Connecting Fields In Access Via Selected Drop Down Menu?

i am new to PHP and finding it interesting difficult at the same time. I need to connect to an access database depending on the selected menus in the drop down box and printing the selected field in a table.
Here is my code

<?php
$ch1 = 'unchecked';
$ch2 = 'unchecked';
$ch3 = 'unchecked';
$ch4 = 'unchecked';
$ch5 = 'unchecked';

[Code]...



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Selected Value In Drop Down Menu

I want the first option ("selected" option) in a drop down menu to be based on the URL a user clicks on. For example, if the user clicks on the link:

layouts.php?submit=true&category=abstract

The first and selected value in the drop down menu should be abstract
Then the rest of the menu will pull from the database.

The code I am using now (below) lists all my categories from my database, but it is not putting the selected value first. How can I get this to work?? Code:

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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:

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SQL Query Selected From A Drop Down Menu

I've got the following query which gets values $search and $furn from a form.

SELECT * FROM properties WHERE description LIKE '%$search%' AND furniture='$furn' ORDER BY id ASC LIMIT $start, $display"

This works almost fine. The only problem is that because $furn value is either 'yes' or 'no', selected from a drop down menu, sometimes when is not selected the query doesn't work. Code:

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I have a drop-down menu with the states. I don't necessarily want to have the menu dynamic with a database, but I want to be able to control what is selected using a variable. Code:

<select name=state id=state>
<option value="" selected>--</option>
<option value="AL">Alabama</option>
<option value="AK">Alaska</option>
etc...
</select>

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Show The Drop Down Menu With The Selected Options?

First, I am building a form with multiple drop down boxes. The options of the second drop down box will depend on what the user chooses on the first, options of the third drop down box depend on what they choose on the second, etc. The values of each option are inside a database which I can query for and put into a php array.

Second, when this form is completed, it should be viewable/editable by the user so it should be able to show the drop down menu with the selected options that the user chose. How would I be able to set the certain options for the drop down box to what the user chose and be able to make it dynamic again so they can make changes to what they chose.

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I am trying to get my drop down menu to use a session variable as the selected menu option. Here's my code:

echo "<option value="">Delivery Area</option>";
$selected = "";
if(isset($_SESSION['postal_area']) && $_SESSION['postal_area'] == "England") {
$selected = "selected";
} elseif(isset($_SESSION['postal_area']) && $_SESSION['postal_area'] == "Wales") {

[Code]....

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Drop Down Menu - Every A Value Of It Is Selected To Reload The Page And Submit That Value.

i have a drop down menu and i want every a value of it is selected to reload the page and submit that value. i'm using something like this: HTML Code:

<select name="menu">
<option selected value="1">GENERAL</option>
<?php
while ($cat = mysql_fetch_array($q)) {
$cid = $cat[0];
$cname = $cat[1];
echo "<option value='$cid'>$cname</option>
";
}

echo "</select>";

i tried using <select name="c" size="1" onChange="MM_jumpMenu('parent',this,0)"> but i got confused so i'm looking for an alternative way..

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Assign Selected In Drop Menu Box From 2 Tables (joint)

3 tables : one events table (each event can have differents types) one types table (each type can have differents events) and one liaison table (link between events and types tables)

I have some problems when I show screen.

<select name="type[]" size="10" multiple="multiple">
<?php
$req_type = "SELECT * FROM type ORDER BY type ASC";
$res_type = mysql_query($req_type);

[Code].....

The selected don't run and don't show correct value.

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I have a drop menu as below and when submitted it sends either no value, male or female to the database.

<select id="gender" name="gender">
<option selected="selected" value=""/> </option>
<option value="Male">Male</option>
<option value="Female">Female</option>
</select>

Now when i go back to the page i want it to show the value that was previously selected/stored in database.Basically the php code i have at the moment is to get the values and it sends to database. I already have a while loop so all i need to do is $row[gender] to get get value but how i implement it into the drop menu html code i don't know.

