PHP - Help Selecting Multi Tables From Mysql

Hi all,

I am trying to get data from MySQL to display in a html table in TCPDF but it is only displaying the ID.

$accom = '<h3>Accommodation:</h3>
<table cellpadding="1" cellspacing="1" border="1" style="text-align:center;">
<tr>

<th><strong>Organisation</strong></th>
<th><strong>Contact</strong></th>
<th><strong>Phone</strong></th>
</tr>
<tbody>
<tr>'.
$id = $_GET['id'];
$location = $row['location'];
$sql = "SELECT * FROM tours WHERE id = $id";
$result = $con->query($sql);

if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {

'<td>'.$location.'</td>
<td>David</td>
<td>0412345678</td>
</tbody>';
}
}
'</tr>
</table>';

Anyone got any ideas?

Hi, basically, here's the deal:
I have a lit of checkboxes that are added by the admin (there's an unlimited amount, just depends on how many are added). Then, those are put in a form, in which the user picks whichever ones need to be chosen and those values get sent to a MySQL table.

Here's the code that displays the checkboxes

Code: [Select]
<?php
$sql="SELECT * FROM category ORDER BY categoryID ASC";
$result=mysql_query($sql); 

while($row=mysql_fetch_array($result)){
    echo "<input type='checkbox' id='". $row['categoryName'] ."' name='licensed[]' value='" . $row['categoryID'] ."' /><label for='". $row['categoryName'] ."'><br>" . $row['categoryName'] ."</label><br>";
}
?>
What I'm making now, is an edit form where whichever checkboxes were checked, will show up checked in the edit form. But I'm not really sure how to go about this, since there is only one actual input tag in the code, I can't select the different ones.

I also have this SQL query which selects whichever boxes were inputted into the DB

Code: [Select]
$sql2="SELECT * FROM category, categoryInfo WHERE category.categoryID = categoryInfo.categoryID AND categoryInfo.parentID = $parentID";
$result2=mysql_query($sql2);
Where $parentID is the ID of whichever form you're editing.

But yes, I'm basically not really sure how to go about this and would like some help figuring this out

Thanks for your time

Currently my code picks a random number between 1-6 then extracts that many (random) entries from a list in my database.

My current code displays the whole array, how can I chop the array into 6 bits so I can output the data separately (up to 6 list items)?

$con = mysql_connect("x","x","x");
if (!$con)
   {
   die('Could not connect: ' . mysql_error());
   }

mysql_select_db("x", $con);

$ran_x = rand(1,6);
$appearance_query = "SELECT * FROM  `x` ORDER BY RAND( ) LIMIT $ran_x";
$get_x = mysql_query($x_query);

echo 'x(s): ';
echo '<br />';
echo "<ul style='list-style:none;'>";

while($row = mysql_fetch_array($get_x))
   {
echo "<li>" . $row['x'] . "</li>";
   }

echo "</ul>";
echo "<br />";

mysql_close($con);

Any help would be greatly appreciated,

Thanks,

otester

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

This portion is kind of stumping me. Basically, I have a two tables in this DB: users and users_access_level (Separated for DB normalization)

users: id / username / password / realname / access_level
users_access_level: access_level / access_name

What I'm trying to do, is echo the data onto an HTML table that displays users.username in one table data and then uses the users.access_level to find users_access_level.access_name and echo into the following table data, I would prefer not to use multiple queries if possible or nested queries.

Example row for users: 1234 / tmac / password / tmac / 99
Example row for users_access_level: 99 / Admin

Using the examples above, I would want the output to appear as such:

Username: Access Name: Tmac Admin
I am not 100% sure where to start with this, but I pick up quickly, I just need a nudge in the right direction.

The code I attempted to create just shows my lack of knowledge of joining tables, but I'll post it if you want to see that I did at least make an effort to code this myself.

Thanks for reading!

We recently upgraded from PHP4 to PHP5 and the below script that was working perfectly in 4 has completely stopped working and I can't figure out why for the life of me. I'm not an experienced PHP programmer--I've done some forms, but this is the first time I've used a database.

What needs to (and was) happen: A user enters their username in the form and gets a readout of their participation so far that month.

The problem(s): I know that it's storing the variable 'user' because it echoes it back properly, but the database is no longer allowing me to select the row based on that variable. I know it's not that I can't connect to the database because if instead of '$user' I change the code to a username I know is in there, I get the proper readout. This all started as soon as I transferred over to PHP5--before that, no problems at all.

The database information is all correct, I just took it out for privacy's sake.

