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PHP - Inserting Values To A Database Table Depending On The Value Of A Drop Down Menu
Hello all
Ok here is the problem... I want when a user inputs the requested data to the text fields , the script to insert those data in the prope table depending on the choise the user does by choosing one option from the drop down menu. Below is the php code (apparently not working) $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } $site_type = $_REQUEST['category_selection']; if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Web_Sites")) { $insertSQL = sprintf("INSERT INTO partner_sites (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Blogs")) { $insertSQL = sprintf("INSERT INTO partner_blogs (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Directories")) { $insertSQL = sprintf("INSERT INTO partner_directories (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } And the html form <form action="<?php echo $editFormAction; ?>" method="POST" enctype="multipart/form-data" name="link_submission" id="link_submission"> <table width="630" border="0" align="center" cellpadding="5" cellspacing="5"> <tr> <td width="76">URL:*</td> <td width="519"><label for="url_field"></label> <span id="sprytextfield1"> <label for="url_field"></label> <input name="url_field" type="text" id="url_field" size="50" /> <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></td> </tr> <tr> <td>Anchor Text:*</td> <td><label for="anchor_field"><span id="sprytextfield2"> <input type="text" name="anchor_field" id="anchor_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>URL Title:*</td> <td><label for="title_field"><span id="sprytextfield3"> <input type="text" name="title_field" id="title_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>Description:*</td> <td><span id="sprytextarea1"> <label for="description_field"></label> <textarea name="description_field" id="description_field" cols="45" rows="3"></textarea> <span id="countsprytextarea1"> </span><span class="textareaRequiredMsg">A value is required.</span><span class="textareaMaxCharsMsg">Exceeded maximum number of characters.</span></span></td> </tr> <tr> <td>Webmaster Name:*</td> <td><label for="textfield2"><span id="sprytextfield4"> <input type="text" name="webmaster_nane_field" id="webmaster_nane_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>Webmaster E-mail:*</td> <td><label for="textfield3"><span id="sprytextfield5"> <input name="webmaster_email_field" type="text" id="webmaster_email_field" size="40" /> <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></label></td> </tr> <tr> <td>Category:*</td> <td><span id="spryselect1"> <label for="category_selection"></label> <select name="category_selection" id="category_selection"> <option>Select An Option</option> <option value="Web_Sites">Web Sites</option> <option value="Blogs">Blogs</option> <option value="Directories">Directories</option> </select> <span class="selectRequiredMsg">Please select an item.</span></span></td> </tr> <tr> <td> </td> <td><label for="select"></label> <input type="submit" name="button" id="button" value="Url Submission" /></td> </tr> </table> <input type="hidden" name="MM_insert" value="link_submission" /> </form> Im begging for your help..... Similar TutorialsHi all, I'm just wondering if there's an easier way of doing what accomplishing the following: I have a value in my database which represents a selection in a drop down menu, i want to read it from the database and have it automatically selected depending on the stored data. I have the following working but just wondered if there was an easier way to get the same result: Code: [Select] <?php //database connection $query = "SELECT id FROM `tablename` WHERE username='$username'"; $result = mysql_query($query); $row = mysql_fetch_object($result); $id = $row->id; $id = (int)$id; ?> <select name="id"> <option value="">Select your option...</option> <option value="1" <?php if (($id - 1) === 0) { echo 'selected="selected"'; }?>>Selection 1</option> <option value="2"<?php if (($id - 2) === 0) { echo 'selected="selected"'; }?>>Selection 2</option> </select> Sorry if it's not very clear, i'll explain best i can if anyone can help. Thanks Hi guys, I am building a website with basic e-commerce functionality, using php and using xampp to test it. I am having issues when attempting to submit a quantity (into table orders) using a form and validating it against an existing value (from table products), giving a response on whether there is sufficient quantity in the second table. I am then, in another page (same one performing the validations), attempting to then show a result based on the initial quantity entered, with a summary of the order details and calculation of the quantity * price to display a total as well. This has all been built from scratch, however I may have taken the wrong approach for these two pages... any assistance or insight as to where I am going wrong would be greatly appreciated. Here is the page I have placed the products, existing quantity and a text field they are able to enter their desired quantity: Code: [Select] <?php session_start(); require_once "../database/db.php"; require_once "../includes/functions.php"; $page_title = 'Product Catalogue'; include_once "header.php"; $conn = mysqli_connect ($dbhost, $dbuser, $dbpassword, $dbname); $query = "SELECT * from products"; $result = mysqli_query($conn, $query); if (!