PHP - Create Some Formualrs With Php - Problem With Select-field

I want to select the user from super_administrators, administrators, teachers and students, and give the user permission based from what table he "came". But is giving the "Query failed" error...

Code: [Select]
<?php

//Start session

session_start();



//Include database connection details

require_once('../config/config.php');



//Array to store validation errors

$errmsg_arr = array();



//Validation error flag

$errflag = false;



//Connect to mysql server

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$link) {

die('Failed to connect to server: ' . mysql_error());

}



//Select database

$db = mysql_select_db(DB_DATABASE);

if(!$db) {

die("Unable to select database");

}



//Function to sanitize values received from the form. Prevents SQL injection

function clean($str) {



$str = @trim($str);



if(get_magic_quotes_gpc()) {

$str = stripslashes($str);

}



return mysql_real_escape_string($str);

}



//Sanitize the POST values

$email = clean($_POST['email']);

$password = clean($_POST['password']);



//Input Validations

if($email == '') {

$errmsg_arr[] = 'O campo Email nao foi preenchido.';

$errflag = true;

}



if($password == '') {

$errmsg_arr[] = 'O campo Senha nao foi preenchido.';

$errflag = true;

}



//If there are input validations, redirect back to the login form

if($errflag) {

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



//Create query

$qry = "SELECT * FROM super_administrators,administrators,teachers,students WHERE email = '$email' AND passwd = '".md5($_POST['password'])."'";

$result = mysql_query($qry);

$member = mysql_fetch_assoc($result);

$table = mysql_field_table('$result', '0');



//Check whether the query was successful or not



if($result) {



if(mysql_num_rows($result) == 1) {



if($table == 'super_administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/sadmin/index.php");

exit();

}



if($table == 'administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/admin/index.php");

exit();

}



if($table == 'teachers')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/prof/index.php");

exit();

}



if($table == 'students')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_CLASS_ID'] = $member['class_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_REGISTRATION'] = $member['registration'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/aluno/index.php");

exit();

}



} else {



$errmsg_arr[] = 'Suas informacoes de login estao incorreta. Por favor, tente novamente.';

$errflag = true;

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



} else {

die("Query failed");

}

?>

Can anybody clarify the usefulness of "AS" and elaborate on how it operates.

If I select a 'bananas' AS 'yeelow fruit' will it change the TITLE in the table?  Or the table that is being viewed?
If I dump a list of 'FRUITS' into and HTML table, will using AS rename each column for me, or is that handled by MY coing of the HTML table?

Hi Guys,

Can someone please help me before I kill myself!

I have a table holding details of images to be displayed in a gallery.
The table has columns as 'id', 'image_caption', 'file_path', and 'userid'.

The 'id' column is the unique key, and the 'userid' column is obviously id of the user who uploaded the image.

What I would like to do is select and display one image from each unique userid and then display each image along with the details in my gallery. It doesn't really matter which users image is selected aslong as there is only one for each unique user.

I have been trying a number of ways to do this (DISTINCT / subqueries) but I just cant get anything to work.
Could someone please advise me on how I could get this to work.

Cheers in advance.

Hello, i am trying to get my form to list usernames ascending as an option, but the select field just returns blank.

Here is my code:

Code: [Select]
/**
 * editUserForm - Displays the users database table in
 * a nicely formatted field table.
 */

function editUserForm(){
   $q = "SELECT * "
       ."FROM ".TBL_USERS." ORDER BY username ASC";
   $result = $database->query($q);

while ($row = mysql_fetch_assoc($result))
{
echo '<option value="'.$row["username"].'">'.$row["username"].'</option>';
}
}

Code: [Select]
              <form action="/" method="post">

                <div class="column">
                  <p>
                    <select name="user" id="user" placeholder="Select A User To Edit" class="{validate:{required:true}}">
                      <option>Select A User To Edit</option>
                      <? editUserForm(); ?>
                    </select>
                  </p>

                <div class="action_bar">
                  <input type="submit" class="button blue" value="Submit Post" />
                  <a href="#modal" class="modal button">Cancel</a>
                </div>

              </form>

All help is appreciated

Hello All   From the begin i have the following code.