<form id="feedback" method="post" action="index.php">

<input name="user" type="text" value="Enter user name" size="20" maxlength="50" /><br />

<input name="send" id="send" type="submit" value="Submit" />
</form>
<?php
if (isset($_POST['user'])) { 
$_session['user'] = $_POST['user'];
} 
?>

<p>You entered your username as: <strong><? echo $_session['user'];?></strong>. If this is not correct or you do not see your information below, please re-enter your username and click Submit again.</p>

<?php
$con = mysql_connect("database","username","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("database", $con);

$result = mysql_query("SELECT * FROM March WHERE Username='$user'") or die ('Error: '.mysql_error ());

while($row = mysql_fetch_array($result))
  {
  echo "<table border='0'>";
  echo "<tr>";
  echo "<td><strong>Username:</strong> </td><td>" . $row['Username'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 2 Discussion:</strong></td><td>" . $row['Mar2Q'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 2 Poll:</strong></td><td>" . $row['Mar2P'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 9 Discussion:</strong></td><td>" . $row['Mar9Q'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 9 Poll:</strong></td><td>" . $row['Mar9P'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 16 Discussion:</strong></td><td>" . $row['Mar16Q'] . "</td></tr>";
  echo "<tr><td><strong>Mar. 16 Poll:</strong></td><td>" . $row['Mar16P'] . "</td></tr>";
  echo "<tr><td><strong>March Participation To-Date:</strong></td><td>" . $row['Participation'] . "</td></tr>";
  echo "</tr>";
  } 
  echo "</table>";

mysql_close($con);
?> 

ANY help would be greatly appreciated! I've got a couple hundred people who use this on a regular basis and are starting to ask why it's not working.

SELECT * FROM `booking_tbl` WHERE booking_status = 'Check In' AND departure_date_time = NOW()"

I use the datetime datatype for the departure_date_time so i can get data from that data because it checking the date and time   but i want it to check on the date from the datetime datatype   So how can i get the data only with that date without the time in the mysql datetime datatype

Hello, I have been at this for awhile now and cant seem to figure it out.  I have 2 separate arrays like shown below and I am trying to insert each value into my database using one update query. I have tried a for loop inside each array, but only one column populates.  I have tried breaking the values outside their for loop, etc, etc.

Does anyone know a way this can be done?

Code: [Select]

//Cars
$cars = array('ford', 'toyota', 'bmw');

//Places
$places =arrray('Orlando','California','Texas');

 foreach ($cars as $car){
          foreach ($places as $place){
          $place .=$place;
         }
 mysql_query("INSERT INTO vehicles  (cars, places) VALUES ('" . $car. "', '" . $place . "'')");
}

I am currently creating a multi-user login system, and I have created the database in MySQL, but the problem is that my PHP script is not working as expected. I would like it so that if the user enters in a correct username + password combination, then they are redirected to a webpage called "congrats.php" for example. However, with my current PHP code, if I include a redirection instruction based on the correct input, then when I run the script, then the user is instantly taken to the congrats.php page without filling in any login details. I'm quite sure that the database has been connected to, as due to the layout of my script at the moment, the text that reads "You did it!" appears 4 times, the same number of rows in my MySQL database. My PHP coding for this is:

Code: [Select]
<html><head><title>Multi-User Log In Form</title></head>
<body>
<?php
$self = $_SERVER['PHP_SELF'];
$username = $_POST['username'];
$password = $_POST['password'];
?>
<form action = "<?php echo $self; ?>" method = "post">
Username <input type = "text" name = "username" size = "8">
Password <input type = "password" name = "password" size = "8">
<input type = "submit" value="Submit">
</form>
<?php
$conn = @mysql_connect("localhost", "root", "") or die ("Err: Conn");
$rs = @mysql_select_db("test3", $conn) or die ("Err: Db");
$sql = "select * from users";
$rs = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($rs)) {

$name = $row["uname"];
$pass = $row["pword"];
if ($username = $name && $password = $pass)
{
//CODE FOR REDIRECTION SHOULD GO HERE
//header("location:congrats.php");
echo "You did it!";
}
else
{
echo "Invalid username and password combination";
}
}
?></body>
</html>

Any help in trying to get it so that my PHP script will redirect only if a correct username + password combination is entered would be greatly appreciated

Hi I am trying to select and order data/numbers from a colum in a mysql data base however i run the code and it returns no value just a blank page no errors or any thing so i think the code is working right but then it returns no result? Please help thanks

Here is the code:

<?php
$host= "XXXXXX";
$mysql_user = "XXXXXX";
$mysql_password = "XXXXXX";
$mysql_database = "XXXXXXX";

$connection = mysql_connect("$host","$mysql_user","$mysql_password") or die ("Unable to connect to MySQL server.");
mysql_select_db($mysql_database) or die ("Unable to select requested database.");

$row = mysql_fetch_assoc( mysql_query( "SELECT XP FROM Game ORDER BY number DESC LIMIT 1" ) );
$number = mysql_result(mysql_query("SELECT XP FROM Game ORDER BY number DESC LIMIT 1"), 0);

echo "The the highest XP is $number";

