PHP - Php And Xml "database" Trouble.

Hello. Many thanks for your help.

I am writing a PHP/MySQL dating-site and have hit a programming impass.

I have a database full of members and a search form consisting of checkboxes.

So to search, a member ticks say...gender: female; age: 21,22,23,24,25,26; height: 5'4",5'5",5'6",5'7"; county: cornwall,devon,somerset

How can a run a check on the database selecting all entries that fall into the selected criteria. For example a 23 year old female of 5'5" living in Cornwall and a 26 year old female of 5'4" living in Somerset?

The key index of my database is 'id' and the fields a age,height,county
The names of the form checkboxes a
Gender: male, female; Age: 21,22,23,24 etc; Height: 5_4,5_5,5_6 etc; county: cornwall,devon etc

I have a textbox which i want users to be able to search the database for zipcodes, I can't seem to make it work my IDE doesn't show any errors but when I test the code everything goes blank, the mysql credentials are all correct the code is, any suggestions would help

<form action="index.php" method="post">
   <label for="search2" style="color:#FFF;background-color:#0A016D"><b>Search Homes by Zipcode</b></label><br>
 zipcode:  <input type="text" name="zipcode"><br>
   <button type="submit" name="submit" value="submit"/>
   </form>
<?php
    $zipcode= @$_post['zipcode'];
    require("myconn.php");
    mysql_select_db("phplogin")or die(mysql_error());
   $sql= mysql_query("SELECT * FROM users WHERE zipcode='$zipcode' ");
   $numrow= mysql_num_rows($sql);
   while($row= mysql_fetch_assoc($sql))
   {
    echo "zipcode: ".$row['zipcode']." city :".$row['city']."<br/>";
      
   }
   ?>

I need help trying to figure out why my form won't write the database it is supposed to - i checked the connection to the database and it works and the user seems to have permission to edit database - the error I get is "Error: User not added to database." from "register.php". Can someone please look over my code and see if the problem is coming from somewhere within?   I created a connection file (connect.php)

<?
session_start();
// Replace the variable values below
// with your specific database information.
$host = "localhost";
$user = "master";
$pass = "hidden";
$db   = "user";
// This part sets up the connection to the 
// database (so you don't need to reopen the connection
// again on the same page).
$ms = mysql_pconnect($host, $user, $pass);
if ( !$ms )
{
echo "Error connecting to database.\n";
}
// Then you need to make sure the database you want
// is selected.
mysql_select_db($db);
?>

Then there is the php script (register.php):

<?php
session_start();
// connect.php is a file that contains your
// database connection information. This
// tutorial assumes a connection is made from
// this existing file.
	require('connect.php');
    // If the values are posted, insert them into the database.
    if (isset($_POST['email']) && isset($_POST['password'])){
        $firstname = $_POST['firstname'];
        $lastname = $_POST['lastname'];
		$email = $_POST['email'];
        $password = $_POST['password'];
 
        $query = "INSERT INTO `member` (firstname, lastname, email, password) VALUES ('$firstname', '$lastname', '$email' '$password')";
        $result = mysql_query($query);
        if ( !mysql_insert_id() )
  {
  die("Error: User not added to database.");
  }
 else
  {
  // Redirect to thank you page.
  Header("Location: surveylanding_no-sidebar.html");
  }
    }
    ?>

Here is the HTML form:

<form name="htmlform" method="post" class="form" action="register.php"> 
   
    <p class="firstname"> 
        <input type="text" name="firstname" id="firstname" /> 
        <label for="firstname">First Name</label> 
    </p> 
     
    <p class="lastname"> 
        <input type="text" name="lastname" id="lastname" /> 
        <label for="lastname">Last Name</label> 
    </p> 
     
    <p class="email"> 
        <input type="email" name="email" id="email" /> 
        <label for="email">Email</label>
    </p> 
     <p class="Password"> 
        <input type="password" name="password" id="password" /> 
        <label for="password">Password</label> 
    </p> 
       
    <p class="submit"> 
        <input type="submit" value="Register"/> 
    </p> 
   
</form>

Hello Everyone,

I am pretty new to the forums and was curious if i could get some help here. Basically, in a nutshell, i have PayPal integrated into my website. I will use this to collect money from clients. when a client logs into his/her account they see their balance (which is pulled from the database to correspond with the user that's logged-in). Now, everytime a payment is submitted a notify_url is contacted after payment has been verified, that notify_url is the code written below. What I am trying to execute here is when this notify_url is called the current balance is reduced from the amount paid through paypal.

