PHP :: Php Dropdown

Remember Selected Value DropDown
I am attempting to have a drop down menu that displays url from 1 table called links the same table also contains a name for the url. i.e. google.com = name and http://www.google.com = url The value of just the url from links is stored in a separate table called users in a field named search after form is submitted by user. Everything Posts correctly except when go back to edit the record and submit the form the drop down box does not retain the value of what had been previously selected although it is the correct record until form is updated when whatever value is currently in the drop down list will update the record.

I know the code is doing exactly what it is told to do, because I have no functionality in it to recall the value from the field search in the users table. I don't know if it is something I would have to add to the array or option value. I can get it to work if it is all hard coded, but I am attempting not to do this. I am pretty new to php mysql and any help is greatly appreciated.

This is the code I am working with that came mostly from a tutorial. PHP Code:

Selected Value In Dropdown Dependent On Database Value
i need a way to make the selected value of a drop down dependend on a result from a database query. PHP Code:

Dropdown Menu <selected Name="...
I have a dropdown menu like this one:

<select name="country" tabindex="6" id="select_country">
<option value="">- Country List -</option>
<?php
$query = "SELECT * FROM country ORDER BY country ASC";
$result = mysql_query($query) or die('Error, query failed');
while ($row = mysql_fetch_array ($result)) {
$country = $row['country']; ?>
<option value="<?php echo $country ?>"><?php echo $country ?></option>

After I have selected one country and I hit the submit button it go back to the default value in the box, what I want is to show the selected value in box until I selected another value.

If the default value is Australia when I start this page, I then select USA and hit the submit button it reset to the default value "Australia", but I want it to show USA as long I not select anything else. I have tried to use session in selected="<?php echo $_SESSION['country']; ?>" but it didnt work.

Print All Selected Items From Dropdown??
I have a program that allows users to select a 'New Graduate' membership, and for each one selected, they get a free tshirt. I've got that working using a 'is_new_grad' function that I created. So, if they select 2, they see 2 dropdowns to select the sizes of them. It then sends an email w/ the membership info and the shirt sizes selected.

However, when the email is being sent, it's only returning the size of the LAST item selected in from the drop-downs. Example: if I sign up for four of these memberships and select four different tshirt sizes (1 Medium, 1 Lg, 1 XL, and 1 XXL), the info in the email is all XXL: Code:

How To Pass The Value Of Selected Dropdown Menu To Another File?
there is a php-myql script that list the mysql databases in the drop-down menu, use wil select one and the submit button should pass the value or variable $db_name to the test.php, this script list the databases in the drop-down menu but when i select one and then click on submit cannot pass the db_name to the test.php:

Make Selected Option Stay Selected In A Combo After Submit
I have 3 comboboxes, one for the day, other for the month and another one for the year. Every time i select a value for the three of them and press the "Submit" button, they reset to the default option. Anyway this can be solved?

Dynamic Dropdown And Hardcode Dropdown In Select Form
I have plenty of examples of dynamic dropdown choices but none of hardcoded dropdown choices. The ultimate goal is to have a job with various tasks and to track the status of those tasks for a given job. I've used one of the tutorials here to begin the process. Below is the code I have to add work on a given task. Perhaps I actually need to "Update" a job as opposed to "add".

But the problem of the moment is I can't seem to hardcode one set of my options. This is the code I have: PHP Code:

Multiple Selected="selected" In A List.
I have a database with 1 to many and the many is a list with multiple
selects in a list. When I click on a master record I have as part of
my form the select statement for the multiple choice list. I want the
list to highlight the multiple chosen values. here is part of the
code.

$dresscat = specdresscat($dresstypeid); //this is the query to get the
values in the detail table
$drcatrow = pg_fetch_array($dresscat);// the array for the detail
table.
<select name="stages[]" multiple="multiple">
<?php
$allstages = allstages(); // this is a function to query my database
to get the value for the list.
while ($allstagerows=pg_fetch_array($allstages)) {
$eachstage= $allstagerows["stageid"];
$specstage= $drcatrow["stageid"]; // this is the point I need help
if ($specstage == $eachstage){ ?>
<option value=<?php echo $eachstage;?> selected="selected"><?php echo
allstagerows["stage"]?></option>
<?php }else{ ?>
<option value=<?php echo $eachstage;?>><?php echo
$allstagerows["stage"]?></option>
<?php
}
}
?>
</select>

No Database Selected
what I've got it a simple system that includes pages the reference to which are in a database. This bit worked a treat. until i tried to put something that did some work in one of the included pages. PHP Code:

Selected Box Action
how to set the respond to action when user choose an option inside a <select></select> box? eg. when user select an option in the <select> box then php search for data inside mysql and then display the required data on the <input type=text>.