$result) { include_once "header.php"; die ("Error, could not query the database"); } else { $rows = mysqli_num_rows($result); if ($rows>0) { while ($row = mysqli_fetch_array($result)) { ?> <form> <br /> <br /> <br /> <table> <tr> <td style="width: 200px">Product Code:</td> <td><?php echo $row['ProductCode']; ?></td> </tr> <tr> <td>Product Name:</td> <td><?php echo $row['ProductName']; ?></td> </tr> <tr> <td>Product Description:</td> <td><?php echo $row['ProductDescription']; ?></td> </tr> <tr> <td>Product Colour:</td> <td><?php echo $row['ProductColour']; ?></td> </tr> <tr> <td>Product Price:</td> <td>$<?php echo number_format($row['ProductPrice'],2); ?></td> </tr> <tr> <td>Product Image:</td> <td><img src="<?php echo $row['ProductImagePath']?>"/></td> </tr> <tr> <td>Quantity in Stock:</td> <td><?php echo $row['ProductQuantity']; ?></td> </tr> </table> </form> <form method="post"action="processQuantity.php"> <table> <tr> <td style="width: 200px">Quantity:</td> <td><input type="number" name="Quantity" id="Quantity" value="<?php if (isset ($quantity)) echo $quantity; ?>"size = "20" /></td> <td><input type="submit" name="Purchase" value= "Purchase" /></td> </tr> </table> </form> <hr /> <?php } include "footer.html"; } } ?> Here is the page that I am using to validate the data as well as show a result based on the entered amount: Code: [Select] <?php session_start(); require_once "../includes/functions.php"; require_once "../database/db.php"; $quantity = $_POST['Quantity']; $productquantity = $_POST['ProductQuantity']; $orderid = $_POST['orderid']; $productcode = $_POST['productcode']; $productprice = $_POST['productprice']; $total = $quantity * $productprice; $error_message = ''; if ($error_message != '') { include_once "displayCatalogue-PlaceOrder.php"; exit(); $conn = mysqli_connect ($dbhost, $dbuser, $dbpassword, $dbname); if (!$conn) { echo "Error"; } else { //sanitise date $scustomerid = sanitiseMySQL($customerid); $sproductcode = sanitiseMySQL($productcode); $squantity = sanitiseMySQL($quantity); $sproductprice = sanitiseMySQL($productprice); $sorderdate = sanitiseMySQL($orderdate); $query = "select productquantity from products where productcode = '$sproductcode'"; $result = msqli_query ($conn, $query); $productquantity = mysqli_num_rows($result); if ($quantity < $productquantity) { $error_message = "You cannot order more than what is currently instock"; include_once "displayCatalogue-PlaceOrder.php"; exit (); } else { $row = mysqli_fetch_row($result); $query = "INSERT into orders (customerid, productcode, quantity, productprice, orderdate) values ('$scustomerid', $sproductcode', '$squantity', '$sproductprice', '$sorderdate')"; $result = mysqli_query($conn, $query); $row = mysqli_affected_rows($conn); if ($row > 0) { include "header.php";?> <h3>Order Confirmation</h3> <p>Thank you, your order is now being processed.</p> <table> <tr> <td style="width: 200px">Order Number:</td> <td><?php echo $orderid; ?></td> </tr> <tr> <td>Product Code:</td> <td><?php echo $productcode; ?></td> </tr><tr> <td>Quantity:</td> <td><?php echo $quantity; ?></td> </tr> <tr> <td>Price:</td> <td><?php echo $productPrice; ?></td> </tr> <tr> <td>Total Cost of Order:</td> <td><?php echo $total; ?></td> </tr> </table> <?php include "footer.html"; } else { $error_message ="Error placing your order, please try again"; include "displayCatalogue-PlaceOrder.php"; exit(); } } } } //this is used to validate the quantity entered against what is available in the database ?> Is it possible to use for loop to insert values into database using for loop? Or what would be the another way to insert these values into the database if the number of values being entered differs from one time to another? Here is basically what I am trying to do: mysql_query("INSERT INTO MaxMillionsNum SET Day='$weekday', DrawDate='$CompleteDate', DateTime='$timestamp', for($i=1; $i<=$SetNumber; $i+=1){ for($j=0; $j<7; j+=1){ $NumberName="Number".$i.$v[$j];//Generating name of database field $$NumberName=$Match[0][$j];//Here I would store the $Match value into the fieldname name created above } } Tries='$RunCounter'"); Hello there,
I'm really new at PHP and I've been reading several beginner tutorials so please accept my apologies for any stupid questions I may ask along the way.