In the second If condition, you will see that the word success is being entered into the paypal.txt file, which is working perfectly fine. Now, you will also see the variable


$update_balance;  


which is suppose to update the original balance with the balance paid through PayPal

BUT IT'S NOT!! WHY?? LOL

Thank You in advance!


<?php
ob_start();

session_start();

include_once ('/home/rdewebde/public_html/includes/paypal.php');

$myPaypal = new Paypal();

$myPaypal->ipnLog = TRUE;
        
include_once "/home/rdewebde/public_html/includes/_config.php";

$username = "".$_SESSION['username']."";

$users_data = mysql_query("SELECT * FROM `members` WHERE `username`='".$username."'");

$user_info = mysql_fetch_array($users_data);

$current_amount = $user_info['balance'];
     
$deduct_amount = $myPaypal->ipnData['payment_gross'];

$new_amount = $current_amount - $deduct_amount;

$update_balance = mysql_query("UPDATE `members` SET `balance` = '$new_amount' WHERE `username` = '".$username."'");

if ($myPaypal->validateIpn()) {

    if ($myPaypal->ipnData['payment_status'] == 'Completed') {

         $update_balance;  

         file_put_contents('/home/rdewebde/public_html/lounge/paypal.txt', 'SUCCESS');

            

    }

else { file_put_contents('/home/rdewebde/public_html/lounge/paypal.txt', "FAILURE\n\n" . $myPaypal->ipnData); }

   }

?> 

I am very new to php and am trying to create a simple application that uploads a PDF file to a database. I have one field for the Volume Number and on file field for the PDF to be uploaded. My issue is i can't get the PDF to upload or insert the name of the pdf (eg volume1.pdf) into the data base.

I would also like to point out that I know i have a low post count, but i only seek help when i truly need it and have exhausted all other resources...

Here is what i have, please go easy on me this is my first round at php:
Code: [Select]
<?php
if(isset($_POST['submit'])){
$vol_num = $_POST['vol_num'];
$pdf = $_FILES['pdf']['name'];
$path = '../pdf/'.$_FILES['pdf']['name'];
move_uploaded_file($_FILES["pdf"]["tmp_name"], $path);
        mysql_query("INSERT INTO volumes set vol_num='$vol_num', vol_link='$pdf'") or die (mysql_error());
echo "<script>location.href='add_volume.php'</script>";
}
?>
what am i doing wrong here?

Thanks in advance

Hi All,

First time posting here. I've googled the problem, but can't seem to find a response that's the same.

All I want to do is have a list of id numbers and for each id number in the array, submit a MySQL query to retrieve information relating to the id number.

When I execute the code below however, I end up with only the last item in the array being printed in the echo statement.

Any clues?

Thanks,

Code: [Select]
// get array of ids
$ids = getIDs($ids);

// loop through input list
foreach ($ids as &$id) {
getVarDetails($id);
}

function getVarDetails($local)
{
$con = mysql_connect('localhost:3306', 'root', '********');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

// set database as Ensembl
mysql_select_db("Ensembl", $con);

$result = mysql_query("SELECT * FROM variations WHERE name = '$local' LIMIT 1");
$row = mysql_fetch_array($result)

while($row = mysql_fetch_array($result))
{
echo $row['name'] . " " . $row['id'];
echo "<br />";
}

// close connection
mysql_close($con);
}

I've been trying to simplify this code, let alone make it work.  It's a loop to change the different question checkboxes to "checked" if certain degree_ids are found.  All of the examples i have looked at are very strange and I'm not sure how to go about this, thought it seems to be a common question.  Everyone seems to have their own way.  Any advice would be greatly appreciated.

for($i =1; $i <= $arraycount; $i++){
$query = "SELECT degree_id FROM `user-degree` WHERE `user_id` = '".$user_id."' AND degree_id = '".$i."'";
$result = $conn->query($query) or die(mysql_error());
$row = mysqli_fetch_array($result) ;

echo $row['degree_id'];

if(!$y){
$y = 0;
}

if ($row['degree_id'] == 1){
$q4[0] = 'checked';
}
if ($row['degree_id'] == 2){
$q4[1] = 'checked';
}
if ($row['degree_id'] == 3){
$q4[2] = 'checked';
}
if ($row['degree_id'] == 4){
$q4[3] = 'checked';
}
if ($row['degree_id'] == 5){
$q4[4] = 'checked';
}
if ($row['degree_id'] == 6){
$q4[5] = 'checked';
}
if ($row['degree_id'] == 7){
$q4[6] = 'checked';
}
if ($row['degree_id'] == 8){
$q4[7] = 'checked';
}
if ($row['degree_id'] == 9){
$q4[8] = 'checked';
}
}

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

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.

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&nbsp;(<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