How To Give A Value To Non Selected Checkbox
I would like to get a value for a non selected checkbox in a form.

Imagine my form contains :

<input type='checkbox' name='t[0]' value=Ƈ'>
<input type='checkbox' name='t[1]' value=Ƈ'>

When I analyse $t, if first checkbox is not selected and second is, I have :
$t[0] has no value. $t[1] equals 1
and count($t) equals 1

I would like to find a way to get :
$t[0] equals 0. $t[1] equals 1
and count($t) equals 2

Resize Only Selected Images
I have an image upload script that is resizing every image that a user uploads.
What I would like to do is have a check box added to the upload form that a user can check to resize the image if they want to or leave it unchecked to just upload the image.

So if the box is checked the original image gets uploaded and a resized image is created as well If the box is not checked just the original image uploaded. I tried adding a check box but every image still was resized - checked or not - Im assiming because it was in the foreach loop so I dont know what to do. Code:

PHP And A Pre-selected Option In A Form
I've got a little text editor thing on my site, and to open a file, you input the file name and choose the folder from a drop-down list and click the Open button. The default filename is "index" and that file keeps track of all the files contained in that folder. It works okay. What's annoying is that regardless of what folder you're messing around in, the drop down list always goes back to showing the default folder name as the page is reloaded. I want it to be preselected to the folder it's in.

It seems simple enough--the folder you're in is $_GET['dir']. But I've read some HTML form how-tos, and I can only see that you can preselect using the tabindex="" or selected attribute.

Radio Button Selected
What I wish to do is have a radio button pre-selected depending on what is pulled from the database.

For example if I had 2 radio buttons:

o Yes o No

One has the value 1 and the other has the value 0

If 1 is selected from the database then Yes is pre-selected, else 0 is pre-selected. How would I do this?

Setting Selected Option Using PHP
I have a drop down list which I want to be set to what was selected when the form is submitted. As the form is processed by the same page, it returns to the preset default. How would I go about making this selection box set the option selected on the next page?

Looking For Dropdown Help
I have a dropdown in which i want it to open a soundfile when selected. So, how do I go about this. The form will get submitted, and through a series of conditional statements it will point the right selection to the right if statement. but what do I put in the if statement to get it to open a realplayer with my ram file?? Here's what I got so far.
Code: <FORM ACTION="<?=$_SERVER['PHP_SELF']?mode=changedropdown?>" METHOD="POST" NAME="sermondropdown">
<P>
<SELECT NAME="sermons" SIZE="1" onchange="this.form.submit();">
<OPTION VALUE="pleaseselectasermon">Please Select a Sermon
<OPTION VALUE="082403Daniel_1_1-8">August 24, 2003 Daniel 1:1-8
<OPTION VALUE="083103Daniel_1_1-8">August 31, 2003 Daniel 1:1-8
</SELECT></P>
</FORM>

<?
if('changedropdown' == $_GET['mode']) {
$selection = $_POST['sermons'];
if (?Daniel_1_1-8' == $selection) {

} elseif (?Daniel_1_1-8' == $selection) {

}
?>

Dropdown
I have a dynamic drop down that needs to be populated from different columns in a database. The problem I'm running into is that sometimes fields do not have a value. So I would rather those didn't print out. The code below prints out the blank fields.

I know I need to loop through and check for an empty value, but I'm drawing a blank on the how part. Code:

Dropdown Possible With PHP ?
i'm looking for a way to do the following :

1. selecting data from a MySQL database
2. putting one field into a dropdown
3. after selecting a value from the dropdown follow some more php code

step 1 is no problem, step 2 is no problem with the html select statement however this leaves me on the client side. After this i'm not able to return to the server side to execute some more php code, with the onchange attribute i can start some javascript but i want ( read need ) to use php code. Concrete question : is there a way to create a dropdown in pure php code ?

Dropdown Box
I currently have a dropdown box on a page. What I would like is that when a user selects something from the dropdown box it automatically changes the display of another box.

Value In A Dropdown
I have a piece of code where I get the player_id value from the url at the top of the page. Rather than this I want to get the value from variable named $player_one. Code:

Export MYSQL Selected Rows
Any ideas on how to export selected rows from a mysql table into a schema.sql file?

I want to use mysqldump --opt database table but I don't wanna dump the whole table only results of a select.

PHPMyAdmin does it but I need to do it in shell script code.

Get The Info Depending On The Selected Option
I have a page that is intendend to update a table field on DB i have a drop down menu wich displays all the position fields available on db and a text area where i will insert the new data.

now the problem: I'm using the query SELECT * FROM table WHERE position = $position
the $position is the choosen option from the drop down menu. then have echo "<textarea>$functions</textarea>" where it will show to me the present data available on DB after all this, i have another query UPDATE table SET functions = $functions WHERE position = $position

All is ok and working except the fact that he doesn't displays $functions
when selecting a option from the drop down menu?