I've gotten as far as installing XAMPP, set up a database plus PHP form and I'm struggling to figure out how to insert values from an array into my database.
I've learnt the code in one particular way (see beginner tutorials) so I was wondering if you could help me keeping this in mind. I know there'll be a million better ways to do what I'm doing but I fear I will be bamboozled with different code or differently structured code.
Anyway the tutuorials I'm reading don't see to cover how I can insert an array of values into my database, just singular values.
In the attached file, I have 10 rows of 2x text inputs (20 text inputs total). Each row allows the user to enter a CarID and CarTitle. I've commented out the jQuery which validates the inputs so I can build a rudimentary version of this validation with PHP.
I thought that because the line $sql="INSERT INTO carids_cartitles (CarID, CarTitle) VALUES ($id, $title)"; is inside the foreach, means that for each pair of values from the form it'd insert to the database.
It doesn't do this. If I enter two or more CarIDs and CarTitles, only one pair of values gets saved to the database.
I'm sorry if I haven't explained this well enough, any questions please let me know.
Many thanks for your help in advance.
Attached Files
form.php 4.43KB
5 downloads I have posted one set of values into my database and it worked fine but when i input another set they wont go inside unless i changes the value of the primary index colum. I want to be able to insert a new values regardless of the primary index value. Any idears...? I want to take data from one table and insert it into another in the same database, the problem is the two tables have different values so it wouldn't be as simple as using an INSERT INTO script. From table 1 I want to extract a row ID as well as a field called systems. I then want to insert that into the second table which is structured, Id (null) Name (games) Value (the data given from Systems in the other) ID2 (the ID from the other table) Category (news) Any ideas how I can go about this? im trying to write a script takes an xml files with tv show info, splits it into the show and the eppisode info and the place it into a table, i have got it to proccess all the info and print it out on a web page, but i cant, for the life of me, get it to insert said data into the table, it seems to be just ignoring the code and prints out the data as if nothing happens, no errors or anything. here is my code (abit messy but im only just starting and its my test.php) Code: [Select] <?php $tvdb_mirror = "http://www.thetvdb.com/api/"; $tvdb_time = "http://www.thetvdb.com/api/Updates.php?type=none"; $dbname = "mediadb"; $dbuser = "root"; $dbpass = ""; $dbserv = "127.0.0.1"; $rss = simplexml_load_file('sample.xml'); $showName = "Show Name = ".$rss->Series->SeriesName; print $showName; print "<br />Show Discription = ".$rss->Series->Overview; print "<br />"; mysql_connect('127.0.0.1', 'root', ''); @mysql_select_db('mediadb') or die("Unable to select database"); foreach ($rss->Episode as $item) { $seasonnum = $item->Combined_season; $EpisodeNumber = $item->EpisodeNumber; if($EpisodeNumber < 10){ $EpisodeNumber = "0".$EpisodeNumber; }; $EpisodeName = $item->EpisodeName; $Overview = $item->Overview; $airdate = $item->FirstAired; $tvdbid = $item ->id; $query = "INSERT INTO eppisodes VALUES('', '1', ".$EpisodeName.", ".$Overview.", ".$airdate.", '1', ".$tvdbid.", '-1', ".$seasonnum.", ".$EpisodeNumber.")"; mysql_query($query); print "<br />".$showName." - ".$seasonnum."x".$EpisodeNumber." - ".$EpisodeName." Overview:<br />".$Overview; } mysql_close(); ?