<?php
$sql = "SELECT id,phonemodel FROM iphone";
$rows = $conn->sqlExec($sql);

$nr_row = $conn->num_rows;
$meniu ='<ul>';
if($nr_row>=0) {
  foreach($rows as $row) {
     $meniu .= '<li><a href="iphone.php?id='.$row['id'].'">'.$row['phonemodel'].'</a></li>';
	 }
  }
  $meniu .= '</ul>';
  
  echo $meniu;
	 
	 if(isset($_GET['id'])) {
	    $id =  (int)$_GET['id'];
		$sql = "SELECT * FROM iphone WHERE  id = $id";
$rows = $conn->sqlExec($sql);

$nr_rows = $conn->num_rows;
if($nr_rows>0) {
  foreach($rows as $row){
   echo 'Name Tel: '.$row["phonemodel"].' Title : '.$row["titlereparation"].' 
   Pret : '.$row["price"].' Message : '.$row["msj"].' ID : '.$row["id"].'<br />';
   }
  }
  else {
  echo '0 Results';
  }
}
?> 

Basicaly i have a website for phone repairs.And i want to create in iphone.php a menu from DB and when i acces the menu with _GET variable, when i press Iphone 5s (iphone.php?id=id page) the code have to display to me,all reparations for iphone 5s. Until now,i succssed to create the Menu but the code keep add same line in the menu when i add for example a second reparation for 5s.And i don't know how to select all reparation for 5s and display them in a single link like above Iphone 5s (iphone.php?id=id page). Now the script working like this.Create a menu with all phone names. Iphone 5s Iphone 3 Iphone 3s Iphone 4 Iphone 5s Iphone 5s Shoud be like Iphone 5s Iphone 3 Iphone 3s Iphone 4 Other new devices.. And display the reparation in every link from list By ID.i try it to select by phone names,not working. Any body with any ideea please? Thx so much

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=318572.0

I created a drop-down menu using a MySQL statement in php for a form.
My drop down menu works fine, but I want to assign a default value to it (normal text) the value will never change,
thus I do not need to extract the default value from the table, I already know the value.

Here is the basic snippet from the script:

<?php
   include("../includes/xxx.php");
   $cxn = mysqli_connect($host,$user,$password,$dbname);
   $query = "SELECT DISTINCT `plant_id` FROM `plant` ORDER BY `plant_id`";
   $result = mysqli_query($cxn,$query);
      while($row = mysqli_fetch_assoc($result))
      {
      extract($row);
      echo "<option value='$plant_id'>$plant_id</option>\n";
      }
?>

I want to make a table for each paddler_id (Field) which exists into pushup Database

My code for paddler_id = 1 is


require "../sql.php";

$result = mysql_query("EXPLAIN pushup");
$r = mysql_query("SELECT * FROM paddlerinfo t1 LEFT JOIN pushup t2 on t2.paddler_id=t1.id LEFT JOIN practicedate t3 on t3.practice=t2.practice_id ORDER BY t1.firstname ASC ");
$r1 = mysql_query("SELECT * FROM pushup  where paddler_id =1 ORDER BY practice_id ASC ");

echo "<table width='30%' border='1' cellpadding='0' cellspacing='0' style='font-family: monospace'>";
echo "<tr>";
echo "<td>DATE</td>";
while ($row = mysql_fetch_array($result)) {

if (in_array($row["Field"], array("paddler_id", "practice_id", "p_id"))) continue;

echo "<td>",($row["Field"]), "</td>";
}
echo "</tr>";

while (($data = mysql_fetch_array($r1, MYSQL_ASSOC)) !== FALSE) {
   unset($data["p_id"],$data["paddler_id"]);
echo "<tr>";
   foreach ($data as $k => $v) { 
echo "<td>";
      echo"$v";
echo "</td>";
}

echo "</tr>";
}


echo "</table>";

mysql_free_result($result);
mysql_free_result($r);
mysql_free_result($r1);
mysql_close();



The problem is on line $r1 = mysql_query("SELECT * FROM pushup  where paddler_id =1 ORDER BY practice_id ASC ");
I want to put something which makes it paddler_id = X where x = 2,3,4,5... and x exists in database and do not print twice the x (because that database may have rows where x = same id


How do i do that??

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

Hi. I will try and explain simply.   I have a list of item codes eg.  size1 Acolor1 size1 Bcolor3 size1 Bcolor1 size1 Acolor3 size1 Acolor4 size2 Acolor1 size3 Bcolor1 size3 Bcolor2 size3 Bcolor2 size3 Acolor1   I want to take these and split them into parts:   size1 Acolor2 size1 Bcolor3 etc.   Then I want to create a form with 2 dropdowns. The first option would just have all unique sizes so: size1 size2 size3   Then when one is selected, option 2 would show the colours available in that size.   Is there a standard way of doing this? Can I use PHP to create the array and then jquery to display and hide the options?   I'm totally stuck and i'm not just looking for somebody to code it all for me. any help would be greatly appreciated   thanks Sam

Or something like that... I am not sure how to put this..

