Assigning A Dropdown Menu
how i can assign a drop down menu value to a session variable.
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[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.
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Dropdown Menu
i seem to be having some problems on finding how this ajax drop down menu i seen at Anime-Eden.com. it has a nice effect on how it appears. Now i am familar on how PLAIN drop down menu works, but not how this works.
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Use A Dropdown Menu
I have a list of contacts which are sorted by name. I was hoping to create a dropdown menu in which someone can sort the list of contacts by office, city, state, and contact type. Is this possible? (I am sure it is, but I am not sure what the concept is called so I may research how to do it).
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Advanced Dropdown Menu
I want to populate a dropdown menu with countries that are in one column of my table. I know how to do that, but I can't find how to eliminate multiple items. Some countries appear several times in different rows, but I only want each available country to be shown once in my dropdown list.
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Mysql / Dropdown Menu
what i am trying to do is to make a system with a few input pages and edit and read pages.. so right now i got the input/edit/read pages working. (yay) But now I want to have on 1 input page a dropdown menu that reads its information from the database. and that if i go to the page i can select a value out of the db and submit that inside my form and that it reminds the value if i view or edit the page. So what i got now is a little system that is able to add/remove/edit a value inside a field called "drop01" Well now i got a other input page called "input01.php" inside this script i want to have the dropdown menu that reads its values from the "drop01" field. PHP Code:
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Autoselecting A Value In A Dropdown Menu
I'm revamping my sites administration panel and I put in a dropdown menu to select the section to administrate (it was a LOT better than the list of links I had before, I'll probably update it some more though) and I want to know how I can autoselect whatever page I'm on in the dropdown menu, for example, here's the menu: Code:
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Dynamic Dropdown Menu
with help from the PHP Gurus here, I have a dynamic menu script: PHP Code: <?php $time = time() + 28800; $time_plus2 = strtotime('+ 6 days'); echo '<select name="select">' while ($time <= $time_plus2) { Â Â Â Â echo '<option value=>' . date('l, F d', $time) . '</option>' Â Â Â Â $time = strtotime(date('Y-m-d', $time) . ' + 1 day'); } echo '</select>' ?> how do I now get the days of the week to link to different pages?
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Populating Dropdown Menu From Db
Okay, I am trying to populate a drop down menu with items from a table in my db. I have a table called "teacher" which has the following info: I use the following code to grab the info from the teacher table. Code:
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Dropdown Menu Updates
I have seen this on lot of sites, where say you goto a car site, and you click the make of the car from a drop down menu, and than after u click it, it will load up the appropriate models for that make.
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2 Questions. Dropdown Menu And # Of Rows.
1. I have a table named pics and a column named name, just to let you know for use. I want to take the data from each row in the name column and insert it into a dropdown menu. I know how I could do it, but I would have to go: Code: <SELECT NAME="name" SIZE="1"> <OPTION SELECTED><?php $name[10 ?> <OPTION><?php $name[1] ?> <OPTION><?php $name[2] ?> <OPTION><?php $name[3] ?> <OPTION><?php $name[4] ?> <OPTION><?php $name[5] ?> <OPTION><?php $name[6] ?> </SELECT> but then I could make too many or too less. I want it to automatically adjust to the number of rows. 2.What is the code needed to find out how many rows are in a table?
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Unwanted URLs In URL Dropdown Menu
I'm working on a PHP script and it has a user login, and the variables (username, password, id number) are passed via the URL bar. Is it possible to somehow not allow the visited page show up in the dropdown list of URLs in the browser? The problem is that people can view the list and easily see people's usernames and passwords. Example: http://www.somesite.com?username=john&password=12345 What I only want is: http://www.somesite.com/ I know that Hotmail.com doesn't show all of the pages that have been visited, all it shows in the list is hotmail.com, so I think it is possible to do this, but I don't know how.
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Make A Dropdown-menu Filter?
I have a dropdown menu which is supposed to filter a table to show results for professionals, students or all. But I don't really know what I'm doing. This is what I have so far. Code: Initially, I was able to filter for students or professionals, but since trying to have an all option (students and professionals) it has gone to pot, and I keep getting errors. I know that my quotation marks are not right, but I can't think of a solution. I would be more than happy to see any ideas, even if they are completely different to mine!
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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.
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Trying To Create Dropdown Menu From List
I am trying to create a dynamic dropdown select menu from a directory containing other directories and files. This is to select a certain page from the list so I can edit it. I only want the files in the menu. Here's what I have so far but it doesn't populate the dropdown. Code:
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Selecting Item In Dropdown Menu
Im having problems trying to select a specific option in the drop down menu which has been populated with a list pulled from the 'category' table in the db. The option that needs to be selected is pulled from an id in a 'item' table, relating to the aformentioned category table.
