PHP :: How Do I Display Radio Button Results?

Passing Radio Button Values To A Submit Button (gif)
How do i pass the value of the radio button to be put of the action
that includes the value as part of the URL being submitted - "/
picked.php?pick_num=". Any help with be appreciated. Thanks

<form method="post" action="/picked.php?pick_num=".<php? $value ?
<input type="hidden" name="cmd" value="0">

<table width="100%">
<td align="center">
<INPUT TYPE='RADIO' NAME='pick_1' VALUE=Ƈ'>Day;
<INPUT TYPE='RADIO' NAME='pick_2' VALUE=ƈ'>Week;
</td>
</tr>
<tr>
<td align="center"><input type="img src="/images/pick_now.gif"
BORDER="0" name="send"></td>
</tr>
</table>

Radio Button
What i want to do , is when someone click a radio button , a text will load up at the bottom at the page , without the whole page be reloaded, i ave 0 idea how do it , just the readfile maybe...

Radio Button
I have a radio button:

<input type = "radio" name = "id" value = "">

The info that needs to be put into value field is $id. I know that I can not put $id there. If I put 7 there, it works with no problem. I am getting the $id value from a database. Any ideas on how to handle this.

Radio Button Problem
from one radio button on Cilck is it possible to get 3 values
as the attached image having folder jpg's but i want to use radio button in place of that so that i can use in my PHP file.

Passing Var For Radio Button
I passed as variable (temp) from one php page to a php preview page - radio buttons "hot" and "cold".

Now, I want to pass it back to the first page so the radio button can be checked when someone wants to go back to change it.

The plan is to check if it's page 2 value is "hot" or "cold".
If hot, echo an input line in the form going back to the first page. If cold do the same but with a diff name.

Then on the first page check if "hotchecked" is true. If true then php puts "checked" in the input of the radio button form making it checked as before the preview. If "coldchecked" the same would happen making cold button checked.

This code below crashes when I try to get the value for comparison...

-------------------------------------
<?php if ($HTTP_POST_VARS["temp"]=="hot") {
echo "<input type="hidden" name="hotchecked" value="sss">"; }?>

<?php if ($HTTP_POST_VARS["temp"]=="cold") {
echo "<input type="hidden" name="coldchecked" value="sss">"; }?>
---------------------------------------

Radio Button Output
i have a little form with some radio buttons rate 1 to 5 kinda thing. my script mails me when submitted but it is blank even though i selected the buttons. i have done this with regular text fields with no problem but i think the out of the radio buttons needs to be different Code:

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?

Validate Button Radio In Form
I use php and javascript on a form.
My validate script doesn't work with radio button. What's wrong? I
want to be sure that one of the button is press. M or F

I get on my first page:

Setting Radio Button State
I am trying to set the state of a radio button. I do not see what I am
doing wrong. Sorry, I am new at this.. I need another set of eyes to
look at this snip of code. I am trying to set the radio button with this
link of code:

echo 'SCRIPT language=JavaScript
setCheckedValue("'.$_SESSION['abbr_letter'].'");</SCRIPT>' //?

<snip of code>

<?php
session_start();
if(!isset($_SESSION['abbr_letter'])) {
$_SESSION['abbr_letter'] = 'A'
}
require_once 'includes/config.inc.php'
require_once 'includes/header.inc.php'

if (xdebug_is_enabled())
echo "enable";
else
xdebug_enable();

//
// Start a Form
?>
<HTML>
<HEAD>
<link href="quotesCss.css" rel="stylesheet" type="text/css">
<SCRIPT TYPE="text/javascript">

Radio Button/Checkbox Value Passing
I have a simple 3 page registration form. One form, one "data
validation" script and one "insert into database" script. The customer
bounces back and forth from the form to the verification script until
the data is all correct. (When I say bounces, I mean logically. All the
customer ever sees is the registration form, plus or minus error
messages.)

The problem I am having, is that if verification fails, the customer is
redirected to the form. I am able to pass the input values from the
"data validation" script back to the registration form using this
method:
<input id="Name" name="Name" type="text" value="<?php echo $Name ?>>

It works well, but I don't know how to pass the variable of a radio
button or check box back. I remember in javascript and other similar
languages you could do something like:
checkbox.value=checked

Is there a way to do this in PHP? What I am attempting to do, is make
is so that if the customer chooses YES on a radio button and the form
is submitted,; ifthey are sent back to the form to correct an error,
the radio button YES is still checked. I am not just trying to pass the
variable, but to also make it select the right option when it goes
back.

(Or is this just more of an HTML question, at which point I will ask
elsewhere)

Radio Button Not Showing Preselected In FF
I havea a PHP form that is working as expected, when there is an error the user is prompted to fix it before they can move on. I have a series of radio buttons generated from an array. I use a quick if check to see if the array value matches the post value so the radio button can be turned on or not.

