Dropdown List - Insert Into Mysql
I simply want to add a dropdown list where, when an option is selected, its value will be inserted into a dbase table. More specifically, asume there is a page where the user sees his order.
He will be able to select the order status (pending or delivered). So, for example, if his ordered is deleivered, by selecting "delivered" option, "delivered" will be inserted into the 'Status' column. Code:
View Complete Forum Thread with Replies
Related Forum Messages:
Mysql Dropdown List
Im trying to automatically fill a dropdown list with data from MYSQL table. Sizes Table -------- ProductId | size1 | size2 | size3 | size4 | size4 | size6 | * the 'size' rows hold a numeric value for the quantity I have of each I was hoping to only list the sizes that have a greater number than 0. I have used an auto fill select code on another project but I'm just not sure how to fill in the blanks. Code:
View Replies !
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:
View Replies !
Dropdown List Box
I am able to populate data in a dropdown list box using php and mysql. However, I a wondering if i can add inside some icons such as close or More records, where the drop down acts to the specified request, ie., closing the drop down or fetching some more data from backend.
View Replies !
Populating Dropdown List
I am wanting to know how to populate the <SELECT> dropdown menu with all existing values from the DB and also have the value associated with that id selected. This is an updating area of my admin. Any ideas? Here's the code so far. At the moment it only retrieves the value assigned to that id. PHP Code:
View Replies !
Smarter Way Of Doing Dropdown List?
I've got a dropdown list that is acting as a filter on a table, so the user can select a value and the page is reloaded only showing the matching values from the database. Now when the page reloads the dropdown was set to 'all' as it was first in the list, even if I'd selected a different value, this was confusing for users, so I added an initial <option> field which pulled the correct value from $_GET (code below) - however this is a little odd as it means the value appears twice! Does anyone have a genius solution to this? I've seen a <select selected="yes"> sort of thing around, but I don't know how I can tell the php to add this value dynamically? <select size='1' name='occupation'> <option>".$_GET['occupation']."</option> <option value='All'>All</option> <option value='Student'>Students</a></option> <option value='Professional'>Professional</option> <option value='Hobbyist'>Hobbyists</option> </select>
View Replies !
Check The Dropdown List
i have a drop down list in my php code. i need to check the drop down list is set or not. for that what is the code. i heard that there is no php code for that. javascript is needed. my code is here. PHP Code: <?php $database="sms";                    /* DB name       */ $host="localhost"; $user="root";                        /* Set DB Username */ $pass="";                            /* Set DB password*/ mysql_pconnect($host,$user,$pass);  /* DB connect....*/ mysql_connect ($host,$user,$pass); mysql_select_db($database) or die( "Unable to select database"); /*DB Select*/ function orgresult() .......
View Replies !
Add Variables To Dropdown List
I have a list of variables: $variable[0] $variable[1] $variable[2] The amount of variables changes so i've used $total = count($variable) to find the amount there are. How would i tell php to create a dropdown menu with the amount of options=$total and then fill the options with $variable[0] etc. I know i can use a loop to increase the value of say $i by one each time to that i can use $variable[$i] each time, i'm just stuck with adding the whole thing to a dropdown menu with the right number of option...
View Replies !
Autopopulate Dropdown List From Database
An events-registration page with a form containing a dropdown list for upcoming events, the events will be pulled from a table which is updated by...? what i've gathered so far is that this can be done with PHP. I am a little unclear on the concept of how this is done. How does the updated information get input into the table? 1) can this be done (updating database) through a web interface by the client, or is it something i would have to do locally and then upload. If done by the client, can he delete/edit existing information--assuming he is not familiar with code. 2) what other things aside from html(good), sql(so-so), and php(newb) will i have to know? i've seen a lot of good tutorials here and will have access to the necessary software. I just want to know if there is anything else i should know before i dive in head first without testing the PHP water, which i assume is frigid.
View Replies !
Dropdown List Of Files In Folder
I am preparing a form that includes a drop-down list consisting of the names of files in a certain folder on the server. I only need to trap the file name (jpg), not the path and I'm not trying to upload the files or anything - simply allow the user to select one of a number of different image files (they're actually location maps).
View Replies !
Dropdown List - Trying To Get Default Populated From Db.
Nooby question but when I try and get this to work (it should identify which rows in the database have GK, MID, DEF, FWD against them and then put SELECTED into the option value to give a default of what the entry is currently) it just adds SELECTED to all the results as if $row["Position"] is returning true against all four conditions? What am I doing wrong? CODE:..
View Replies !
Cascade Dropdown Or Conditional List
I would like to create a conditional cascade form. based on the selection, it can display or not display an input form for the user to enter an string of text. then at the bottom of the page (or when a user submit) connect all the sections including the input field all together (it's like a story maker?)
View Replies !
