PHP - Retrieve Information From Mysql Through Drop Down List
I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information.
How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> Similar TutorialsI'm doing this activity where a user chooses a base timezone and when the user clicks convert the current time in the selected GMT will be converted to GMT-11 to GMT+13. As of now, I have these codes: act09_view.php: Code: [Select] <?php session_start(); $s="GMT "; echo "Select the base time zone:</br>"; for($n=-11;$n<=13;$n++) { if ($n>=0) $s="GMT +"; $gmt[]=$s . $n . "</br>"; } echo "<select name='gmt'>"; foreach ($gmt as $value) { echo '<option value="' . $value . '">' . $value . '</option>\n'; } echo '</select>'; //$_SESSION['value']=$value; ?> <form action="act09_process.php"> </br><input type='submit' value='Convert'/> </form> act09_process.php: Code: [Select] <?php session_start(); //$value=$_SESSION['value']; //echo $value; date_default_timezone_set('Asia/Manila'); $gmttime=date('M j, Y g:i:s A'); echo "The current date and time at" . " is " . $gmttime; ?> I've tried using session variables. I think I executed them incorrectly. The output is supposed to look like this: Hi, Is it possible to retrieve all the information from a posted value without knowing the input field? My page is taking information from a database and then creates a page with checkboxs so that people can select certain fields. There are 1300 fields in the database so I don't know what fields will be submitted when a person saves the page, is there a simple way to check what fields have been submitted without writing a check for each checkbox? I hope this makes sense, thanks in advance for any help. I am working on a web application in php for ipad. I need to retrieve the ipad's unique identifier programmatically. Is there any way i could check the unique identifier of an ipad in the php application. I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly. There are two tables, one that has the category ID and category name. The second table has the data and the catid which is referenced to the category id in the first table. Code: [Select] -- -- Table structure for table `categories` -- DROP TABLE IF EXISTS `categories`; CREATE TABLE IF NOT EXISTS `categories` ( `id` int(11) NOT NULL AUTO_INCREMENT, `categories` varchar(37) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ; -- -------------------------------------------------------- -- -- Table structure for table `links` -- DROP TABLE IF EXISTS `links`; CREATE TABLE IF NOT EXISTS `links` ( `id` int(4) NOT NULL AUTO_INCREMENT, `catid` int(11) DEFAULT NULL, `name` varchar(255) NOT NULL DEFAULT '', `url` varchar(255) NOT NULL DEFAULT '', `content` varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (`id`), KEY `catid` (`catid`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ; Then is the list records file, I have Code: [Select] <?php include ("db.php"); include ("menu.php"); $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; $id = $_GET['id']; $query="SELECT * FROM links ORDER BY catid ASC"; $result=mysql_query($query); ?> <table width="65%" align="center" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="100%" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="7"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Name</strong></td> <td align="center"><strong>URL</strong></td> <td align="center"><strong>Content</strong></td> <td align="center"><strong>Update</strong></td> <td align="center"><strong>Delete</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td> <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> </td> <td><? echo $rows['catid']; ?></td> <td><? echo $rows['name']; ?