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PHP - Dynamic Xml From Database
Hi,
I'm learning php and I'm trying to get to grips with dynamic xml and php coming from a database. The trouble I am having is I can't get my code to give any output. I don't get an error I message, I just can't get the xml to display. If anyone can give me any pointers as to what I'm doing wrong that would be great... Here's the code I have. Code: [Select] <?php DEFINE ('DB_USER', 'root'); DEFINE ('DB_PASSWORD', 'password'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'flashphpbible'); $link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $query = "SELECT * FROM dynamicxml WHERE id=1"; $result = mysqli_query($link, $query); $xmlData .= " "; $xmlData .= "<store>\n"; while($row = mysqli_fetch_array($result)); { $xmlData .= " <item>\n"; $xmlData .= " <name>".$row['name']."</name>\n"; $xmlData .= " <section>".$row['section']."</section>\n"; $xmlData .= " <price>".$row['price']."</price>\n"; $xmlData .= " <inStock>".$row['instock']."</inStock>\n"; $xmlData .= " </item>\n"; } $xmlData .= "</store>\n"; echo "response=".$xmlData; ?> Thanks for any help offered. Ben. Similar TutorialsHey, I'm currently planning on how i will code some breadcrumbs for a shopping site where the admin can add more options/levels to the database which in turn store products. I understand how to get the url of the previous options/level but am stuck on retrieving the name of the level and can only imagine retrieving it from the database. Though Its possible to do so is there any other method as relying on the database for crumbs seems like a easy way out or am i just picky? the typical url is like so: website.com/shop.php?cat_id=1004, so i cant retrieve the name from it unless i cross reference it to its database name and dont know how to retreive from their <a href> tag. On a side note is there a graph that can show a general outline of how many db queries a server can make per second etc depending on its hardware? Thanks. I'm praying there's a function somewhere out there that: generates a html form based on the fields of a particular table within a DB (namely mySQL) generates the appropriate form input based on the fields type or notes (ie. for field: avatar_img, it knows to generate a file input) uses the POST method to submit generates and submits SQL insert query to database In addition to that, a similar function that generates a form that allows the editing of records. It doesn't take too long to write these each time myself, but if there's a function that you provide the formname, database driver, table name, whether your form is to insert or edit, that would be quite nifty i think. ie. generateForm(addUser, mySQL, tUsers, insert) Seen any such thing? Many thanks. Hiya I was wondering how one would get make dynamic url variable value pairs from a databases fields. I have a database that I want to use to store variable value pairs that will be used on my pages for dynamic content. It will check one thing for certain besides if the variable value pair match. The table looks like this rpg loc_type loc_type template I want it to check if the rpg is in an array and then check if the url variable and value pair match on from the database. I dont know much about auto code that would generate this on its own and I'm not sure if you can say $_GET[$location]; I dont want to hard code it so it only works with what i code I want the urls variable value pair to be based on the database info which determines which template to show based on those values. I'm using some code to create a select menu of a fieldname of data I have in a mysql database : Code: [Select] <?php mysql_connect('localhost' , 'dbname', 'password'); mysql_select_db('dbname'); $result=mysql_query("SELECT * FROM Persons"); if(mysql_num_rows($result)>0) { ?> <select name="Persons"> <?php while($rows=mysql_fetch_array($result)){ ?> <option value="<?php echo $rows['id']; ?>"> <?php echo $rows['FirstName']; ?></option> <?php } ?> </select> What I would like to do is to elevate this into a jump menu form so that if the user selects an item from my form they are taken to a results page showing the full row of data from the database. Basically a search form containing items from the database they can choose to see more details on. eg. In my example you select a persons name and then you are taken to a results page which displays the details of that person from the database. Problem is I don't know how to do this and have been trawling around for a couple of days to find a solution (sorry I'm new to php). I would appreciate some help or a working example would be great of : 1/ A working dynamic jump menu 2/ The page that would process the form 3/ The results page displaying the data I have selected. Thank you for your time... im making a game and i need to show a users money but i dont know how help? Hi everyone,
I'm a complete newbie here. I've looked around for help and tried a few solutions without success, so because time is short I'm asking for help.
I have a registration form which includes address details. The form holds a country field which is a listbox, the values being selected from a MySQL table. How do I assign the selected country option to a field list in the insert statement?