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I have one drop down menu and one textfield. The drop down menu will display the book's id and book's name that retrieve from the database while the textfield will automatically display the book's name based on the user selection from the drop down menu. I try to use the javascript function to do this but since the value for the option is book's id so the textfield is displayed the book's id but not the book's name. What should I do so that the textfield will display the book's name? Below are parts of the code.

<?php $sql = "SELECT * FROM store ORDER BY book_id";
$result = mysql_query($sql);
?>
<select name="book_id" id="book_id" class="font" onChange="updateText();"><option value=""></option>
<?php while($row = mysql_fetch_array($result))

[Code]....

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Heres my page I have a table with rows of ID and its status. Status is a drop down menu with 4 possible options. The default displayed is the value from database. When the user picks a value form drop down menu say "completed" I want this value to be updated fro that particular id to database.

Heres my script to create drop down:

<tdWIDTH=200HEIGHT=40align="center"rowspan="2">STATUS<formid="main_form"name="main_form"method="post"action="userZFN.php"></td>

[CODE]....

This drop down is for each row of the status column. When The user clicks submit by choosing some status value . I receive it in the same page using post.

}elseif(isset($_POST['main_select'])){
$lcSelectedValue=$_POST['main_select'];
echo$lcSelectedValue;

But when I use update I need to use something like this

"Update TABLE set COLUMN = 'progress' where ID = $id";

But How DO I remember the ID???Can I pass the ID through Post/select too?

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What I want is that the text box is only accessible if a certain option is picked from the drop down menu and i have a html form as below:

a.Did any of your staff participate in training or orientation sessions related to any aspect of social performance management, during the reporting year?

Yes
No
No, but planning in future

if not,and not planning explain why not

Now when a user selects the option No, but planning in future then the text box must be enabled otherwise the textbox must be disabled

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Not able to store multiple fields selected from drop down lists in form to variables in php using GET syntax. I am doing a search query list data in field based on the selection.

My Code;

The code is as below;

HTML Form;

<html>
<body>
<form action="search3.php" method="GET">
<Table>
<tr>

[Code].....

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i am currently trying to populate a a drop down menu that holds the fields of a database, the code i am using is a follows,

PHP Code:

<select name="existing"><option>Select One</option><?php // Retrieve all the categories and add to the pull-down menu.$query = mysqli_query("SELECT category_id, category_name FROM category ORDER BY category_name ASC", $dbc) ;$result = mysqli_query ($dbc, $query);while ($row = mysqli_fetch_array ($result)) {echo "<option value="{$row[category_id]}">{$row[category_name]}</option>
";}mysqli_close($dbc); // Close the database connection.?>

I cannot see the problem however when i do or die i get an error saying there is a parse error and unexpected '?' what ever that means?

The main problem how ever is that the menu is not populating.

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This way, I can also have an archive page listed by category. How can I write a "new entry" form with a drop-down list that uses the categories from the category table (and, should I ever add a new category, have it automatically update the drop-down list)? Code:

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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.

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Menu2
- Menu2.1

[code]...

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shown below.

When selecting one of the options, its value is passed to variable
$townsearch.

How do I change the drop down list so the option selected last time is
then the current one shown in the drop down box?

<body>

<?php $townsearch = $_get['town']; ?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="get">

<select size="1" name="town">

<option>ipswich</option>
<option>bury</option>
&nbsp;
</select>&nbsp;

<input type="submit" value="GO" />

</form>

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Code:

[Select] <?php
$result=mysql_query("SELECT*
FROMproduct
GROUPBYRim");
?>
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<?php
if ( ! isset($location) ) {
$location="Bodrum";
}
@mysql_connect ( "localhost", "root", "passwd" ) || die ("MySQL'e baglanamadim");
mysql_select_db ( "ceyhun" ) || die ("Veritabanýný seçemedim");
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........

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PHP Code:

<?PHP
$query=mysql_query("SELECT*FROMclient_masterWHEREclient_status=1");?>
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<?php
do{
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<formaction=""method="post"class="ajax_form">
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