1 Record In Database But 2 Records Were Selected?
i'm retrieving a record from database. below the record, i have a yes and no button, where i can let user click and then count the number of yes and no.

everything works fine, but when i start to click the yes or no button, my record ,becomes two records. i only have one record in my database, but two same record was displayed.
this is the code: PHP Code:

Selected Hyperlinks Activate PHP Script
I have a sitemap page with approximately 20 different pages listed on it, 10 of which are password protected, 10 of which are not. I want the 10 non password protected page links on my site map page to execute the following PHP command when they are clicked:

<?
session_unregister("CustomerID");
session_unregister("UserID");
session_unregister("FirstName");
session_unregister("AccessLevel");
session_destroy();
?>

How do I integrate this PHP function into the following code so that, when one of these links is clicked, the PHP script above is executed:

<li class="content_text"><a href="index.php">Home Page</a> - The home page for #*$! </li>
<li class="content_text"><a href="login.php">Login Page</a> - The login page for #*$! </li>

However, I don't want the PHP script to execute when a user clicks one of these links:

Set The Selected Text For Option Box Of A Menu..
Having major issues with this simple task, and I cant work out why its
not doing as it should/expected.

Basically, ive got a drop down box with the added bit of php:

<select name="title" id="title">
<?php
if($_SESSION['title'] != "") {
echo '<option value="'. $_SESSION['title'] .'"
selected="selected"></option>'
}
else {
echo "<option value="" selected></option>";
}
?>

<option>Mr</option>
<option>Mrs</option>
<option>Miss</option>
<option>Ms</option>
<option>Dr</option>
</select>

Basically, if the user submits a form, the title is stored in the
session, this works fine, and then when returned to the page, the value
should be filled in.

Yet, when they return to the page, there value they selected is not
shown in this box, instead, theres just a gap. The source seems
correct, and this is whats shown:

<select name="title" id="title">
<option value="Mr" selected="selected"></option>

Html Pre-selected List Box Scrolled
Is it possible to have a list box with a pre-selected value automatically scrolled to that value when the form is loaded? i.e. if the size of the list box is 3, and the 4th value is pre-selected, I want the list box to appear starting at the 4th value in the form.

Showing What Was Selected In Drop-down Menu
I'm trying to create a form where the user inputs data, i validate it with php, and if some of the input is missing or invalid then i print out errors with the form filled out with the information they entered/selected previously so they can change/add to it.

The problem I'm having is setting my drop down menus to be selected when the form is returned instead of returning their original state. The menu I have is named A ($A). Here's an example of what I'm doing that isn't working: PHP Code:

Keeping Select Option Value Selected
I'm trying to learn how to keep the value "selected" once a form is set to read the values and then f-write them (all within a loop). The part that's giving me a problem is the part which would return the value of the form as "selected". PHP Code:

Display Selected Record That Has Been Passed To URL
I have a page which can list all records from a table in my local database.

For each record that is displayed there is link next to it. When the link is
pressed another page opens up and the ID (automated primary key in my table)
of the record selected is passed to the URL.

So now the URL looks something like this:

http://localhost/Webpages/newpage.php?recordID=3

How do I display the details of the record that has been selected on this
new page?

Multiple Select Box With Selected Option ...
first of all I have three tables (tournament_game, umpire_game, umpires). Updating tournament_game umpires, each game may have from two to four umpires, and those umpires can work from one to n number of games, that's why I'm using connection table umpire_game. I'd like to get selected multiple select box of umpires in game, it could be two to four options, with all umpires listed as options example:

<select name=umpire_game[] size=10 multiple>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>

PHP Code:

How Do You Do Selected In A Multi-value List Box In A Form?
I am populating a list box from a dictionary table and then picking
muliple values to insert into a detail table. How do I get the list box
to have multiple selections highlighted when I pick one of the master
items?

How To Use Selected Data In The Listbox For My Query..
I have a multiple select listbox and I pass the data using array in php. How can I use these records to my query, or in one select statement. For an example:

I have a listbox of item codes; the user selects an item code/s to view it's inventory. The selected itemcodes will be pass on php using array and these records will be used as a query or search items on my database.

Select Pictures And Show Selected
I need to make a webpage (php) where users can select some
excercises(with picture) out of 40, and then display a page with only
the selected excercises (with their picture) that you can print.