Microsoft Access Dropdown List
l am building a new site using PHP which has a lot of pages pulling from an Access Database. I would like to have a dropdown list that pulls exhibitions from the database and then when the user selects one it shows the rest of the details from DB just below the dropdown list, hope that makes sense. Code:
View Replies !
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:
View Replies !
Dropdown List Selected By Default?
Does any one know how to have an item in the dropdown list selected by default? $city_field = HTML_QuickForm::createElement('autocomplete', 'city', 'City', $cities); $cities is an array of city names and I would like, say 'Los Angeles' selected by default.
View Replies !
Insert Date To Database Using Dropdown Boxes
I be able to insert the date into my database if i have 3 dropdown boxes wherein it contain the month,day & year. i know the basic of inserting data into my database but with this 3 dropdown box i don't have any idea. after submitting, the data selected on these 3 boxes shall be in one field in my database with fieldname Date.
View Replies !
Auto Fill Dropdown List Error
Im using the code below to autofill a dropdown list from my MYSQL DB. The only problem is it is leaving out the first entry (alphabetically) for each $vehicle_make ? Does anyone know why this may be hapenning? Code:
View Replies !
Stuck On Dropdown Menu List, Could Use Some Guidence
I have a form with some drop down list/menus. I do a check for ommissions and if found display a message to re-try. What I need is a way to show which options where chosen when the user submitted the form the first time. If someone could show me how to do the first one, I'm sure it's the same process for the second one. Here's the code:
View Replies !
Auto Fill Date Dropdown List
How would I modify this code to display the years in the dropdown list like so: current year 2006 2005 back 50 years <select name="year" id="year"> <?PHP for($i=date("Y"); $i<=date("Y")+2; $i++) if($year == $i) echo "<option value='$i' selected>$i</option>"; else echo "<option value='$i'>$i</option>"; ?> </select>
View Replies !
Creates A Dropdown List Of Information Belonging To That User.
I have a page that uses php to look up specific fields in a specific table of a specific database based on the username. Using this, it creates a dropdown list of information belonging to that user. The fields in question each have 13 pieces of data. Based on what they click in the dropdown box, I need that specific data to pass onto hidden fields on the page. Code:
View Replies !
Pick From A Drop Down List That Automatically Gives Options In A Second Dropdown Box
does anyone know how to create a form that allows the user to firstly pick from a drop down list that automatically gives options in a second dropdown box. eg. 'category1' is selected from dropdown box 1, this then populates dropdown box 2 with the options 'category1.a' or 'category1.b' or if 'category2' is selected from dropdown box 1 then dropdown box 2 gives the options of 'category2.a' or 'category1.b' i f any one knows how I should code this please help, if not point me in the direction of a tutorial that can.
View Replies !
Dropdown List Which Selects A Post A Title (to Edit)
I want to have a dropdown list which selects a post a title (to edit), then when you click submit I want to use POST to submit the data for that post (post title, username, subject, content) so that the edit pages form fields will already be filled with the posts data.
View Replies !
Insert Into One List Box
I have two tables data. After querying it I have two results. I should insert into one list box where data from table A i should color it in font black and another in red and disable it(from table B).
View Replies !
Insert From An Drop-down List
i have a table in wich i've inserted city's name.then i have an registration form in wich i want to display an dropdown list with the cities from the table, and if one city is selected to be automaticly added in the second table. how can i do that?
View Replies !
How Do You Insert A Value From Drop Down List To SQL Table?
I am a bit new to PHP and SQL so this may seem like a dumb question. I have already created a drop down list as part of a form which is automatically populated with values taken from a separate database. When a user goes onto this page and either leaves the default value or selects a value from the drop down list and presses the submit button, I would like that selected value to be stored into a database which I have already created in SQL. Just to let you know that I can do the above using a text field but just don't know how to do it with drop down list. If your going to explain any coding then it may help if I give you the names of certain items that are involved. Database is called "Company" Field within database is called "Name"
View Replies !
INSERT INTO - Enters List In Reverse
I have a function that creates a list of IP address and enters them into a table. Most of the enteries are poopulated with enteries from a file with the LOAD data infile command. These are entered into the DB in ascending order. The below function enters the information info the table, but my problem in that it's entered in reverse order. Code:
View Replies !
MySQL Dropdown
Is there a way to make a dropdown by getting the options directly from the database? And I'm using this as a form so it has to have a name so I can call upon the value and do something to it. Example to what I want to do: - I have a database table named members and in that table, it has 3 columns - name, age, sex. What I want to do is make a dropdown inside a <form> using the data from the database. And then when the person selects the name of the member and presses submit, it deletes the row with that name from the table and the database.
View Replies !
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:
View Replies !
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:
View Replies !
MySQL, DropDown, PHP, Form!