Anyway, I'll just get started with explaining my problem.

I have an admin-page in which you can delete the comments given on blogs, using checkboxes and clicking on a button with the value 'verwijderenSubmit'. The deletion part works just fine, nothing wrong.

However, I also want to be able to EDIT the comments with an other button called 'bewerkenSubmit', using the same checkboxes that I use for deletion.
Selecting the right CID (CommentID) is no problem, because that works the same as the deletion-part, but selecting the right textarea to update into the database is the problem...

I uploaded a file here with the whole code: http://dhost.info/ddfs/myproblem.html
I escaped the textarea within with square brackets, because otherwise the whole textarea would screw up..
I also added <!-- RELEVANT CODE --> to select the parts that I need to change.

Well, I hope you understand my problem and can help.

Hi I have tried the mysql forum but have had no joy with an answer to my problem so wondered if php would be better.

I want my users to be able to select from 5 different drop down lists where they can chose any combination from 1 up to all 5,  I have attached the front end.  These lists are being populated from mysql tables.

Code for the drop down lists is as follows
Code: [Select]
  <form action="horse-events-devon.php?url_countyid=<?php echo  urlencode ($url_countyid ['url_countyid'])  ; ?>&go" method="POST">
    <table id="searchtable">
      <tr>
        <th>Find By Discipline</th>
        <th>Find By Venue</th>
        <th>Find By Championship</th>
        <th>Find By Organiser</th>
        <th>Equine Association</th>
        <th>Submit Your Selections</th>
        </tr>
      <tr>
        <td><select name="dis_id">
                                    <?php 
                                        $upcomingdis = upcomingdis($url_countyid);
                                        $upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis);
                                    ?>
                                    <?php 
                                    do { 
       ?>   <option value="<?php echo $upcoming_dis_bycounty ['dis_id']; ?>" >
<?php echo $upcoming_dis_bycounty ['dis_description']; ?></option>
<?php 
                                    } while ($upcoming_dis_bycounty = mysql_fetch_assoc ($upcomingdis)); 
                                    ?></select></td>
        <td><select name="ven_id">
                                    <?php 
                                        $upvenbycounty_set = upcoming_venevents_bycounty($url_countyid);
                                        $upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set);
                                    ?>
                                    <?php 
                                    do { 
       ?>   <option value="<?php echo $upcoming_ven_bycounty ['ven_id']; ?>" >
<?php echo $upcoming_ven_bycounty ['ven_name']; ?></option>
<?php 
                                    } while ($upcoming_ven_bycounty = mysql_fetch_assoc ($upvenbycounty_set)); 
                                    ?></select></td>
        <td><select name="champ_id">
                                    <?php 
                                        $championship_set = findchampionships();
                                        $champlist = mysql_fetch_assoc ($championship_set);
                                    ?>
                                    <?php 
do { 
?>   <option value="<?php echo $champlist ['champ_id']; ?>" >
<?php echo $champlist ['champ_description']; ?></option>
<?php 
} while ($champlist = mysql_fetch_assoc ($championship_set)); 
?></select></td>
        <td>
        <select name="org_id">
                        <?php 
                            $uporgbycounty_set = upcoming_organevents_bycounty($url_countyid);
                            $upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set);
                        ?>
                        <?php 
                        do { 
                        ?>   <option value="<?php echo $upcoming_org_bycounty ['org_id']; ?>" ><?php echo $upcoming_org_bycounty ['org_name']; ?></option>
                        <?php 
                        } while ($upcoming_org_bycounty = mysql_fetch_assoc ($uporgbycounty_set)); 
                        ?></select>
       
        </td>
        <td><select name="ass_id">
                                    <?php
                                        $upassbycounty_set = upcoming_assevents_bycounty($url_countyid);
                                        $upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set);
                                    ?>
                                    <?php 
                                    do { 
                                    ?>   <option value="<?php echo $upcoming_assbycounty ['ass_id']; ?>" >
<?php echo $upcoming_assbycounty ['ass_description']; ?></option>
                                    <?php 
                                    } while ($upcoming_assbycounty = mysql_fetch_assoc ($upassbycounty_set)); 
                                    ?></select></td>
      <td><input name="submit" type="submit" /><input name="countyid" type="hidden" value="<?php echo $url_countyid ['url_countyid']; ?>" /></td>
        </tr>
      </table>
  </form>
My search processing is as follows