Here's what I have so far:
Registration page form:
<div class="form-group"> 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 I can not get the values from the javascript add row to go dynamically as a row into MySql only the form values show up as the form below as one row. I made it as an array, but no such luck, I have tried this code around a multitude of ways. I don't know what I am doing wrong, kindly write out the correct way.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR...ransitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Dynamic Fields js/php to MySql need to submit dynamically to the database</title> <?php require ('database.php'); ?> <script type="text/javascript"> var counter = 1; var collector = ""; function addfields(indx) { var tbl = document.getElementById('table_id'); var newtr = document.createElement('tr'); counter = counter + indx; newtr.setAttribute('id','tr'+counter); newtr.innerHTML = '<td><input type="checkbox" name="checkb'+counter+'" id="checkb'+counter+'" value="'+counter+'" onclick="checkme('+counter+')"></td><td><input type="text" name="text1[]"></td><td><textarea name="textarea1[]"></textarea></td>'; tbl.appendChild(newtr); } function checkme(dx) { collector += dx+","; } function deletetherow(indx) { var col = collector.split(","); for (var i = 0; i < col.length; i++) { var remvelem = document.getElementById('tr'+col[i]); var chckbx = document.getElementById("checkb"+col[i]); if(remvelem && chckbx.checked) { var tbl = document.getElementById('table_id'); tbl.removeChild(remvelem); } } } </script> </head> <body> <form enctype="multipart/form-data" id="1" style="background-color:#ffffff;" action="<?php echo $_SERVER['PHP_SELF']; ?>"></form> <table id="table_id" > <tr id="tr1" class="trmain"> <td> </td> <td> <input type="text" name="text1[]"> </td> <td> <textarea name="textarea1[]"></textarea> </td> </tr> </table> <input type="button" value="Add" onClick="addfields(1);" /> <input type="button" value="Delete" onClick="deletetherow()" /> <input type="submit" value="Send" id="submit" name="submit"/> <?php if(isset($_POST['submit'])) { for ($i=0; $i < count($_POST['text1']); $i++ ) { $ced = stripslashes($_POST['text1'][$i]); $erg = stripslashes($_POST['textarea1'][$i]); } $bnt = mysql_query("INSERT INTO tablename (first, second) VALUES ('$ced', '$erg')")or die('Error: '. mysql_error() ); $result = mysql_query($bnt); } ?> </body> </html> What I have so far is a page that produces a list of usernames with checkboxes next to them. From the usernames that I check, I want to update a field for each of those users. My problem is that the number of checkboxes I check is not going to be static so I am pretty sure I have to run a loop based on the number of checked checkboxes ("People chosen"). Here is what I have so far: Code: [Select] <?php if (isset($_POST['submit1'])) { $pick = $_POST["pick"]; $how_many = count($pick); echo 'People chosen: '.$how_many.'<br><br>'; echo "<br><br>"; $count = 0; while ($count < $how_many) { //mysql code probably should go here but I am stuck $count++; } } else { $gender = mysql_real_escape_string($_POST['Gender']); $limit = mysql_real_escape_string($_POST['limit']); echo $gender; echo "<br />"; $count = 0; if ($gender == 'Male') { $result = mysql_query("SELECT * FROM users WHERE gender =\"1\" LIMIT 0, $limit", $connection); if (!$result) { die("Database query failed: " . mysql_error()); } ?> <form method="post"> <?php while ($row = mysql_fetch_array($result)) { ?> <input name="pick[]" type="checkbox" value="<?php echo $row['username'] ?>"><?php echo $row['username'];?> <br /> <?php } ?> <input type="submit" name="submit1" value="Make Offer" /> </form> <?php } ?> I hope this makes sense. Any help would be greatly appreciated. Hi, I had a form which contains, 8 select boxes. In those 8 select boxes, 3 of the boxes are interlinked, like, I am able to generate data in other 2 select boxes, if one box is been selected. i.e. based upon selection of one select box, and i am able to generate data in other select box. So, 3 select boxes are interlinked. And remaining 5 selects are optional, if the user selects, then we need to generate data based upon selection of the user. Here the main issue is how to grab data in the post back form, like, how to capture the number of selects boxes selected and what are those selected, and based upon that we need to generate data. Here one more important thing is that, in the report generation form(postback form), I need to show the data in a table. So , in table I had 9 columns and these would be common to whatever the selection made, but, only the requeired data would be changed based upon the selection of the remaining 5 optional boxes. Hope you got me!!!! Folks, I need help (Php code ) to generate a Dynamic Text on a Base Image. What i want to do is, to make this Image as header on my Site and to make this Header Specific to a Site, i want to Add the Domain Name on the Lower Left of the Image. Got the Idea? Here is the Image link: Quote http://img27.imageshack.us/i/shoppingheader1.jpg/ PHP Variable that holds the Domain name is: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hi all I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option> <option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option> <option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option> </select> Although this works OK, I need it also to show dynamic values like this: select name="category"> <?php $listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC"); while($categoryReturned=mysql_fetch_array($listCategories)) { echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Hi guys, I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific. Thanks in advance! hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda |
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