> i am a noob @ php and mysql, but i have doubke and triple checked the names of the db and table. here is an sql dump of my db Code: [Select] -- phpMyAdmin SQL Dump -- version 3.3.5 -- http://www.phpmyadmin.net -- -- Host: 127.0.0.1 -- Generation Time: Nov 13, 2010 at 12:48 PM -- Server version: 5.1.49 -- PHP Version: 5.3.3 SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO"; /*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */; /*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */; /*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */; /*!40101 SET NAMES utf8 */; -- -- Database: `mediadb` -- -- -------------------------------------------------------- -- -- Table structure for table `eppisodes` -- CREATE TABLE IF NOT EXISTS `eppisodes` ( `id` int(11) NOT NULL AUTO_INCREMENT, `showID` int(11) NOT NULL, `eppname` varchar(255) NOT NULL, `eppdesc` longtext NOT NULL, `airdate` date NOT NULL, `format` int(11) NOT NULL, `tvdbid` varchar(20) NOT NULL, `dohave` tinyint(1) NOT NULL, `season` varchar(2) NOT NULL, `eppisode` varchar(3) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; -- -- Dumping data for table `eppisodes` -- -- -------------------------------------------------------- -- -- Table structure for table `shows` -- CREATE TABLE IF NOT EXISTS `shows` ( `id` int(100) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `description` longtext NOT NULL, `TVDBID` int(100) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Dumping data for table `shows` -- INSERT INTO `shows` (`id`, `name`, `description`, `TVDBID`) VALUES (1, 'higogo', 'some info', 67546); any help would be very much appericiated. fyi im running win7 with easyPHP 5.3.3 with php 5.3.3, mysql 5.1.49 apache 2.2.16 Hi guys, I had a problem before which is similar to the problem im about to describe, however Pikachu helped me solve the last problem so a big thanks to him! However this problem is slightly different and im struggling to find a solution. I will first describe the problem I had which pikachu helped me solve as it would be easier for me to describe this past problem. I have an update form, so a user can edit existing records which are pulled from a database. When the user changes the values in a text field then submits the form, the values in the text fields updates fine. However if they change the values in the text box and then tries to submit the form but there was an error, the values in the text fields go back to the values from the database and the user has to change the text fields again. Well this problem was solved with the help of pikachu and the this is what I did... <?php if(isset($_POST['title'])){echo htmlspecialchars($_POST['title']);} else if (isset($title)) {echo htmlspecialchars($title);}?> Again, the above works fine however the problem I have now is the same problem as described above however it relates to drop down menus and not text fields. I can't seem to figure out how to retain the values in a drop down menu the same way I have for text fields. Below is the code that im using that I thought would work however it isnt working, the values of the drop down menu (if there was an error) are going back to the values from the database. Heres the code... <option value="Psychopathic" <?php if ((isset($_POST['category']))&&($_POST['category'] == 'Psychopathic')) { echo ' selected=selected'; } else if ((isset($category))&&($category == 'Psychopathic')) { echo ' selected=selected'; } ?>>Psychopathic</option> Does anybody know what i am doing wrong and how i can fix this problem? I understand what i need to do but i cant seem to get the code to do what i want to do and thats to retain values of a drop down menu when the page is returned with an error. Any help would be much appreciated. AJay GREAT this forum - JUST GREAT !... Issue: All data entered into my online form was lost (blanked out) and the form returned correctly with message "wrong verification code", when submitted with the wrong verification code. However, going through this great forum I managed to get all - manually entered - data back ! I placed value="<?php echo $_GET['the_field_name'];?