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Dynamic Dropdown Menu Glitch
I wrote this code to dynamically generate an array from files on my server and put it into a drop down menu, but the code cannot be inserted on any page anywhere without it cancelling the loading of the rest of the page. Any ideas? Here is the code: PHP Code: <? $the_array = Array(); $handle = opendir('walrus/strips/.'); while (false !== ($file = readdir($handle))) { Â Â Â if ($file != "." && $file != "..") { $file = substr($file, 0, -4); $the_array[] = $file; Â Â Â } } closedir($handle); asort ($the_array); reset ($the_array);.....
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Network Bar, Dynamic, Dropdown Menu
Basically I have a "network bar" that I place on the top of all my sites, which has a dropdown menu box to interlink all my other sites. Basically I would like to really modify this idea out more. Code:
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Tree View Dropdown Menu
I have such a categories MySQL table: ID | name | parent and have such a function that displays the categories in a SELECT element as a drop-down menu: function DD_ProjectCategories($parent=0) { include("db.php"); $query = "SELECT ID,name FROM categories"; $result = mysql_query($query) or die(mysql_error()); echo '<select name="parentcategory">' echo '<option value="">Choose a parent category</option>' echo '<option value="0">Root Category</option>' while ($row = mysql_fetch_array($result)) { echo "<option value="$row[0]""; if ($parent == $row[0]) {echo " selected"; } echo ">$row[1]</option>"; } echo "</select>"; unset($query,$result,$row); }.....
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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:
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Stuck On Dropdown Menu List, Could Use Some Guidence
I have a form with some drop down list/menus. I do a check for ommissions and if found display a message to re-try. What I need is a way to show which options where chosen when the user submitted the form the first time. If someone could show me how to do the first one, I'm sure it's the same process for the second one. Here's the code:
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Duplicates In A Dropdown Menu Populated From My Database
I've been searching to figure out how not to have duplicates in a dropdown menu populated from my database. This is what I have, but how do I change it to get rid of duplicates. <?php $result = mysql_query("SELECT City FROM residential order by City"); echo "<select MULTIPLE NAME="City_code">"; while($row = mysql_fetch_array($result)){ echo "<option name="City" value="$row[City]">$row[City]</option>"; } ?> </select>
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Pagination And Dropdown Menu Used In Searching Mysql
I am working on building a database made for searching. I have a text search and a dropdown box used for searching different table collumns (lets people search for certan criteria). I have that working correctly. I also only want to show 25 results per page. I have that working correctly. My problem is that when I combine the two it breaks the code. The pagination works fine without the dropdown and vice versa. I think it is due to the variable that is being used to choose the collumn in the sql query. Code:
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Choose From A Dropdown Menu Of Encryption/decryption Cyphers
I've written a php page which allows users to type in a text string and a key, then choose from a dropdown menu of encryption/decryption cyphers, and a method (encrypt, decrypt). The whole thing works, except the mcrypt command doesn't work. here is my mcrypt command: if ($method == 0) { //Encrypt $output = mcrypt_ecb ($algorithm, $key, $input, MCRYPT_ENCRYPT); } else { //Decrypt $output = mcrypt_ecb ($algorithm, $key, $input, MCRYPT_DECRYPT); } Here's a list of what the variables may be: $algorithm = MCRYPT_3DES $key = "plain text" $input = "more plain text"
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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?
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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.).
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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:
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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:
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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:
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Assigning A Value
Hi all, I have a form that when submitted goes to the script below: <? $date_in = date("Ymd"); $db = mysql_connect('localhost', '', ''); if (!$db) { echo "Error: Could not connect to database. Please try again later."; exit; } mysql_select_db("rip"); $query = "insert into securepay(pay_id, name, address, city, postcode, telephone, instructions, ponum, optional_info, amount, merchant_id, date_in) values('$pay_id','$name','$address','$city','$postcode','$telephone','$instructions','$ponum','$opti onal_info','$amount','$merchant_id','$date_in')"; $result=mysql_query($query); if($result) header ("Location: sales_cc.php?pay_id=$pay_id"); echo "<br><a href="membership.php">Back to Members</a>"; ?> My problem is that when the data is inserted into the database and the header function is executed the $pay_id value on the sales_cc.php page always equals "0". How can i assign the correct value for $pay_id corresponding to the form entry? I don't know if i have explained this too clearly but if you reply i can answer your questions. Cheers, micmac
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Assigning A Null Value
I have a set of radio buttons labelled 1-5 and a 'not selected' option which I have coded as 6. I later use responses of 1-5 to calculate a score, but at the moment the 6 is included in that score when it should really be excluded as null. Is there a way to tell PHP that any values of ƌ' should be null, and to use Ɖ' instead?
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Assigning JS Variable To PHP
Suppose I have a variable like this JS var index =1; I need to assign this to a PHP variable like this <?php $index = echo "index"; ?> Somehow this doesnt work and the page fails to load. What wrong am I doing ?