Display The Content That Is Selected From The Dropdow
what im trying to do is, i have a drop down that displays all the files that are in the directory when the page loads. when a item is selected from the drop down list i want that file to load in the text area but i cannot get it to work, can someone tell me what im doing wrong? Code:

Calculating The Selected Prices In A Listbox
I am wanting to know how I can add the all of the selected prices in a
listbox? I need to add the selected prices altogether. The listbox
should have only the names of the products. The listbox should be like
below:

How To Get Form To Remember Selected Option?
I am now trying to make a form remember which color (a select list) they selected before pressing any of the submit buttons. I entered this code into the form:

<select name="color">
<option value="red">red</option>
<option value="blue">blue</option>
</select>

The submit is a GET type. For example, if the user selects blue and then clicks on either one of the "submit" buttons, I want the form to remember/show the blue option. Right now the select list always goes back to red, the first choice. I tried doing a <?php echo $red; ?> and <?php echo $blue; ?> for the option value, but this does not work since the select list needs a value.

Select Option Stays Selected
I have a normal drop down menu:

<select name="status" class="db_list_text">
<option value="All" selected="selected">All</option>
<option value="True">Active</option>
<option value="False">Inactive</option>
</select>
The idea for this is for the user to select either true or false and then to click submit to find data that matches either one. The coding is in the same page so the form basically just reloads the page and grabbes whatever info is requested. How can I with PHP, make it that the dropdown stays on the selected option when the page is reloaded?

List All But Echo In Checkbox Only Selected
There is a 3 table: I want to list all the product groups but to be checked in checkbox only what choose the user.

If from outside (choos.php?tt_reg_id=1) user is nr 1 then in listed checkboxes need to be checked only those checkboxes what the user nr 1 choose / but listed the all product groups. Code:

Echo Error In Selected File
Is there a way of echo'ing a reported error in a selected file, Like if i have a file exists validation can i echo the reported error in my file_exists.php file and place the echo on the page where i want it to show?

Renaming A User Selected File
I have an upload script that allows users to upload images to the server/database. My problem is that if a user uploads a file with the name of a file that already exists on the server, it overwrites the existing file. If I have a script that randomly generates 3 letters/numbers, how would I insert those in between the file name and the file extension? I'm thinking the solution would have something to do with implode, but I'm not sure how I would use it.

I'm using the $_FILES['upfile']['name'] variable.

Associate Radiobuttons With Selected Value From Dropdownbox
I have made the following code to show a dropdownbox filled with data from a mysql-database. It also shows 2 radiobuttons. What I don't know is how to associate the radiobuttons with the selected data in the dropdown.

For example, I select the person 'Bill' from the dropdownbox, then if I select one of the radiobuttons and I submit the values, I want that value to be inserted/updated in the database for the user 'Bill'. Code:

Best Way To Show Selected Option In List Box
I need a way so that we can show selected option in list box without using if condition in loop since some times we have lot of options and we have too much if conditions to check.

Selected Item To Use For Multiple Variables
I've got a list of 'items' and those 'items' have unique 'categories'.
I have 1 table for 'categories' that contains a list of category_ids and their 'category_names'. The other table is the 'item' list. And here's where I'm running into difficulty, I'd like to have the 'item' table contain the category_id from my 'categories' table as a Foreign Key and that category_id's 'category_name' in a 'category' field. I'm getting the Foreign Key, but I can't seem to grab the 'category_name' and assign it to my $category variable in the 'item' table.

I've got a page to insert items and a page to update these items. I was working with the update first. And I'm able to grab the list of categories and select the correct one, but I'm not seeing where to assign the category_name to the $category variable. Here's dropdown list: Code:

Dropdown Box And A Query
Heres my situation:

In a form I want to read the categories listed in table $inv_table and present them in a dropdown window so the viewer can choose and add to a different table. The script below works with one glitch. There is a huge white space on the page above the dropdown box. This will get me by but it isn't too purty.

<td valign="bottom">
<a name="theform">
<select name="catname" size="1">
<?
{
$result = mysql_query("SELECT DISTINCT category FROM $inv_table ORDER by category",$db);
while ($myrow = mysql_fetch_array($result))
{
echo "<option>".$myrow["category"]."</option><br>";
}
}
?>
</SELECT>
</td>

Dropdown List
I have a Dropdown list with Select with 1, 2, 3, 4, after select on of them it bust to refresh the page to what you select but it don't do it. PHP Code:

Dropdown Menu
I want to upload a feild values into a dropdownmenu. can anyone help me how to write in php code.

Dynamic Dropdown?
I am posting a code for dynamic dropdown by Leon Atkinson, it works fine but i have a doubt: I tried to add another functionality to it where i got stuck! Once the city also gets selected from the second drop down i wanted that on clicking the submit button it should go to some html page wrt to the city choice. Code:

Dynamic Dropdown Box
im looking to have two drop down boxs one with makes of car and
the second drop down box to automatically populate with all the model
of that particular make of car. there are about 50 makes and upto 50
models per make.

I just would like to know the best way to do this - javascript or use
a database? it would be a lot of code todo it in javascript wouldnt
it?

Php Dropdown - Selected?