>"/ after each input field. BUT... not so with input fields entered from drop-down menu ! How do I put a similar string for the field "Payment by" in this sample: <tr> <td class="table-inquire" width="47%"> <font face="Verdana" size="1" color="#000042"> Payment by:</font></td> <td class="table-inquire" width="51%" colspan="2"> <font color="#400000" face="Verdana"> <select name="payment" size="1"> <option value="VISA">VISA</option> <option value="MASTER">MASTER</option> <option value="CASH">CASH</option> <option value="T/T Banktransfer">T/T Banktransfer</option> <option selected>Please select</option> </select></font><font size="2" color="#400000" face="Verdana"></font></td> </tr> Any advise greatly appreciated. Thanks. Hi all, i want my list/menu field values to come from my database. how can i accomplish that? thanks i did Code: [Select] <select name="select"> <option value="0">--select below--</option> <option value="1">Me</option> <?php require_once '../konnect/konex.php'; $result = mysql_query("SELECT * FROM is_clients"); while($row = mysql_fetch_array($result)) { echo "<option value ='2'>".$row"['name']</option>"; echo "<br />"; } ?> </select> I'm trying to make it so that someone can add an item to the shop based upon the current items within the database. How do I go about doing that? I know i have to change <input> to <select>, but beyond that, how do I code it so I can run an array to get all the current items from the table `items` and list them by their `name` field? Thanks in advance! Here is what I have so far: function addshopinv($id) { error_reporting(E_ALL); ini_set('display_errors', 1); if (isset($_POST["submit"])) { extract($_POST); $errors = 0; $errorlist = ""; if ($name == "") { $errors++; $errorlist .= "Name is required.<br />"; } else if ($errors == 0) { $dbh=dbconnect() or die ("Userlist read error: " . mysql_error()."<br>"); mysql_select_db("XXX"); $query = mysql_query("SELECT id FROM items WHERE name='$name'"); while ($result = mysql_fetch_array($query)){ $item_id = $result['id']; $query1 = mysql_query("INSERT INTO sale SET shop_id='$id', item_id='$item_id'"); } admindisplay("Inventory Item Added.","Add New Inventory Item"); } else { admindisplay("<b>Errors:</b><br /><div style=\"color:red;\">$errorlist</div><br />Please go back and try again.", "Add New Item to Shop"); } } $page = <<<END <b><u>Add New Inventory Item</u></b><br /><br /> <form action="admin_panel.php?do=addshopinv:$id" method="post"> <table width="90%"> <tr><td width="20%">Name:</td><td><input type="text" name="name" size="30" maxlength="255" value="" />*255 character max</td></tr> </table> <input type="submit" name="submit" value="Submit" /> <input type="reset" name="reset" value="Reset" /> </form> END; $page = parsetemplate($page, $row); admindisplay($page, "Add New Inventory Item"); } 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 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'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> Hi All, I have 2 tables: one CarMake - CarMakeID - CarMakeDesc two CarModel - CarModelID - CarModelMake - CarModelDesc Depending on what the user selects in the first dropdown (carmake) the possible selection in the second dropdown (model) needs to be limited to only the models from the selected carmake. in the second table (Carmodel : the 'CarModelMake' = CarMakeID, to identify the make) How do I limit the dropdown 'CarModel' based on the selected CarMake in the first dropdown. link : http://98.131.37.90/postCar.php code : -- -- -- Code: [Select] <label> <select name="carmake" id="CarMake" class="validate[required]" style="width: 200px;"> <option value="">Select CAR MAKE...</option> <?php while($obj_queryCarMake = mysql_fetch_object($result_queryCarMake)) { ?> <option value="<?php echo $obj_queryCarMake->CarMakeID;?>" <?php if($obj_queryCarMake->CarMakeID == $CarAdCarMake) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarMake->CarMakeDesc;?></option> <?php } ?> </select> </label> <label> <select name="carmodel" id="CarModel" class="validate[required]" style="width: 200px;"> <option value="">Select MODEL...</option> <?php while($obj_queryCarModel = mysql_fetch_object($result_queryCarModel)) { ?> <option value="<?php echo $obj_queryCarModel->CarModelID;?>" <?php if($obj_queryCarModel->CarmodelID == $CarAdCarModel) { echo 'selected="selected"'; } ?> > <?php echo $obj_queryCarModel->CarModelDesc;?></option> <?php } ?