></td> <td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td> <td><? echo $rows['content']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td> <td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> So, how do I get this code Code: [Select] $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> to give me an output that will be something like if catid exactly matches categories.id echo categories.categorie ??? so far everything I have done produces either a default category of the last category, the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing. How do I get just the category name? I will keep reading and try to figure this out, but any help would be greatly appreciated. Thanks in advance Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me Hi everyone, I am having trouble passing/displaying the values inside of a selected list. I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute. Please kindly let me know what should I do in order to get the values and I really appreciate your help. <?php $selected = $_POST['selectto']; if(isset($selected)) { echo "something in selected<br />"; for ($i=0;$i<count($selected);$i++) echo "selected #1 : $selected[$i]"; foreach ($selected as $item) echo "selected: item: $item"; } ?>This is the form As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> Hi folks, Complete No0b here when it comes to PHP and MySQL. I am in the middle of creating a PHP website. What I want to do is have the contents of a page in a MySQL table and have PHP gather the page content and display it on the page. Here is what I have in my home.tpl file: Code: [Select] <?php $query = "SELECT home, FROM $database_name"; $result = mysql_query($query); ?> And, in my index.php I simply call that home.tpl to display the data by using: Code: [Select] <?php include 'templates/default/home.tpl'; ?> Now, all I get is a blank page on index.php. Is there something else I should be doing? Remember, I am a complete No0b! Thanks folks. Hi, Im trying to retrieve HTML from a mysql database but nothing i've tried seems to work. The HTML im trying to retrieve is an iframe with a link and styles (code from amazon associates). Im trying to display links to specific products on amazon from the product page on my site. All data about the product is retrieved from the database so i have code to select the amazon link row in my database table but i cant get it to display. It says the html isnt a string so i cant echo it, fair enough. I have tried using the following code: Code: [Select] $get_buylink_sql = "SELECT mobo_buylink FROM mobo WHERE mobo_id = $mobo_id"; $get_buylink_res = mysqli_query($mysqli, $get_buylink_sql) or die(mysqli_error($mysqli)); while ($buylink = mysqli_fetch_array($get_buylink_res)) { echo"<iframe src=\"".$buylink."\" style=\"width:120px;height:240px;\" scrolling=\"no\" marginwidth=\"0\" marginheight=\"0\" frameborder=\"0\"></iframe>"; } mysqli_free_result($get_buylink_res); mysqli_close($mysqli);I have also tried putting the whole iframe code in the database which didnt work either. The mobo_id variable works fine for retrieving the rest of the data and i need to get the amazon link from the same record. I hope i've put this in a way you can understand, but if not i'll try and explain better and give you a link to my site if needed. Thanks, Alex I want to use session to do a query and will I be able to do this? I have a session that was gathered from login and now i was to use this session to do a query If Yes, How? Hello everybody!
I am trying to figure out how to obtain all the data related to a key, but I've got no results so far and I am becoming really frustated, let's see if any of you could help me out with this.
Imagine I have a table with several columns, but we bother about two of them, let's say we have serial numbers of some product, on the left Incoming serial number (we can repair or swap the unit), on the right the outcoming serial number (same if we have repared the unit, different if we swap it for another unit).
Then we have, for example:
A -> A (Unit A enters and we repaired it)
A -> B (Same unit came another day for some reason and we couldn't repair it, so we swap it by giving B to the customer)
B -> C (Unluckily B was defective so we have to change it again)
C -> C (C had another problem and we repaired it)
We have that in the database from different days and the such, so now, we want to know the historical and we know "C". If we perform a SELECT * FROM... WHERE incoming/outcoming serial number = "C" we'll get:
B -> C
C -> C
So we should seek now for B and keep going... but I cannot proceed correctly, 'cause if I SELECT using B I'll get again B -> C (and A -> B, what I want), but when do I know I have to finish? How could I implement this as a function or whatever? showing every not repeated line from the beginning.
Could your minds help mine? Thank you very much in advance!.
