PHP :: Populating A Select Box

Populating A Select Box From A Database
I have a subscription database that allows you to edit a subscribers data. For example, you enter a search term like "Bob" and it returns a list of everyone named Bob. You click the one you want and it goes to a subscriber detail page, where all of his data is populated to a form. You can directly edit this form and click save to overwrite it.

The only issue is the select boxes (drop downs) - say I have four options in the drop down box, red, blue, green and black. On my details page for bob I end up with five, I have those four PLUS whatever was saved for his color, so it shows up twice. If Bob is green, my select box looks like:

Green
Red
Blue
Green
Black

How can I make it so that Green only shows up once?

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 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:

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.

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:

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 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 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 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 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:

Populating City,state,country Drop Down Menu
In the registration form I have city, state, country fields. I was
wondering if there was a database available on the net which has the
list of states in each of the countries. That way when a user selects
a country I could automatically populate the state drop down menu ...

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();

Auto-populating Multiple Boxes... Php And Pg_fetch_array Problem?
There is a nice tool using jquery/php to populate multiple select boxes out
there, it must be a very stupid question but I?m trying to apply this tool to get data from postgreSQL, but I always get an empty [] value. It means that the...

Dynamically Populating Listboxes Based On Previous Selection
How do I populate the second listbox in this form based on the selection of the first listbox?

The first listbox is populated by this query:
$year_query = "SELECT distinct year FROM bench_data";
The second listbox should essentially be populated by this:
$industry_query = "SELECT distinct comp_desc FROM comp_desc, benc_data where $_POST['year']=bench_data.year and bench_data.comp_key=comp_desc.comp_key";

I know this is alot to ask, but please EXPLAIN the code. I am new and need the explanation elucidated and not just supplied code that will accomplish the goal. Code:

Select Statement - Select Random ID #
I have made up a page that pull info on 13 branches of stores onto one main page. It's basically there to display all the branches on one page... then you click on the one you're interested in and then you go to a more detailed page.

Back to the main page. On the top, there's kind of a featured branch that has a bigger picture and bigger title but it's still getting pul pulled from the same table etc etc etc.

In order for me to avoid favortism to one branch (and for other branches to start complaining) I want to ramdomized the order every time someone came to page but I have no idea how.

right now I have a super simple select statement.

Populating "Month" Dropdown Menu
I can get the current month to display in a text box by inserting
value="<? print strftime("%B"); ?>"> into the code for that
particular text box on a form but how can I populate a drop down with
names of all the months with the current one being 'default' as it
would be in the text box example?

Populating Form In Parent Window With Value From Popup Window
I have a text box in the parent window.. now i want to populate this text box with a user_id that i will get from query within a popup window.. How do i get the value from the popup window to be passed to the main window form field in php. I know how it can be done in javascript but when i tried combining javascript and php i dont get the desired result.

Populating A List Box Or A Dropdown List Using Php And Odbc
I need to populate a list box and/or a dropdown list on a form. I have all
the bits and pieces together, all bar the code which takes the result of a
query and creates a list box.

SELECT *
If I have a lot of articles, all with a unique IDs.

First I would like to search for that ID, I could do

SELECT * from ARTICLE where ID = xx

But I also want to display the 5 articles before and after that article.

SELECT * from ARTICLE where ID > xx LIMIT 0, 10
and
SELECT * from ARTICLE where ID < xx LIMIT 0, 10

(I choose a limit of 10 in case the ID is one of the first one or one of the
last one, that way I will always have at least 10 articles).

Is it possible to do the above in one single query?

Should I even bother doing that? does it make the whole operation faster to
do it that way?

Select Box
I have a problem that I'm not quite sure what algorithm will help me on. I'm
bringing in records from MySQL using PHP and need to spit this records into
a select box. Fairly simple, here's my code for the first select box:

<option selected
value="<?=$currentCatID?>"><?=$currentCategoryName?><option>
<?php
while($rowGetCat = mysql_fetch_array($resultGetCat)){
if($rowGetCat['pCatID'] != $currentCatID){
echo "<option
value="".$rowGetCat['pCatID']."">".$rowGetCat['pCatName']."</option>";
}
}
?>
</select>

Problem is - based on their choice with the first select box, I need to
query the DB and populate a second select box. So basically, I need a 2nd
select box that is populated on the fly based on their choice from the first
select box. The only solutions I have found so far use their own form, and
the form I have now has numerous other options and values that I need to
save the state of for editing.

I was thinking AJAX might help, or maybe just regular javascript, but
without having the entire page reload I'm not sure.

Sub Select
I have this line of code:

$result = mysql_query("SELECT jobnumber, projectname, description FROM timesheets WHERE description IN(SELECT distinct jobnumber, projectname FROM timesheets)",$db);

But this returns nothing. What I want it to do is display the 3 fields jobnumber, projectname and description where the jobnumber,projectname combination is unique.

Get A Value From Select Box
I am having difficulties getting the current value of a select box to pass to a function Code:

Select The Id And The Name
I have this code that when the user provide his email address he will participate in a raffle, my problem is that i want to select the id and the name but my selection is not workin Code:

Select From
Is there a way to select from multiple mysql tabels?

Value To Use To Select All
I have done the following query:

$query = sprintf ("SELECT * FROM users WHERE country LIKE '$selCountry'");
$result = mysql_query($query,$link);

I want to obtain the users that are from a certain country but when $selCountry is equal to "All" I want to obtain all of them.
I have tried the query in mysql webpage using '%' and it works but when using this character from the php file it doesn't work.

SELECT *
What would I have to put within the $sort variable to have that request display all rows starting with a number? right now I have to put 1,2,3,4,5,6,7,8,9 seperate. Anyway I can include it all in one variable that would work below?

$q = 'SELECT * '
. ' FROM `company_details` '
. ' WHERE `cname` LIKE '' . $sort . '%''
. ' ORDER BY `cname` ASC';
$r = mysql_query($q);

Populating A Select Box

What I want to do for members of my website is to allow them to click a button that will add their name to a list that populates a select box. I can do this on my own, but I was curious if I have to use a database in order for this to be possible.