> </select> </label> I'm a novice.. and appreciates all the help ! Hi. Maybe a tricky question? How do I reflect the content of a column from a database table in a roll down select menu in the browser? Let's say that the content of the table column is: Anna Michael These names should be reflected in this select menu like this: <select name="friends"> <option value="Choose a name">Choose a name</option> <option value="Anna">Anna</option> <option value="Michael">Michael</option> So visitors can choose a name, and thereby turn it into a variable, for reuse in the database. Best regards Morris I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database. Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too. Thanks in advance! just a little confused where ( brackets go in this statement . im just wanting to to display the information differently depending on which message is returned from the database. thanks Code: [Select] if ($info['message'])="<a href=\"freindacept.php?username=$myusername\">Has sent you a friend request </a>";{ Echo "<a href='viewprofile.php?username={$info['username']}'><img src='http://datenight.netne.net/images/".$info['img'] ."' width='30' height='30''></a>".$info['from_user']." ".$info['message']; }elseif ($info['message'])="You have sent a contact request to"; Echo " .$info['message']<a href='viewprofile.php?username={$info['username']}'><img src='http://datenight.netne.net/images/".$info['img'] ."' width='30' height='30''></a>".$info['from_user']."; } } Alright, so meanwhile i wait for answer on the other thread i started another project, to fix the menu... but it was a bit confusing so now i ask you again!... Heres my current menu.php Code: [Select] <?php include "config.php"; if (!$_SESSION["valid_user"]) { echo '<a href="login.php">' . 'Login <br/><br/>' . '</a>'; } if ($_SESSION["valid_user"]) { echo '<a href="logout.php">' . 'Logout <br/><br/>' . '</a>'; echo '<a href="members.php">' . 'Control Panel <br/><br/>' . '</a>'; } echo '<a href="memberlist.php">' . 'Members <br/><br/>' . '</a>'; echo '<a href="ranking.php">' . 'Rankings <br/><br/>' . '</a>'; ?> Now it works perfectly when logged in, but when you logout the session dies, leaving me with this: Notice: Undefined index: valid_user in C:\wamp\www\kawaii\menu.php on line 5 Login Notice: Undefined index: valid_user in C:\wamp\www\kawaii\menu.php on line 10 Members Rankings Now, how can i make a login that depends on if your logged in if this happens when you logout..., i hope someone can find a solution that can work out Hey guys, I am kind of new to php but I am stuck on this question. I have a mySQL database and table set up called members. I have some data in the table and I want to be able to add certain things to a paticular row in the table based off the ID number of the row. One of the values of the table is an auto incremeted ID number. I want to add a text value called message to a specified ID number. How do I go about doing that. I have this code already but it doesn't seem to work. Code: [Select] $sql = "INSERT INTO members (message) VALUES ('$_POST[message]',(SELECT id FROM members WHERE id='$_POST[id]'))"; Any Ideas? Thanks Hi People i'm a newb at this so bare with me but i currently have a php file called Newkpi.php which has a select statement in. this selects data from a table called "StaffList". this then populates the page with a html table with 14 records (this will increase/decrease over time) i then have some extra text boxes to enter more detail into like service amaount service date and so on. when i click the submit button i want it to cycle through each row and insert the data into a separate table called "Services". however i cannot for the life of me get this to work and need some help with it. people find attached the code for Newkpi and see if you can help me with this. in total i want it to take the names from stafflist and populate Services with the names and the extra detail which is entered on the page Much Thanx in advance SLOWIE |
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