I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks Hey Everyone... First off, I am only a young web developer and i'm working on a school project and am making a text-based game online... Now what i'm having trouble with... I want a drop-down list that has a list of characters classes Clubber Mixer Sauceror Tamer And I want whatever is selected to be placed into the database along with the username/password (THIS ALL WORKS FINE JUST NOT THE DROP DOWN LIST) All help appreciated I m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php include("db.php"); $query=mysql_query("select * from table where id='3' "); $row=mysql_fetch_array($query); $r=$row['image']; header("content-type:image"); echo $r; ?>" i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance... I wonder whether someone can help me please. I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page. Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be: Code: [Select] <?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?> The problem I have, is that when I try and run this, I receive the following error: Quote Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";. I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem. I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong. Many thanks and kind regards Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago. I started out with some tutorials and feeling im getting the hang of it. Enough of me, lets get to the point: <?php $con = mysql_connect('localhost',$user,$pass)or die(mysql_error()); $selectdb = mysql_select_db($selectdb)or die(mysql_error()); $sql = "SELECT * From table"; $result = mysql_query($sql); $num = mysql_num_rows($result); $myarray = array($result); $i =0; while ($i < $num){ echo $myarray[$i]; $i++; } ?> Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results. The output in the browser is this: Resource id #3 I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple". Thanks in advance! Dan-Levi Hello guys, I am a new programmer and i am building a new website. I would like to give me your advice for the following problem: I want to build a webpage, in which there will be a <div id="book-content"> ...............</div> part. Inside this div i would like to dynamically display pages from a book. Each page will have text, scripting code blocks, blocks with the output of each scripting code, and images. There will be a bar on the left of the webpage in which the user can select which page of the book he wants to load. For example...My web page will look something like this... Code: [Select] <HTML> <HEAD> </HEAD> <BODY> <DIV ID="PAGE-HEADER"> //LOGO OF THE WEB PAGE </DIV> <DIV ID="PAGE-MENU"> //PAGE MENU </DIV> <DIV ID="BOOK-INDEX" WITH FLOAT:LEFT > //HERE A WILL SHOW THE CHAPTERS OF THE BOOK AND THE PAGES OF EACH CHAPTER </DIV> <DIV ID="BOOK-PAGE"> //THE PAGE SELECTED FROM THE PREVIOUS 'BOOK-INDEX' MENU WILL BE DISPLAYED HERE WITH A MYSQL QUERY ****** </DIV> </BODY> </HTML> Then in a mysql database, i would like to have records with a text field, that will contain for example the followng: <h1> Chapter 1: bla bla </h1> <p> in this chapter we will speak about bla bla bla.... </p> <div id="code"> int main() { int x,y; x=2; y=3; x=x+y; } </div> <p> this will outpout the following:</p> <div id="code output"> x=5! </div> I would like to get this html code from the database, and then show it in the <div id=BOOK-PAGE> div. But i dont want to use php eval(). Also, if i store the code to a file and then include it, i will have too may files(equal to the book's number of pages etc 100). Any ideas? Hi, My output in html is : Code: [Select] <uL><li id="B1"></li> <li id="B2"></li> <li id="B3"></li> <li id="B4"></li> <li id="B5"></li> <li id="B6"></li> <li id="B7"></li> <li id="B8"></li> <li id="B9"></li> <li id="B10"></li> <li id="B11"></li> <li id="B12" class="active"></li> <li id="B13" class="no"></li> <li id="B14" class="no"></li> <li id="B15" class="no"></li> <li id="B16" class="no"></li> <li id="B17" class="no"></li> <li id="B18" class="no"></li> <li id="B19" class="no"></li> <li id="B20" class="no"></li> </ul> If MySQL query result is equal to `6`, then `<li>` tag with `id` equal to "`B12`" should have class "`active`". All the `<li>` elements occuring after this active element should have class "`no`". This shows images of horizontal rating between `0` and `10` and between 0.5 Example : 0 .5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 In the example above elements from `0` to `5` will be blue, element `6` will be white and elements from `7` to `10` are black. How could I generate this using PHP and/or MySQL? Thanks, XXXX chirstmas XXXX I am having a lot of trouble with this code, and I have no clue how to fix it. Right now, I have a GUI for a fictitious car dealership that has 5 populated drop down menus called Make, Model, Year, Color, and Mileage. What I want the code to do is read the selections made by the user with the drop down menus once the user hits the submit button and then filter the tables that I have in a mysql database to meets the choice requirements of the user. The code will bring up the GUI, but once I hit the submit button, I get the following errors Please if anyone can help me that would be fantastic. I really have no clue |