PHP - Show New Data From Mysql On The Same Page Without Reloading

Hi again PhpFreaks, yet again I got a problem regarding coding.

Hi,

How can I get others informations on another page by clicking on one row's value

I have this code, but it doesn't work:


while($row = mysql_fetch_array($result))
  {
 echo "<tr><p>";
 
  echo "<th><p>" . $row['fname'] . "&nbsp;&nbsp;&nbsp;" . $row['lname'] . "</p></th>";
  echo"<th><p><a href='student.php?id='".$row['topic']."'\'>".$row['topic']."</p></a></th>";

  echo "</tr>";
  }
echo "</table>";


Student.php


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

mysql_select_db("college", $con);

$result = mysql_query("SELECT * FROM students");
$topic = $_Get['topic'];
while($row = mysql_fetch_array($result))
  {
  
 echo "<tr><p>";
  echo "<th><p>" . $row['month']."&nbsp;" .$row['datetime']."</p></th>";
  echo"<th><p>". $row['topic'] ."</p></th>"; 
  echo "<th><p>" . $row['gender'] . "</p></th>";
  echo "<th><p>" . $row['fname'] . "&nbsp;&nbsp;&nbsp;" . $row['lname'] . "</p></th>";
  echo "<th ><p>" . $row['id'] . "</p></th>";
  echo "</tr>";
  
echo "</table>";

mysql_close($con);
  }
?> 

I have a php generated image:

(number.txt contains a number 1 or 2)

pic.php
Code: [Select]
<?php
// Set the content-type
header('Content-type: image/bmp');
// Create the image
$im = imagecreatetruecolor(80, 80);

$count_my_page = ("number.txt");
$hits = file($count_my_page);


if ($hits[0] == 1) {
$im = imagecreatefrompng("image1.png");
$hits[0] ++;
$fp = fopen($count_my_page , "w");
fputs($fp , "$hits[0]");
fclose($fp);
}else{
$im = imagecreatefrompng("image2.png");
$hits[0] --;
$fp = fopen($count_my_page , "w");
fputs($fp , "$hits[0]");
fclose($fp);
}

// Create some colors
$white = imagecolorallocate($im, 255, 255, 255);
$black = imagecolorallocate($im, 1, 1, 1);

// Using imagepng() results in clearer text compared with imagejpeg()
imagepng($im);
imagedestroy($im);
?>
And I have another page

index.html
Code: [Select]
<img src="pic.php">
<img src="pic.php">

Basically, I need one of the images on the index page to be image1 and then the second image to be image2, but it must be contained within the php file because I am unable to edit index.html. Is there anyway to do this? The method I have now where it reads a file to see if it's value is 1 or 2 doesn't seem to compute each time the image is generated. I have tried
header("Cache-Control: no-cache");
header("Pragma: no-cache");
in the image code to force it not to cache but to no avail. Can anyone solve this?

hey guys im sure this is possible im wondering how a user can execute a mysql query by a click of a button but without the page reloading...thanks guys

Hi everybody

I am a PHP rookie for now, and
I want to reload the same dynamic page - it's content in several languages - by clicking little flags which are located at some point on the page - AND ITS NOT WORKING.

Clicking on the flag triggers a PHP script which basically says

Code: [Select]

]$_SESSION['lang'] = <Some_language>;
require ('webpage.php');


where webpage.php is the dynamic page, in different languages, with it's content drawn out from a database (which contains the content in all languages I chose to display, blah, blah)
Funny thing - when I look at the page html (View Page Source) - it seems OK (identical from one page to the next) ! But WHAT is displayed after the first attempt to change the language is not.
Is something fundamentally wrong with my approach ?

Mike

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=305930.0

Hello,
Im getting some data from a database. Im showing them in a php page.
well its like this... Its a song and lyric site. Im displaying the whole song titles and lyrics when a user search.
there's separate page to list all the songs available. Which displays all the TITLES of the songs.

How can I make those Titles LINKS to view the lyrics when clicked??
A tutorial regarding a thing like this would be much helpful..
Thanks...

I used to be good at this but I changed servers and everything is different...  Heres my code so far:

Code: [Select]
<?php
$rated=$_REQUEST['rated'];
echo $rated;
$rating=$_REQUEST['rating'];
echo $rating;

// Make a MySQL Connection
mysql_connect("localhost", "********", "********") or die(mysql_error());
mysql_select_db("*********") or die(mysql_error());

$result = mysql_query("SELECT * FROM main WHERE username = '$rated'")
or die(mysql_error());  

$row = mysql_fetch_array( $result );

$votes = $db_field['$rating'];
$newvotes = $votes + 1;

echo $newvotes;

mysql_query("UPDATE main SET $rating = '$newvotes'
WHERE username = '$rated'");


?>
Whats going on here is the colomb that I want to update comes as a variable $rated (That works) and then the database selects the  row to update with $username (That works) and gets the variable $newvotes by taking the original value of the data its about to update and add 1 to it (That works)  Then it updates the field to $newvotes....

I don't know why the update won't go through... there are no errors....

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=314750.0

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=326913.0

Ok i am stuck, i created a php script that draws data from a mysql dB into a php page. How can I send the page as the body in email? I do not need to preview the page, but can send it without ever viewing if that helps. When I put my code into a variable it does nothing. Is it caching? Should I load the results into mysql then send?please advise..

Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way.