In the rendered HTML you can see the value "checked" in the radio button field, and the radio button displays checked in IE, but not in FF. Any thoughts? PHP Code:

Multiple Radio Button Groups - Please Help! :)
I'm having a problem with multiple radio button groups within a while
loop. I have the page made so it pulls from a database the row values
- one of which is a Y or N value. I have a radio button group for the
Y and N value for each row of the database that gets pulled but only
the last row will populate the value retrieved from the db. I know
that I have to create a unique name for the radio buttons for each row

Radio Button To Post Variable ?
how do i post the below code of a radio button as the name is a variable called $button_name?

<td align="center"><input type="radio" name='.$button_name.' value = '.$orderId.' unchecked> Selected</td>

The line below doesnt seem to be takin the value to the other page, iv tried " " around the variable but dont work? Code:

Radio Button Values Won't Post To SQL DB
I have a form with radio buttons. I typically don't use them and thought I would take the time to figure them out this time. The values won't post to the database. Please let me know what I'm doing wrong here. The joint field is a yes/no question with 2 radio buttons: Code:

Default Selection Radio Button
I have a table where users input a date, time, and quantity for a reservation. On my page ... because I have a junk calendar and coding, every time you click a different month on the calendar or click the "book now" button without entering all the necessary data, the page refreshes and erases all the data.

One significant problem is that the times display as radio buttons. Some tours have one time and some have multiple times. On the page, if there is only one time, then to make it easier for users to book tours, I inputted CHECKED into the html code. But unfortunately if the tour has multiple times, then CHECKED doesn't work for the first radio button, it defaults the last radio button because of the php loop. Code:

Radio Button Group Value Caption
I have 2 radio buttons in a group named Gender one being male and one being female, now how would i get the value from the user's choice out of the two radio buttons and assign it to an $Gender variable..?

Passing Radio Button Form Data
I am running php as a sapi module on winXP with Apache. I am passing the data from a radio button from one page to another to be processed. For some reason it seems to be passing no data - for example in the 1st page I would have 3 radio buttons in a loop:
<input type='radio' name='radio' value="<?php echo $val ?>">

I would then attempt to read the data in page 2 : <?php echo $radio ?> but this comes up empty. I tried using "get" in my form, and although the name/value pair appears in the URL string, it does not pass itself to the $radio variable in the next page. What the??? Does anyone know of any reason why php would NOT allow a radio button (or any other form object) it's data to pass on. I am not using sessions.

Radio Button - Reading Pre Selected Values From Db
I am able to make a selection of information when entering a record, with
radio buttons giving the option of yes, or no.

Which stores their answer as yes, or no in the database. I then have an
update record page that reads information in the database, and presently it
will not retrieve the previously selected option from the database. I have
tried all sorts of if else statements, and can't get anything to work. I
want the previously selected option to be selected when they are entering
the update page. Thus, they have the option to change it, but want their
previous setting selected so they are aware of their currently selected
option is.

<td width="50%"> <input type="radio" name="ud_digdocs" value="Yes">Yes
<input type="radio" name="ud_digdocs" value="No" checked>No

Mapping Radio Button Set To Multiple Fields
I've been staring at a form for about a day now and hope a fresh set of eyes can shed some light on it. It has a couple of issues, but here's the first:

I'm trying to map a set of four radio buttons to eight fields in a MySQL database. If you select Radio Option 1, for example, that value should be passed into SQLField1, the now() value passed into SQLField2, and SQLField3 through SQLField8 would be passed no value. Code:

POST Values Of Radio Button And Checkboxes
i am looking to get the values of my radio buttons and checkboxes and have the values added to my $msg variable and sent in an e-mail. this is my code:

PHP Conditional Setting Of Radio Button Default
I'm working on an edit inventory form, and I'm pulling a field from my database that's going to be a 'Y' or 'N'. Based upon this, I want to default the 'customizable' radio button to either checked="Y" or checked="N". Code:

Create Radio Button Group WHERE Other Fields Can Disappear?
I'm trying to create radio buttons on my form where if one radio button is selected, then other fields disappear. Here is classic example:

Do you have a passport?

o NO

o YES, if so Date issued: [date pull down menu]

See what i mean? If the user clicks NO, then the "yes" selection as well as the Date pull down menu will turn grey so that the user cannot select it anymore.
Is it possible doing this without reloading the page?

How To Perform IF Stement Depending On Value Of Form Radio Button
I have a form with a radio button and depending on the value of it's 'yes' or 'no' I want to perform 2 differnet SQL queries. Code:

Select From Radio Button + Enter Username = Open User Profile
I have 4 radio buttons. each button to determine which page to go to. But
before page open user enters username in a text box. So when submitted, Page
opens relevant to that user displaying selected page.