?>

i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel

<div class="nyie-outer">    
<div class="container">
    <div  id="nyieFull" class="carousel slide carousel-multi-item" data-interval="false" data-ride="carousel">
        <div class="outer-controls-top">
        <div class="nyie-title"><h5>THIS IS NYIE</h5></div>
        <div class="controls-top">    
            <a class="btn-floating" href="#nyieFull" data-slide="prev"><i class="fa fa-chevron-left"></i></a>
            <a class="btn-floating" href="#nyieFull" data-slide="next"><i class="fa fa-chevron-right"></i></a>
        </div>
        </div>
        <div class="carousel-inner" role="listbox">
            <div class="carousel-item active">
                 first row

                 second row

                 third row

                 fourth row
            </div><!--/.First four rows here-->
            <div class="carousel-item">

                  fifth row

                 sixth row

                 seven throw

                 eighth row

            </div><!--/.second four rows here-->
        </div
    </div>    
</div> 
</div>

 

 

sql code   

CREATE TABLE  product(
prodId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, 
image VARCHAR(255) NOT NULL,
prodName VARCHAR(255) NOT NULL,
prodPrice VARCHAR(255) NOT NULL,
description TEXT NOT NULL
PRIMARY KEY (prodId) 
);

php code

<?php 
include_once('db.php');
$sql = "SELECT image,prodName, prodPrice, description FROM product ORDER BY prodId ASC LIMIT 4";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
    // output data of each row
        while($row = mysqli_fetch_array($result)) {
            $activeprod =  '
            <div class="col-lg-3">
            <a href="more.php" class="newAdded">    
            <figure class="figure">   
               <img src="images/'. $row['image'] .'" alt="" class="img-responsive" >
                <figcaption class="figure-caption">
                    <span class="prodName"><b>Name: </b>'. $row['prodName'] .'</span>
                    <span class="prodPrice"><b>price: </b>'. $row['prodPrice'] .'</span>
                    <span class="desc"><b>description: </b>'. $row['description'] .'</span>    
                </figcaption>    
            </figure> 
            </a>
            </div>
            ';
    }
} 
?>

how can i echo that result in those rows 

 

 

 

just trying to see if i have this right in my head here..

if i have a page that has a topic on it and many replies,  the tables would look like this..


would this be the right code to pull out the topic AND all responses for say.
www.mysite.com/view?id=1

$topic = 'SELECT * from topics where id = $id'

$replies = 'SELECT * from replies where replies.topic_id = $id'

Okay so I have 2 tables in my database. One called user and one called messages. A user logs in to the message board and leaves a message (eg nice website). They write in the author name and the message then after the message is posted it says

"Nice website"
Posted by (author) on (date).

All is good so far. It works. However if you look at my code you will see I have a session started. This session is storing the username of the logged in user. From the column username in the users table. (This table has has an id for each user). Ive played around with the code trying to make it so the user doesnt have to fill in the author box. I want rid of that box So the logged in user just leaves a message then it says "posted by (username) on (date).

Im missing something from my code.
Can anyone tell me what? Please?

<?php
session_start();
mysql_connect("*************", "*****************", "***************");
mysql_select_db("***********************");
$time = time();



//this checks to see if the $_SESSION variable has been not set 
//or if the $_SESSION variable has been not set to true
//and if one or the other is not set then the user gets
//sent to the login page
if (!isset($_SESSION['username'])) {
    header('Location: http://***************.com/login.php');
}

$query = "INSERT INTO messages VALUES( NULL, '". mysql_real_escape_string($_POST['message']) ."', '". mysql_real_escape_string($_POST['username']) ."', '$time' )";if( $result = mysql_query($query) ) {
        if(mysql_affected_rows() > 0 ) {
                echo "Message Posted.<br><a href='messageboard.php'>Return</a>";
        } else {
                echo 'There was an error posting your message. Please try again later.';
        }
} else {
        echo "There was a database error.";
        // comment out next line for live site.
        echo "<br>Query string: $query<br>Returned error: " . mysql_error() . '<br>';
}


;

Heyyzarrh, iv been having troubles trying to make it so my PHP script will echo out the username by corresponding the posters ID with the ID in the members table. Heres what iv got so far:

Code: [Select]
<?php
if ($urlid == "") {
}

$sql = mysql_query("SELECT members.username,
posts.mem_id,
posts.post_date,
posts.post
FROM members
JOIN posts
ON members.id = posts.mem_id
WHERE posts.mem_id='$urlid' ORDER BY post_date DESC LIMIT 20");

while ($row = mysql_fetch_array($sql)){
    echo '<b>' .$row['$username']. ':</b> '.$row['post'].'<br />At: '.$row['post_date'].'<br /><br />';
}
?>

Please help!! :S so far only the actual posts data comes out not the username from the table "members"...