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Assigning Variable
I'm a newbie when it comes to PHP, I've hit a brick wall trying to figure this out. I want to look up "user_id" from a table and assign it to the variable $user_id. I keep getting an undefined index notice and I need it to be defined. $query = "SELECT user_id FROM users WHERE last_name = '{$last_name}'"; if(!($result = @mysql_query($query, $connection))) showerror(); $row = mysql_fetch_row($result); $user_id = $row[user_id];
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Assigning A $_SESSION
I need to be able to pass a very large array between PHP pages. $_SESSION['holder_array'] = array(); and then assign to the $_SESSION['holder_array'] as though it were just created as $holder_array = array();
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Assigning A Variable
I'm trying to multiply $total before I write to the database. $total is always the number of tickets times 25 ($25 per ticket) plus any donations this is what I have: PHP Code: <?php if ($submit) { Â Â Â Â Â Â Â Â {$total = $tix*25+$donation} Â Â Â Â { Â Â Â Â Â Â Â Â Â Â Â $sql = "INSERT INTO vday (name,address,city,state,zip,email,vp,day,tix,donation,total,ccname,ccnum,ccmonth,ccyear,cctype,notes,entered) VALUES ('$name','$address','$city','$state','$zip','$email','$vp','$day','$tix','$donation','$total','$ccname','$ccnum','$ccmonth','$ccyear','$cctype','$notes', Ɔ')"; Â Â Â Â Â Â Â Â $result = mysql_query($sql);Â Â Â Â Â Â Â Â } ?>
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Assigning Places
Im trying to figure out how to designate places to teams and how it should be coded. For example, I have 5 teams. After a tournament each teams results are input one at a time. How should I code the places (1st, 2nd, 3rd, etc.) Do I need to add a field to the results table for places? And if so, do all of the results get input and then click a button to designate each teams place or have it check the result getting added to the database against whats already there and then designate the appropriate place.
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Assigning Values To Arrays
I am performing a MySQL SELECT (which returns multiple rows) to $result and then extracting the results with mysql_fetch_array($result). I then want to build a number of arrays using the fields of each row. Here is the relevant code: <snip> $result = $i = 1; while ($row = mysql_fetch_array($result)) { extract($row); /* break the row into its fields */ $array0[$i] = $row[0]; /* 1st Field of Row */ $array1[$i] = $row[1]; /* 2nd Field of Row */ $array2[$i] = $row[2]; /* 3rd Field of Row */ $i++; } </snip>
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Assigning Values Within Href
I'm having a hard time finding information (probably due to how I'm phrasing my searches) on how to assign a variable to an href. The big picture is I have a bunch of photos I'm trying to put on my website. I have an equally large bunch of thumbnails with links, all to the same page (pictures.php) that has a snippet of php (seen below). In my links I am attempting to assign a value to a variable that corresponds with a picture. Code:
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Assigning Function To Variable
I've been trying to pass a function name to a variable to store it and run it later.. here's a snippet: $this->timer(1,"ping",$this->action->ping(),REPEAT); Not putting "$this->action->ping()" in quotes runs it, however putting it in quotes results in the error: Code:
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Assigning Perl's Output
I have a perl script that retrieves country name based on an ip. I need to pass it on to my php code as a variable. I tried php's Perl extension but that won't work. The "virtual" function almost does its job, however it does not work with ob_start that looks very simple: I only need to assign perl's output to php variable.
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Assigning Id Value From Database Table
I have a products table and an uploads table. uploads stores image and product stores product info. okay heres the prob. Within my form im posting all the neccessary details but i also need to pass the upload_id from the uploads table to be stored as a value in the products table. Currently my script seems to be logging the same image id and ultimately pulls the wrong image. Ill include the code: ....
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Assigning Unique Names
I'm creating an upload page where I use dreamweaver to create a multi form to upload multiple Images. The page I'm working on has 6 images. I currently have it so when the user hits submit, it uploads the image into my ftp's images folder and renames it to 1_image1.jpg. I need to have the images in a named scheme such as 1_image1.jpg, 1_image2.jpg, 1_image3.jpg, 1_image4.jpg etc. I was wondering if there was a way to have dreamweaver name it, (between the form tags), so when it arrives in my ftp folder, it's already named, and I wouldn't have to code php to change the name.
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Assigning Values To Variables
I was attemping to make an array of configuration variables for Smarty when I ran into this little problem. Here's some example code first of all: <?php class Testclass { var $testvar = 'something'; } $test = new Testclass; echo "before foreach: $test->testvar "; # echos 'something' $arr = array('testvar', 'something elseeeee'); foreach ($arr as $key => $value) { $test->$key = $value; # set $test->[variable] to $value } echo "after foreach: $test->testvar "; #echos 'something' Problem. This outputs "something" in both places, while I expected it to echo "something elseeee" after the foreach is done. Setting $test->testvar manually and outputting it workes fine, but breaks in the foreach. To me this seems weird, but might just be how PHP workes. If that's the case, how do I work around this?
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