Page 1: <radio button> >>>>>>>>>> this page to go to page1.php (this is a
table of user records 1)
Page 2: <radio button> >>>>>>>>>> this page to go to page2.php (this is a
table of user records 2)
Page 3 :<radio button> >>>>>>>>>> this page to go to page3.php (this is a
table of user records) 3
Page 4: <radio button> >>>>>>>>>> this page to go to page4.php (this is a
table of user records 4)

Enter Username: [text box] >>>>>>>> any user with their username

[submit]

Result should be :

If User FAB5 selects radio button 1 >>>>>>>>>. to display FAB5's page1.php

Setting Radio Buttons Via A Tables Results.
I'm generating a dynamic table based on a SQL call to a database. For each row returned I need four radio buttons in a group which somebody can make a decision on. The above is no problem but what I want to do is at the same time set the radio buttons to the correct figure based on the data returned. PHP Code:

Form / Results / Navigator's Back Button
I have a research form script, a result list script and an item page. Just
classic!

form page => result list page => item page

When I'm at the item page and I click on the Navigator's back button I have
the following message:

Avertissement : expiration du délai pour cette page

La page requise a été créée en utilisant les informations fournies dans un
formulaire. Cette page n'est plus disponible. Par sécurité, Internet
Explorer ne renvoie pas automatiquement les informations vous concernant.

Pour envoyer vos informations à nouveau et afficher cette page Web, cliquez
sur le bouton Actualiser .

The normal use in fact .

Is there any way to get to the list again?

Back Button Doesnt Display Previous Page
Here is an example of my code. PHP Code:

Display Files Inside A Directory? With Download Button.
My first version of simple code, display files and directories but when i tried only to display pdf files and subdirectories inside the given path, with pagination and downloadable option, everything went wrong. Code:

Display Results
I have created a form where the number of rows are displayed based on a value from a select box. Everything works excellent this far. however I need to take it to the next step where I display the information back into the form based on values in a mysql table. Code:

Categorical Results Display
I want to take these results and display them by category. I.e. each of these results would fall into different categories like "insurance", they are maked in the database simply with a "Y" for each category that they would appear in. I could make a page like this for each of the 20 categories, but I am sure there is a better way.
<?
if(!IsSet($start))
$start = 0;
$result = mysql_query("SELECT site_name, site_address, site_image, publish FROM sites WHERE publish = 'Y' ORDER BY site_name ASC LIMIT $start,144") or die(mysql_error());
?>
<table width="100%" border="0" cellspacing="4" cellpadding="4" align="center">
<tr>
<?
while($row = mysql_fetch_object($result))
{
echo '<td width="25%" align="center" valign="top"><a href= "',$row->site_address,'" target="_blank"><img src="samples/',$row->site_image,'"width="100" height="75" border="0"></a><br><a href= "',$row->site_address,'" target="_blank">',$row->site_name,'</a></td>'
if(++$i % 4 == 0)
echo '</tr><tr>'
}
?>
</table>

How Do I Display The Results Of A Test?
i just started learning php and have a quick question. this is my code:

$a=4;
$a==7;

what i want to know is how to display the results (either true or false). i tried print($a); but it only outputs the original variable value (in this case 4).

Display Results Vertically
I am creating a catalogue-like application and I want to display the items vertically instead of horizontally. Being somewhat new to PHP, the only thing I can think of are seperate queries for each field. Is there an easier way? Here's an example of how I want it to look.

Category 1
field1: result
field2: result
field3: result

Category 2
field4: result
field5: result

and so on and so forth....

Display The Number Of Results
How can I set the classical search engine message? Found ### results. How can I count the returned results and display them in the results page? so now has the Found 5 results or any number that the result contains.

Display Results According To Distance
I have a php/mysql driven website. On my search bar I have the option of searching by zip code. I have the function to find the radius of the surrounding zip codes. But I want to display the result along with my information from the other table.

An example would be:

I want to search for ads around my zip code for a range of 20 miles. The class I have will find all the zips and get their distances plus will sort them by closest to farthest. My problem is I want to display the records according to the distance provided. I heard of doing a sort with php but I am not sure which route to go.

I hope someone has found out how to do it, because this is the only thing left to do in order to launch the site.

Display Query Results In A Table
$sql = "SELECT name FROM employee_tb;
$result_query = @mysql_query ($sql);// run the query
while ($row = mysql_fetch_array($result_query,MYSQL_NUM)){
$name = $row[0];}
// assuming i have 8 results

how can i display the results from this query in a table with 3 rows each column?

Search & Display PEAR DB Results
I'm only a novice in PHP & PEAR DB, and currently stuck in an attempt to create a "search & display query PEAR DB results" page. Here's my pseudocode:

1) Display page with a searchbox and a search button.

2) Upon hitting the search button, run sql query ... Select all fields from TableA and selected fields from TableB where field1 like "searchbox" order by field1

3) If searchbox is blank
-> prompt user to enter search value
Elseif search result NOT found
-> alert user >> Record NOT found
Else
-> display results in a table (with alternating colours for each row)

4) Able to paginate the results in something like display 10, 20, 30, etc records on the same page .... the amount of records to be displayed can be manually selected by the user (via a dropdown box) ... with Previous Page, Next Page links as well.

>> How would I do that entirely in code from scratch? I've found other tutorials on doing it but it is tailored for MySQL ... but I'm using PEAR DB... the syntax is different.

>> Since I am querying two tables based on the same search value, how am i able to concatenate both the tables search results and display it inline in the display results page?

Display Search Results In Pages
im doing a query on a table that searches articles from the database, and i was wondering how i could display the results as groups of ten, so the next ten search matches are displayed when the user presses a page number.

3 Comlumn Display Results With Pagination
I have the following code that works perfectly. I would like to add an easy pagination feature that limits the number of items, counts the number of records and makes links and pages for the next page. Code:

Display Number Of DB Results Beside A Link?
For an example, if you look at www.menupages.ie they give you a preview of how many results are found for each link.

ie:

Ballsbridge Restaurants (52)
Blackrock Restaurants (17)
Clondalkin Restaurants (18)
Clontarf Restaurants (17)

How is this achieved. I take it that it must be a select count on the DB for each row but that seems pointless, is there an easy way to achieve this?

Display Results Of Survey In A Table.
Is it possible to have them fill in a survey with about 10 form fields, then display
the results of everyone's survey in a table that can be sorted by different criteria
so new users to the site can instantly see which one of our services is better for them.

Display Only Mysql_num_rows Results #-# (or Similar)
I have this mysql database. It works great, the stuff outputs how I want. But I'm making a news website and I need to change the articles each month.

I'm putting all the articles into one database, then into each page as tables (ie: news, opinion, sports, etc.), then the articles each have their own row. On the homepage here it outputs all 5 articles from the news table, but I only want it to output result numbers (blah) through (whatever). Code:

Display Results From Multiple Tables
I have two tables, one table (article_cat) has four fields ("id", "title", "feature1", and "feature2") and the other table has six fields (articles) has ("id", "cat_id", "title", "content", "feature1", and "feature2").

I am looking to display the "title" from the "article_cat" table that has "feature1" with a value of 1. And I want to display beneath that all the "title" results from the "articles" table where "feature1" has a value of 1. I have tried this a thousand ways with no luck. Code:

Limiting Results, And Display Page Numbers
How do I make it so only 50 items display per page. And if there's more than 50, it displays a link "page 2", "page 3" and so on? I know how to limit results with MySQL but how do I generate pages for the next 50 within the PHP code?

Select And Display Multiple Queries / Results
Howdy all, i've been looking through various sites / tutorials and through this php book I have here to try figure out whats causing this problem but haven't had any luck yet.

The problem is that its displaying the Designation as the link url instead of the jcode. Hope i explained this well. Above this bit of code has calls the individual Catagorys, if theres data in that catagory it will display it with whats below. PHP Code:

Display Results From Query In HTML-table
I'm working with PHP3 and Oracle 8. Can anyone tell me how to display the results from a query into my HTML-table?

Checkboxes In A Form To Determine Whether To Display In Results
I am making an effort to put checkboxes in my form that will determine whether to display that column or not. This is my first time doing this with PHP and I am hoping you gurus can assist. I have gotten some extremely great assistance in the past, so I am looking forward to any recommendations.

currently my form is as follows:

Display Results From A Database On Multiple Pages
I am new to PHP and Mysql but am really determined to learn these techniques. I have some script that reads and displays results from the databse onto a page. However, I really wish to limit the number of results that are displayed to 10 per page and then have the next/privious links under the results. With these results, the results under the email field should be hyperlinked so that the person viewing the results should double click on the email link and send the person an email. How can i do that? I have to learn this technique.

The code is:-

Getting Results To Display In Two Columns Of A HTML Table
How can i get the following script to display in two columns of a HTML table? At the moment i have it displaying in one column. If there are 5 results from the DB then the left HTML table column will display the 1st, 3rd and 5th result. The right HTML table column will display the 2nd and 4th results. Code:

How Do I Display Radio Button Results?

I have a form which uses several radio buttons to enter data. It works fine as far as entering goes, but I can't figure out how to display which button is checked when I view the record. I could do a if-then routine but it seems like there should be an easier way.

I have a form which uses several radio buttons to enter data. It works fine
as far as entering goes, but I can't figure out how to display which button
is checked when I view the record. I could do a if-then routine but it
seems like there should be an easier way.