PHP - Simple Script To Show Written Text On Different Page

Hi,

I want to show part of the text from a page containing my article on my main page. The article resides in a database in mysql. How do I limit the amount of text that is shown without setting up a separate excerpt box for the article. I want to then provide a link so the user can go to another page that displays the entire article. Thanks for any suggestions.

Hi Guys,
I wonder if you can help me before I go crazy? I have the task of updating a site by adding text to every page, at the moment there are three pages on the live website that show both images and text, I am to update the remaining 6 with text to go with the images.... I thought this was all fine and dandy, added the text in, checked the code, uploaded to the FTP site and nothing, only 1 page worked. I am stuck as I have used the same method to add text to all the pages yet those elusive pages refuse to show text! This is driving me crazy and i am hoping that one of you guys has come across this before and will be able to highlight a missing piece of code or have a solution?

I do appreciate your time and any suggestions that you have. I have attached the page below. Thanks guys!

Hi,   In reference to my first attached image, I have a form which displays two SELECT/drop-down fields (labeled "Store Name" and "Item Description".....and both of which pull-in values from two separate lookup/master tables, in addition to providing an additional option each for "NEW STORE" and "NEW ITEM").   Now, when first-run, and/or if "NEW STORE" and "NEW ITEM" are not selected from the drop-down's then the two fields in green ("New Store Name" and "New Item Name" are hidden, by means of the following code:  

<div class="new-store-container" id="new-store-container" name="new-store-container" style="display:none;">
    <div class="control-group">
        <div class="other-store" id="new_store_name">
            <?php echo standardInputField('New Store Name', 'new_store_name', '', $errors); ?>
        </div>
    </div>
</div>

  Conversely, if "NEW STORE" and/or "NEW ITEM" are selected from the two drop-down's then one (or both) of the "New Name" fields are unhidden by means of the following two pieces of code, one PHP and the second JS:    

<select class="store-name" name="store_id" id="store_id" onclick="toggle_visibility('store_id','new-store-container')">
    <?php echo $store_options; ?>
                            
    <?php
        if($values['store_id'] == "OTH")
            {
                echo "<option value='OTH' selected> <<<--- NEW STORE --->>> </option>";
            }
        else
            {
                echo '<OPTION VALUE="OTH"> <<<--- NEW STORE --->>> </OPTION>';
            }
    ?>
</select>

 

function toggle_visibility(fieldName, containerName)
{
    var e = document.getElementById(fieldName);
    var g = document.getElementById(containerName);
            
    if (e.value == 'OTH')
        {
            if(g.style.display == 'none')
                g.style.display = 'block';
            else
                g.style.display = 'none';
        }
}

  All  of that is working just fine. The problem I'm having is that when I click the "Create" button, after having left any one of the form fields blank, the two "New Name" fields are hidden again, which I don't want to happen i.e. I want them to remain visible (since the values of "store_id" and/or "item_id" are "OTH"), so that the user can enter values into one or both of them, without havng to click on the drop-down a second time in order to execute the "on-click" code.   The second attached image shows how the fields are hidden, after clicking "Create".   How can I achieve that? It would be greate if someone could cobble-up the required code and provide it to me, since I'm relatively new to this.   Thanks much.    Snap1.png   26.14KB   0 downloads    Snap2.png   149.47KB   0 downloads      

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
}

 

Hello, im very green to php and I am having trouble creating a simple log in script. Not sure why this is not working, maybe a mysql_query mistake? I am not receiving any errors but nothing gets updated in the members table and my error message to the user displays.  any help is appreciated! here is my php:


<?php
session_start();

$errorMsg = '';
$email = '';
$pass = '';

if (isset($_POST['email'])) {
   $email = ($_POST['email']);
   $pass = ($_POST['password']);
   
   $email = stripslashes($email);
    $pass = stripslashes($pass);
   $email = strip_tags($email);
   $pass = strip_tags($pass);
   
if ((!$email) || (!$pass)) {
   $errorMsg = '<font color="#FF0000">Please fill in both fields</font>';
}else {
   include 'scripts/connect_db.php';
   $email = mysql_real_escape_string ($email);
   $pass = md5($pass);
   
   $sql = mysql_query("SELECT * FROM members WHERE email='$email' AND password='$pass'");
   
   $log_check = mysql_num_rows($sql);
   
if ($log_check > 0) {
    while($row = mysql_fetch_array($sql)) {
      
      $id = $row["id"];
      $_SESSION['id'];
      $email = $row["email"];
      $_SESSION['email'];
      $username = $row["username"];
      $_session['username'];
      
      mysql_query("UPDATE members SET last_logged=now() WHERE id='$id' LIMIT 1");
      
   }//Close while loop
   echo "You are logged in";
             exit();
}
else {
   $errorMsg = '<font color="#FF0000">Incorrect login data, please try again</font>';
}


     }
}
?>



and the form:

<?php echo $errorMsg; ?>
<form action="log_in.php" method="post">
Email:<br />
<input name="email" type="text" /><br /><br />
Password:<br />
<input name="password" type="password" /><br /><br />
<input name="myBtn" type="submit" value="Log In" />
</form>

Hi,

I would like to do the following but not sure how.

If the you/user is on index.php of http://www.domain.com/ show one page
If not show another

How would I do this?

Thanks

ok, I am hoping the code is self explanatory... but if not, I'm creating a script which I will be using for something else. Basically it says if OS and browser show code under...

