in 'is_material' contains multiple values for each
record in 'is_details'. Because of this I have used
'is_material_lookup' as a reference lookup table containing the
'style_code' and 'material_code' which refer to their full details in
the respective tables.

Currently I have got the script outputting all the details and one
material then in the next block of data, repeating the details with a
different material. What I would like to achieve is having 1 block of
data with a list of all materials in that, instead of the repeat, but
sadly I can't know exactly how to do it.

Populating Dropdown With Mysql Entries
I would like to create a combobox in Flash which is populated with mysql data and programmed with php. For example: There are 3 entries in database .ie. apple, bannana, peach. Now these I want in combobox in Flash MX/flash5.

Populating Table With Text File
My web hosting service uses phpMyAdmin and at the bottom of the screen
iis an area where I can upload a text file to populate a table.

I have a table named groups with two fields:
groups_idauto-increment primary
groups_name

So how do I create my text file to populate this table?

I'm going to use MS Notepad.

If it's set to auto-increment, do I need to include the number? If
so, does it start at 0 (zero)?

Would the file look like:

0,students
1,faculty
2, staff

or just

students
faculty
staff

PHP Not Populating Form Variables Passed Through POST Or GET
I"m running:

Windows Xp SP2
Apache2
PHP5
MySQL4

Everything seems fine and dandy, and I can do scripting on server side, but for some reason I cannot get data from a form to be put into a variable.

I have a simple form in an html page which submits, and in the php script that handles the input, none of the data in the form is going into the variables.

My php.ini file does have all the GPC's in place(ie variable_order, the way they get processed).

Auto Populating Multiple Drop Downs.
im trying to create two dropdowns, i need the first one to be the category and the second one to be the subcategory. The category drop down autopopulates with the correct info from the database. and uses the table "category", the value of each drop down is represented by the "cat" field in the table (cat is basically and integer id number) and "Category" is used as what the user actually sees in the drop down (category is the actual word of the category).

Once the category is selected i would like to have the sub category auto populate with everything that has the same values as the selected category (cat)

Here is a break down of how the tables work.

Table 1 Name: "category"
Fields for Table 1: "cat" (the id number), "category" ( the actual name of the category)

Table 2 Name: "subcategory"
Fields for Table 2: "cat" (corresponds with the cat id from table 1 to pull the correct data), "subc" (the basic id of the subcategory), "subcat" the actual name of the subcategory.

so the way i see it, have a normal drop down populated by a php query. then on change, populate subcategory drop down where cat = cat and display sub category.

Populating The Drop Down And Pressing The Submit Button
1. The first is I have two drop down menus. The first is "year" and the second is "mfr". When a user selects a year from the drop down it then populates the second drop down, mfr, from the MySQL database. Theat is working fine. But the problem I am having is the "submit" button (which I have labeled as "browse"). When I click it. Nothing happens, no action tacks place. I have looked over the code and I can't figure it out. (See Code Box 1 Below).

2. Right now the "mfr" drop down is populated by the MySQL database and reads with a list like "Acr", "Alp", etc. These are abreviations. I need to set up an array to have them instead read the entire mfr name. Example: Instead of "Acr" it needs to be "Acura". Instead of "Alp" it needs to be "Alpine". I need these full names to appear in the drop down. I know I need to do something like this (See Code Box 2 Below) but I can't get my finger on it. Code: