Tracking Forums, Newsgroups, Maling Lists
Home Scripts Tutorials Tracker Forums
  Advanced Search
  HOME    TRACKER    PHP


SuperbHosting.net have generously sponsored dedicated servers to ensure a reliable and scalable dedicated hosting solution for BigResource.com.





Jump Menu Selected Value


I have a set of 2 menus where the values of the 2nd are dependent on the value selected in the first. Yes, I will (eventually) convert this to a javascript process, but for now I am content reloading the page.

I have created a jumpmenu that reloads the page with the state passed in the url (ie. community.php?state=NY). This determines the values displayed in a second (dependent) menu which then lists the counties available for that state. Code:




View Complete Forum Thread with Replies

Related Forum Messages:
Using Jump Menu
I am using a jump menu to select multiple drop down boxes in a form which are related to the choice you make in the first drop down box.  The problem is that the page reloads and everything before this tag is cleared.  I want to be able to repost the data.  I know that I need to use sessions in my code but I dont know exactly how to use them for this case. Code:

View Replies !
Jump Menu
<select name="menu1" onChange="MM_jumpMenu('parent',this,0)">
<option value="www.site1.com">google</option>
<option value="www.site2.com">metu</option>
<option value="www.site3.com">mynet</option>
<option value="www.site4.com">asa</option>
</select>

above code ; i want to get the selected value and post as named $url into a php code.
How can i call the selected value?

View Replies !
Dynamic Jump Menu
I'm looking to create a dynamic jump menu/list menu that will enable me to navigate to a particular record by selecting a value from the list, however the list is dynamically generated from the recordset. Code:

View Replies !
Header - Jump Menu Does Nothing In IE
I have a php header and php footer that are being called into my content pages. In the php header, I have a jump to menu system. The jump to menu works great in Firefox, however, the jump menu does nothing in IE. Have any of you experienced problems with jump to menues in php headers and whatnot?

View Replies !
Jump Menu Call Page
Is it possible to use a jump menu to call a page within a page? To be clearer. I can place an " include ('whatever.php'); " within my page which is fine. But i would like this include to be a variable set by the options within a jump menu. So when I select from the menu the include will swap from its default to the newly selected one. Am I making sense?

View Replies !
How To Create A Jump-menu Using Dreamweaver?
I have 50 names and links I need to show up alphabetically in a jump-menu. I'm pretty green with php and databases. Is there a quick and easy way to accomplish this? I know how to create a jump-menu using Dreamweaver, but how would I set it to read the text file.

View Replies !
Script That Is A Bit Like A Jump Menu But Without The Pull Down List.
I am looking for a script that is a bit like a jump menu but without the pull down list. I want users to be able to enter a post code (4 digits) and then be redirected to a web page based on that post code as each web page is different per code.

the jump menu provided in dreamweaver is almost there but as there are thousands of codes it is impractical to have them drop down.

View Replies !
How I Can Make A List That Acts Like A Html Jump Menu?
how I can make a list that acts like a html jump menu (for a regular list I have to use a "Go" button in order to achieve the desired behaviour and jump menus I cannot use because they work only with links to other pages).

For example,I want such a list filled with questions from my db and when I select one I want to see its possible answers; the page stays the same, only its data differs based on the selection. Or you can have a look at such a jump menu on this website, the one with options for geting to various sections of the forum.

View Replies !
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:

View Replies !
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>

View Replies !
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:

View Replies !
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.

View Replies !
Select Menu + Selected Issue
I have 2 SELECT MENUS. When the FIRST MENU is selected, the PHP code gets data for the SECOND MENU. Then the user selects from the SECOND MENU to display other data.

This works fine. However, I need the select menus to display their selected values.
At the moment they revert to their default value. How do I fix this?

Below is my code attempt...

1/ SELECT MENU 1:

PHP Code:

View Replies !
Sticky Select Menu - Selected Value
I have a drop down menu populated with customer names from a MySQL table like so. Code:

View Replies !
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:

View Replies !
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:

View Replies !
Using Variables To Control What Is Selected In An Drop-down Menu?
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>

View Replies !
Automatic Selected Option + Select Menu
When the below menu is selected from it is processed by PHP code(PART II) I want the selected option to then be displayed automatically. However, the bottom option is automatically displayed. Code:

View Replies !
Showing Selected Items In A Multiple Select Menu
I have a form that has several multiple select menus. I would like to save the multiple selections but be able to show them as selected items when the form data is updated.

View Replies !
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..

View Replies !
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?

View Replies !
Chain Menu But The Menu Should Be Multiple Select Menu/list
I am looking for codes to be able to do the same thing with multiple select menu/list.

PHP codes or javascripts codes are both fine.

1) javascript + php approach: Prefer to be javascripts codes to display the chain menu. I can use the php to get the all three levels menu data from the database, and assign these values to the javascript.

2) php codes only: Or use the pure php codes (the problem is if just using php codes then I have to submit the form to the server every time, when top level menu select changes to create new menu.).

View Replies !
Detect What Option The User Selected And Then Direct That User To The Selected Page.
Quick question regarding HTML forms and select fields. I want to be able to detect what option the user selected and then direct that user to the selected page.

<select>
<option name="1">1.html</option>
<option name="2">2.html</option>
</select>

when the user selects, 1.html, it'll direct them to http://www.domain.com/1.html same with 2.html. I wasn't sure if this was done with PHP or Javascript, if its possible on both, which is better?

View Replies !
[PEAR:QuickForm] Dynamically Change A Dropdown Menu According To Another Menu?
I've been using QuickForm for a few months now and I am now given a
new challenge:

I've got a search form with a dozen of dropdown menus, the first
dropdown menu being "Brand". If you select either Brand A, B, C, D...
Z, the second dropdown menu "Model" must be dynamically changed to
model AA, AB, AC, AD..., according to the models manufactured by the
brand selected in the first box.

I've seen that done on quite a few sites, but never found if QuickForm
had a quick & clean way of doing that.

View Replies !
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>

View Replies !
Jump Frame?
I want to make a jump frame (i think is the name), when people click in an image or video, appears a frame in the top (or left), but i want to put an id for each video and image.

There are a simple and better way, but i don´t know how to do, where we can put only the link and it redirects with the frame in php (out.php) like this one (this is an adult link, not spam, only is an example what i want): Code:

View Replies !
JUMP Out Of The Script
"mysql_connect establishes a connection for the duration of the script that access the db. Once the script has finished executing it closes the connection. The only time you need to close the connection manually is if you jump out of the script for any reason. "

http://www.webmasterworld.com/forum88/903.htm

What does he mean by jump out? I used header redirection, is this "jump out", but after header redirection, I always added exit;, does it mean that I need to close the connection.

View Replies !
Unable To Jump To Row
I have a tariff which is atm 44 rows with 15 columns of numbers. Now i am trying to pull as individual price from the tariff based on user input: Code:

View Replies !
Unable To Jump Row 0
I am getting the following error:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 2 in /home/content/i/n/n/**/html/tracker/connect.php on line 35

Here is my coding from line 34 on down. Code:

View Replies !
Site Jump During Submission
1)I'am collecting form data on my server from a HTML form.

The next steps Iam trying to figure out is:

2) I want to then go to another site (payment site)

3) then back to my site.

4) extract original data gathered from step 1

I know the simple method would be to go to pay site first then to my site. However I would like my form first! I also do not want to write the data into mysql in step 1. Not until step 4.

View Replies !
Jump From One Part Of The Script To Another
Is it possible to jump from one part of the script to another part? Meaning the script would jump from line 7 to line 20, ignoring lines 8-19.

View Replies !
Header('Location:') Jump Ignored?
I'm seeing a problem that has me flummoxed. The only thing I can
think of is that I'm violating some rule I don't know about.

I have some code that does some processing and then does a
header('Location: ...) jump to page A on success or falls through to
the jump to page B. This is the code:

if ( mysql_query( 'LOCK TABLES tableX WRITE', $link ) )
{
mysql_query( $q, $link ) ; // store the record
$ID = mysql_insert_id( $link ) ; // save the new id

if ( mysql_affected_rows($link) == 1 )
{
unlock_tables() ;
$_SESSION['ErrMsg'] = 'New ID is ' . $ID ;
header( 'Location: PageA.php' ) ;
$_SESSION['ErrMsg'] .= ' Error: ignored jump' ;
}
else // the store failed
{ error handling }
}
// second chance (for debugging) to do the right thing
header( 'Location: PageA.php' ) ;
$_SESSION['ErrMsg'] .= ' and it ignored it AGAIN!' ;

// and we shouldn't get here at all -- but we do!
header( 'Location: PageB.php ) ;


Unfortunately, as I know from the telltales I stuff into session, the
store works but the interpreter ignores both header calls that would
jump to A and finally jumps to B instead. Unless I'm more tired than
I'm aware, or more ignorant, this doesn't make sense.

View Replies !
Jump To Section Of Same Page
PHP Group,

How can I in PHP, based on a result, move the cursor to a certain location
on the page? I want the same result as Anchor tags in HTML. I want to go to
a certain section of the same page based on a return result in PHP. What
this is for is an order forms page - when a user fills out a section
incorrectly, I want the page to display at that location in the form after
the user presses a submit button.



View Replies !
Jump Out Of A Multidimensional Array?
trying to get out of an array dynamically.

$transport = array('foot', 'bike', 'car', array('test1', 'test2'));
i'm in test2( $transport[3][1] ), how do I go into $transport[3]? Is there a function capable of doing this?

View Replies !
Jump To Html Page
I am a php newbe and I am trying to figure out how to simply display an existing html page from my php script. Here is the code I use but I get the normal "output started" errors.

$sql = "INSERT INTO author SET
firstname='$name1',
lastname='$name2',
workphone='$workphone',
email='$email',
loginid='$loginid',
password='$password'";
if (@mysql_query($sql)) {
header("Location: formSubmit.htm"); <--------HERE

View Replies !
PHP/MySql Jump Script
I have also read in a affiliate forum (quote below)that putting &afsrc=1 at the end of a link will help click through, how might this also be implemented? "Recently, I put the &afsrc=1 on all my links across my largest sites. Damned if sales have not increased by 10-20% since then. (it has been about 2 weeks at about 13000 clicks a day)"

<?php
$link = mysql_connect ("localhost", "user", "pass");
mysql_select_db ("datebase");
$result = mysql_query ("SELECT link FROM id WHERE number=$locn");
$row = mysql_fetch_array($result);
header("Location: " . $row["link"]);
mysql_close ($link);
?>

View Replies !
Jump Script And Wordpress
I created a very simple jump script, and then edited a page on my blog so the links use the jump script links. I followed a template, and triple checked all the code, so I am sure that part is right. My question is, where do I put the actual .php file so the links in the jump script actually pick up the code and go where they are supposed to go?

I tried a variety of folders (public_html, themes folder, actual theme folder, etc.), but none made the script work. unless I am (quite possibly) missing something else?

View Replies !
Pagination :: Unable To Jump To Row 0
I am running a script to find the "Previous" and "Next" records, the problem is when I get to record 1 I get the error below, same with when I reach the last record in the db.
PHP Code:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 18

Here is the script running it, do I need to add to the query? or how can I use IF/ELSE?PHP Code:

<?php
$id = $_GET['id'];
// Retrieve your record according to id
// Display Record
// Links to Previous Records
$query = "SELECT id FROM jokes WHERE category = '$category' AND id < $id ORDER BY id DESC LIMIT 1" ; // For Previous
$result = mysql_query($query) ;
..........

View Replies !
Collapsing Menu - Code A Menu For A Webpage
I am trying to code a menu for a webpage, the menu is broken up to groups i.e:

Home
Group1
Â*menuitem1
Â*menuitem2
Â*menuitem3
Group2
Â*Â*menuitem1
Â*Â*menuitem2
Â*menuitem3
And so on.

What I want to achieve is that the groups are shown in a collapsed state depending on the page which is currently being displayed i.e so it starts out like this Code:

View Replies !
Jump To Letter In Paginated Results
I'm on a project that has paginated results coming from MySQL, say ten records on a page. Let's say that the results contain a song title. What would be the best method for me to write a script that could take a user to whichever page has the first song title starting with B?

Basically, the client wants a list of letters at the bottom that take the user to the page that records beginning with the selected letter begin on. Advice? Am I going to have to loop through the results and count the positions of the letters, and then calculate which page that position would be on?

View Replies !
Unable To Jump To Row 0 On MySQL Result
im making a login program, and i get this error.

"Warning: Unable to jump to row 0 on MySQL result index 2 in /var/www/html/login.php on line 16"

Code:

View Replies !
Possible To Abort A Script And Jump To The Next Block Of Code?
I was wondering if it is possible to create a button/link that would let a user stop the execution of that piece of code and then resume execution starting with the next 'block' of code. 

View Replies !
Child Menu Name Changes When Parent Menu Changes
THIS IS FOR A SEARCH RECORDS IN A DATABASE TO MATCH SELECTED CRITERIA OF A DYNAMIC OPTIONS LIST

I have a set-up whereby the <select name=" "> of the List determines which field of the datbase is searched. I've create a dynamic drop down menu whereby the Parent menu changes the values of the child menu - that works fine. Code:

View Replies !
Header Problem - Insert The Info Below And Then Jump Back To A Page.
i want to insert the info below and then jump back to a page. Here is insert and the header statement: PHP Code:

View Replies !
Warning: Mysql_result() [function.mysql-result]: Unable To Jump To Row 0
i want to grab the value of a sql entry. but the thing is, sometimes this value is <NULL>. i thought, no biggy, i'll just have an if statement:

if (is_numeric(mysql_result($query, 0))) {
//do something
} else {
//do something else using mysql_result($category_id_query, 0);
}

however, this doe not work. i get this error: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 is there a way to check to see if mysql_result() is <NULL> without throwing an error.

View Replies !
Warning: Mysql_result(): Unable To Jump To Row 0 On MySQL Result Index
i got this message:

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 6 in c:inetpubwwwrootDreamWeaverSiteswebadminwa.php on line 2296

for these lines of code:

$sql11="delete from Booking where Booking_Id='$Booking_Id'";
$result11=mysql_query($sql11);.

View Replies !
Selected Value
Trying to get the below to have the SELECTED value chosen but it is not working. What am I doing wrong?

while ( $buyerline= mysql_fetch_array($getresult, MYSQL_ASSOC))
{
print "<option value='$buyerline[id]'".($buyerline[id]=='$buyer'?' SELECTED':'').">$buyerline[buyer]</option>
";
}
print '</select>'

View Replies !
$selected
I'm trying to generate a select box through php. the problem comes when I try to dynamically add in the 'selected' attribute to one of the options. My script is basically the following: Code:

foreach($values as $key => $value) {
if ($key == $checkVar) { $selected = ' SELECTED' } else { $selected = '' }
$options .= '<option value="'.$key.'"'.$selected.'>'.$value.'</option>'
}
print '<select>'.$options.'</select>'

View Replies !
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>.

View Replies !
Php Dropdown - Selected?
i'm using this to pull information from a database and display it in a dropdown list. The problem is I am using the name code when I goto the "Update" page, but i'm not sure how to add the "selected" field, that way what ever I saved in the database will load.

Example: I goto add and it add's the "id" of 7 to the field, when I load the page again it auto's back to 1 instead of clicking in-to 7 as thats whats already there... = ) Code:

View Replies !
Know Out Of All Checkboxes How Many R Selected And Which Are They?
i have a checkbox array of some unknown size.its a dynamic array created.
i want to know out of all checkboxes how many r selected and which are they?
and then insert the data in the table based on that. i hav written the code as follows:-

<?
$no=0;
$friendname= $_POST['friendname'];
$storecount=$_POST['storeid'];
while($no<$storecount)
{
   if($friendname[$no])
   {
      $sql = "INSERT INTO groupfriends (groupid, username) VALUES ('$stid[$no]', '$friendname[$no]')";

      $result = mysql_query($sql) or die('Query

failed. ' . mysql_error());
   }
   $no=$no+1;
}
?>

i dont think this will work.it will add all the checkoxes data to the table. i want to insert only of those which r selected.

View Replies !
Remember Selected
How does a script remember selected qty when the user needs to edit an order form.

echo"<select name ='blar'>";
for($num;$num<$num2;$num++)
{
echo"<option value = '$num'";

........................... // here needs to be selected = 'selected' at the correct place??

echo">$num</option>";....

View Replies !

Copyright © 2005-08 www.BigResource.com, All rights reserved