Well I have this mysql database with some passwords. Now I have a form where a dropdown list is populated with the name of the users (there are repeated names). Well what I'm trying to do is after selecting an user in the dropdown list, the post will be directed to a result form where I'll have again the name of the user and now his password (for that exact row). Code:
View Replies !
Dropdown & Mysql-tables
I know how to create tables in mysql, thats no problem. I want to use a table with users-information; name, class, email, id and so forth. Then I would like a table with info about events; name, data... What I want is this; when registered users come to the site they want to signup to these events and they want to select events via a dropdown-box. They chose an event and clicks a button and its stored in the db that they want to go to that particular event. Now, in php and mysql, how is that done? I dont know how to setup the layout for the tables. And I think Im missing some in the php to.
View Replies !
Populating Dropdown With Mysql Entries
I would like to create a combobox in Flash which is populated with mysql data and programmed with php. For example: There are 3 entries in database .ie. apple, bannana, peach. Now these I want in combobox in Flash MX/flash5.
View Replies !
Mysql Dropdown And Text Box Scenerio
I have on my web site a facilty where a user can change their trainer name. So 'bill' can change trainer from 'trainer1' to 'trainer4'. that kind of thing. Currently bill would select the trainer from a dropdown menu where the menu is populated from a mysql statement of available trainers. How would I create a facility where bill can either select the new trainer from the list or enter an entirely new one in a text box.
View Replies !
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:
View Replies !
Mysql 3.5 And Mysql 4.1 Data Insert
I am not able to insert data in mysql 4.1 by using same php script i used for mysql 3.5 . only id field and date field which are not VARCHAR get inserted but not VARCHAR PLease help here is my insert statement Code:
View Replies !
Insert Into MYSQL Using PHP
I'm having problems with a form I've created to insert or update into a MySQL database using PHP. The fields (some are varchar and some are text) sometimes contain single or double quotes or slashes or '¿' and some other special characters. These seem to throw off the INSERT process in one way or another. Either that or when I got to display this information, it doesn't come out right. Are there functions that I need to run on the information to make this database safe before I INSERT (or UPDATE) or before I display it?
View Replies !
PHP-MySQL Insert
$Insert_query = mysql_db_query( "INSERT INTO $Table_result VALUES('$Item_Name', $Item_Price)" ); Why do I get a parse error on this line of code???
View Replies !
Insert Into MySQL
I have a system from a company called ProTx they provide online credit card transactions. I have their PHP code, and have had to change some parts of this to work correctly but am getting stuck on this part. This code, is part of a page that sends off some variables to their server and expects a response. I am able to send the variables, and get the response but I want to put them into a database so that I have all refunds seperate from all orders. PHP Code:
View Replies !
Cannot Insert In Mysql Using Php
i have a problem about inserting data into the table. I dont know why the output always said cannot be added to the database! Anyone has idea for me? i check the connect is ok, but dont know why. The following is a function i will call to insert the database. Thx!!! function AddPart($Partnumber, $Partname, $Partdescription) { //Set the variables for the database $DBName="krista"; $TableName = "parts"; $Link = mysql_connect("localhost","","") or die ('I cannot connect to the database because: ' . mysql_error()); mysql_select_db($DBName,$Link); $Query = "INSERT into $TableName (Part_Number, Part_Name, Part_Description) values ($Partnumber, $Partname, $Partdescription)"; if (mysql_query($DBName, $Query, $Link)){ print("Your information is successfully added to the database!! <BR> "); } else{ print("Your information cannot be added to the database! <BR> "); } //close MySQL mysql_close($Link); }
View Replies !
Insert Value In MySQL
I've got a small problem. I'm trying to insert some values into a DB. Now I'm using phpMyAdmin, which I haven't used for a while. But what I've done is echoed out the insert statement and copied into phpMyAdmin and I get this error: Duplicate entry Ɔ' for key 1 What does this mean?
View Replies !
Mysql Insert.
This is the code I am having a problem with. As you can see I have added the values to the database insert but for some reason it still wont insert. $root = "http://$_SERVER[HTTP_HOST]" . ereg_replace('/$', '', dirname($_SERVER['PHP_SELF'])) . '/' $action = $_GET['action']? $_GET['action'] : ($_POST['action']? $_POST['action'] : ''); $connection = @mysql_connect(mysql_hostname, mysql_username, mysql_password) && @mysql_select_db(mysql_database) ¦¦ die('<code>' . mysql_error() . '</code>'); switch ($action) { case 'generate_url': $parsed = @parse_url($_POST['url']); if ($parsed && strlen($_POST['url']) && ereg('.', $_POST['url'])) { if (eregi('^[a-z0-9-]+$', $_POST['tag'])) { $tag = $_POST['tag']; $sql = "SELECT * FROM `urls` WHERE `url_tag` = '$tag' OR `url_id` = '$tag'"; $q = mysql_query($sql); $n = mysql_fetch_assoc($q); ..........................
View Replies !
|
TRACKER