First set up everything for DB:

<?php
//connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//create the main database if it doesn't already exist
$query = 'CREATE DATABASE IF NOT EXISTS moviesite';
mysql_query($query, $db) or die(mysql_error($db));

//make sure our recently created database is the active one
mysql_select_db('moviesite', $db) or die(mysql_error($db));

//create the movie table
$query = 'CREATE TABLE movie (
        movie_id        INTEGER UNSIGNED  NOT NULL AUTO_INCREMENT,
        movie_name      VARCHAR(255)      NOT NULL,
        movie_type      TINYINT           NOT NULL DEFAULT 0,
        movie_year      SMALLINT UNSIGNED NOT NULL DEFAULT 0,
        movie_leadactor INTEGER UNSIGNED  NOT NULL DEFAULT 0,
        movie_director  INTEGER UNSIGNED  NOT NULL DEFAULT 0,

        PRIMARY KEY (movie_id),
        KEY movie_type (movie_type, movie_year)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die (mysql_error($db));

//create the movietype table
$query = 'CREATE TABLE movietype (
        movietype_id    TINYINT UNSIGNED NOT NULL AUTO_INCREMENT,
        movietype_label VARCHAR(100)     NOT NULL,
        PRIMARY KEY (movietype_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

//create the people table
$query = 'CREATE TABLE people (
        people_id         INTEGER UNSIGNED    NOT NULL AUTO_INCREMENT,
        people_fullname   VARCHAR(255)        NOT NULL,
        people_isactor    TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,
        people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,

        PRIMARY KEY (people_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Movie database successfully created!';
?>



********************************************************************
***********************************************************************

second file to load info into DB:



<?php
// connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//make sure you're using the correct database
mysql_select_db('moviesite', $db) or die(mysql_error($db));

// insert data into the movie table
$query = 'INSERT INTO movie
        (movie_id, movie_name, movie_type, movie_year, movie_leadactor,
        movie_director)
    VALUES
        (1, "Bruce Almighty", 5, 2003, 1, 2),
        (2, "Office Space", 5, 1999, 5, 6),
        (3, "Grand Canyon", 2, 1991, 4, 3)';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the movietype table
$query = 'INSERT INTO movietype
        (movietype_id, movietype_label)
    VALUES
        (1,"Sci Fi"),
        (2, "Drama"),
        (3, "Adventure"),
        (4, "War"),
        (5, "Comedy"),
        (6, "Horror"),
        (7, "Action"),
        (8, "Kids")';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the people table
$query  = 'INSERT INTO people
        (people_id, people_fullname, people_isactor, people_isdirector)
    VALUES
        (1, "Jim Carrey", 1, 0),
        (2, "Tom Shadyac", 0, 1),
        (3, "Lawrence Kasdan", 0, 1),
        (4, "Kevin Kline", 1, 0),
        (5, "Ron Livingston", 1, 0),
        (6, "Mike Judge", 0, 1)';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Data inserted successfully!';
?>



**************************************************************
****************************************************************

MY CODE FOR THE QUESTION:


<?php
$db = mysql_connect('localhost', 'root', '000') or
die ('Unable to connect. Check your connection parameters.');

mysql_select_db('moviesite', $db) or die(mysql_error($db));

//get our starting point for the query from the URL
if (isset($_GET['offset'])) {
   $offset = $_GET['offset'];
} else {
   $offset = 0;
}

//get the movie
$query  = 'SELECT
   movie_name, movie_year
FROM
   movie
ORDER BY
   movie_name
LIMIT ' . $offset . ' , 1';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_assoc($result);
?>
<html>
<head>
<title><?php echo $row['movie_name']; ?></title>
</head>
<body>
<table border = "1">
<tr>
<th>Movie Name</th>
<th>Year</th>
</tr><tr>
<td><?php echo $row['movie_name']; ?></td>
<td><?php echo $row['movie_year']; ?></td>
</tr>
</table>
<p>
<a href="page.php?offset=0">Page 1</a>,
<a href="page.php?offset=1">Page 2</a>,
<a href="page.php?offset=2">Page 3</a>
</p>
</body>
</html>

I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith.

The cookie doesn't show if I go to that url page. But it does show up once I reload the page.  So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page?

Here is my code.

// This is in the header
$url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null);

if(!empty($url_ref_name)) {

  $number_of_days = 365;
  $date_of_expiry = time() + 60 * 60 * 24 * $number_of_days;
  setcookie( "ref", $url_ref_name, $date_of_expiry,"/");

} else if(empty($url_ref_name)) {
  if(isset($_COOKIE['ref'])) {
    $user_cookie = $_COOKIE['ref'];
  }
} else {}

// This is for the sign up form
if(isset($_COOKIE['ref'])) {
  $user_cookie = $_COOKIE['ref'];
  ?>
  <fieldset>
    <label>Referred By</label>
    <div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div>
    <input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" />
  </fieldset>
  <?php
}

 

create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";

$result = mysqli_query($conn, $sql );

$mimi = mysqli_fetch_assoc($result);

$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';

$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?

i want to download pdf or text document after clicking button or link how to do that

I have a php tutorial that I followed for creating, inserting, selecting and updating a MySQL database with php.  Everything works fine so I wanted to put it into Wordpress.  I took the code and placed it into the wordpress page and everything worked just fine except the update function. 

Here is the tutorial I used. http://www.phpsimple.net/mysql_insert_record.html  It is also the part I am having trouble with.  I am able to create a record and I am also able to select a record but when I choose "update" the form doesn't load the data.  It will outside the website but not inside wordpress.  I am hoping this is not vague but because of my inexperience I am not sure what else to say.  I will be more than happy to provide any other information you need.