Code: [Select]
  <?php
if (isset($_POST['submit'])){
if (isset($_GET['go'])){
$countyid = $_POST['countyid'];
$ven_id = $_POST['ven_id'];
$dis_id = $_POST['dis_id'];
$champ_id = $_POST['champ_id'];
$org_id = $_POST['org_id'];
$event_id = $row['event_id'];

$sql = "SELECT DATE_FORMAT (events.startdate, '%a, %d, %b') 
   AS stdate, events.event_id, events.title, events.ven_id, events.org_id,
   venue.county_id, venue.ven_id, eventdisciplines.event_id, eventdisciplines.dis_id, county.county_id, discipline.dis_id \n"
. "FROM events \n"
. "LEFT OUTER JOIN eventdisciplines \n"
. "ON events.event_id = eventdisciplines.event_id \n"
. "LEFT OUTER JOIN discipline \n"
. "ON eventdisciplines.dis_id = discipline.dis_id \n"
. "LEFT OUTER JOIN venue \n"
. "ON events.ven_id = venue.ven_id \n"
. "LEFT OUTER JOIN county \n"
. "ON venue.county_id = county.county_id \n"
. "WHERE events.ven_id = ({$ven_id} OR events.org_id = {$org_id})\n"
. "AND events.startdate > NOW()\n"
. "AND venue.county_id = {$countyid}  \n"
. "ORDER BY startdate ASC";

$result = mysql_query ($sql, $connection);
?>
How am I best to do this please?  my OR within the mysql does not work, should I not be doing this with php in the search processing?  someones help would really be appreciated, just to point me in the right direction.

[attachment deleted by admin]

I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly....
I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url---

its supposed to search a matching value and then output the matching value to url...

Here is the droplist code:


            $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'"  onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />';

this outputs a input text box--- i want to have a droplist of options to populate this text box...
If you must know this is the third day im at it...

Hello, I am having a problem with this code. I know it's messy, but it works.

The Problem is that you can register a username more then once, It detects there is already a username of that in the database, but it still inserts it, and then there would be two of the same username.

My Code:
<?php
    if($_SERVER['REQUEST_METHOD'] == 'POST') {
  if($_POST['username'] == "" || $_POST['password1'] == "" || $_POST['password2'] == "" || $_POST['email'] == "" || $_POST['rights'] == "" ||  $_POST['ipaddress'] == "")
  {
    echo '<p class="info" id="warning"><span class="info_inner">You left one or more fields blank.</span></p>';
  }
  else
  {
    $rr = mysql_query('SELECT * FROM users WHERE username=\'' . realEscape($_POST['username']) . '\'') ;
    if(mysql_num_rows($rr) > 0)
    {
    echo '<p class="info" id="error"><span class="info_inner">ERROR: The username is already in use!</span></p>';
    }
    $rrr = mysql_query('SELECT * FROM users WHERE email=\'' . realEscape($_POST['email1']) . '\'') ;
    if(mysql_num_rows($rrr) > 0)
    {
    echo '<p class="info" id="error"><span class="info_inner">ERROR: The email is already in use!</span></p>';
    } else {
      if($_POST['password1'] == $_POST['password2'])
      {
        if(preg_match('/[A-Za-z0-9-\s]{3,13}/i', $_POST['username'], $matches) && strlen($matches[0]) === strlen($_POST['username']))
        {
          if(preg_match('/[a-z0-9]{3,13}/i', $_POST['password1'], $matches) && strlen($matches[0]) === strlen($_POST['password1']))
          {
            if(is_numeric($_POST['rights'])) {
              mysql_query("INSERT INTO users (username, password, rights, ipaddress, email, date) VALUES ('". realEscape($_POST['username']) ."', '". encrypt($_POST['password1']) ."', '". realEscape($_POST['rights']) ."', '". realEscape($_POST['ipaddress']) ."', '". realEscape($_POST['email']) ."', NOW())");
              echo '
              <p class="info" id="info"><span class="info_inner">The account has been created.</p>';
            } else {
              echo '<p class="info" id="error"><span class="info_inner">ERROR: Undefined</span>';
            }
          }
          else
          {
            echo '<p class="info" id="error"><span class="info_inner">ERROR: Invalid password. Your password can only contain Numbers and Letters, and be 3-12 characters in length.</span></p>';
          }
        }
        else
        {
          echo '<p class="info" id="error"><span class="info_inner">ERROR: Invalid username. Your password can only contain Numbers and Letters, and be 3-12 characters in length.</span></p>';
        }
      }
      else
      {
        echo '<p class="info" id="error"><span class="info_inner">ERROR: Passwords do not match.</span></p>';
      }
    }
  }
  }
?>