I am looking for two things.

1. Is there a way to make it so that it can pull from a list and show that if HTTP_USER_AGENT comes back with a specific output, to say you are using X OS with Y browser?

2. The last bit, it is meant to say, if not any of the options show the info below.  For some reason it's not doing that... can someone please tell me why?

Thanks.

Code: [Select]
<?php

//Operating Systems
function xp(){return(eregi("Windows NT 5.1", $_SERVER['HTTP_USER_AGENT']));}
function vista(){return(eregi("Windows NT 6.0", $_SERVER['HTTP_USER_AGENT']));}
function win7(){return(eregi("Windows NT 6.1", $_SERVER['HTTP_USER_AGENT']));}
function ubuntu(){return(eregi("Ubuntu/", $_SERVER['HTTP_USER_AGENT']));}


//Web Browsers
function chrome(){ return(eregi("Chrome/", $_SERVER['HTTP_USER_AGENT']));}
function safari(){ return(eregi("Safari/", $_SERVER['HTTP_USER_AGENT']));}
function firefox(){ return(eregi("Firefox/", $_SERVER['HTTP_USER_AGENT']));}
function ie9(){ return(eregi("MSIE 9.0", $_SERVER['HTTP_USER_AGENT']));}
function ie8(){ return(eregi("MSIE 8.0", $_SERVER['HTTP_USER_AGENT']));}

// do something if XP and Chrome
if(xp() && chrome()){echo 'You are using Windows XP with a Chrome web Browser';}

// do something if XP and IE8
if(xp() && ie8()){echo 'You are using Windows XP with a Internet Explorer 8 web Browser';}





// do something if Windows 7 and IE9
if(win7() && ie9()){echo 'You are using Windows 7 with a Internet Explorer 9 web Browser';}

// do something if Windows Vista and IE9
if(vista() && ie9()){echo 'You are using Windows Vista with a Internet Explorer 9 web Browser';}

// do something if Windows Vista and IE8
if(vista() && ie8()){echo 'You are using Windows Vista with a Internet Explorer 8 web Browser';}

// do something if Ubuntu and Firefox
if(ubuntu() && firefox()){echo 'You are using Ubuntu with a Firefox web Browser';}

if(!ubuntu() || !xp() || !vista() || !win7() || !firefox() || !chrome() || !safari() || !ie9() || !ie8()){
echo'<strong>';
echo '<br />' . $_SERVER['HTTP_USER_AGENT'] . '<br /><br />Administrator someone in your work force is using an unsupported browser/OS combination, please email the information above to the developer of the NCMR software you are using. It will allow your browser/OS combination  to be used correctly. Sorry for the inconvenience.</strong> <br /><br />Please copy and paste the text above and send it to your web administrator. It will explain everything he/she needs to do.';}


?>

Hi can someone pls help, im tryin a tutorial but keep getting errors, this is the first one i get after registering.

You Are Registered And Can Now Login
Warning: Cannot modify header information - headers already sent by (output started at /home/aretheyh/public_html/nealeweb.com/regcheck.php:43) in /home/aretheyh/public_html/nealeweb.com/regcheck.php on line 46

Hi,
In my application I have not used any folders to define the hierarchy. All file s are in a single folder.

What is the easy way of creating the Breadcrumb.
I am rewriting the URL from
www.mysite.com/main.php/page=pagename&id=10
to
www.mysite.com/main/pagename/id/10

Is it possible to use the re-written URL to create the breadcrumb? If yes I can define hierarchy in the URL and use the same to create breadcrumb.

I can't get images to show up in my script. They show up as broken images and I can figure it out. I think it's a problem with (') or this ("). Any help would be appreciated.

$query = 'SELECT * FROM images';
$stmt = $dbh->prepare($query);
$stmt->execute();
$counter = 0;
$numberperrow = 3;  
   
    while($row = $stmt->fetch()) {
    if($counter < $numberperrow)
    {
    echo '<img src='.$row['image_th'].' width=200>';
    
    /**$counter = $counter + 1;**/
    echo ' ';
    $counter = $counter + 1;
    }else{
    
    echo '<br>';
    $counter = '1';
   echo '<img src='.$row['image_th'].' width=200>';
  
    echo '    ';
    
    }
    }
	 
}
catch(PDOException $e)
{
	echo $e->getMessage();
}

Hi guys. I have a problem in my web based examanation system written in php.

Here's the scenario:

when a user takes an exam. 1 question is displayed 1 at a time. my problem is if he/she is at the 2nd question then he/she pressed the back button he/she is redirected back to the 1st question and my system just go crazy. How can i prevent the user from pressing the back button or returning to the previous questions? Please I really need help on this. Can anyone give me suggestion for possible solutions on my problem?

Thanks in advance

by the way here's my code, you can also download it from attachment.

------------------------------------------process_exam.php------------------------------------------------------------

<?php
   
   //start session
   session_start();

   //declare variables
   //iCorrect= variable for score
   $_SESSION['iCorrect']=0;

   //a=variable for iteration kang index kang mga array or serves as an current item number
   $_SESSION['a']=1;

   //connect to database
   include('mysql_connect.php');

   $result = mysql_query("select * from questions where exam_id ='$_SESSION[exam_id]'");
   
   //count the number of found questions
   $_SESSION['num_of_questions']=mysql_num_rows($result);
   
   //declare the iterator
   $i=1;
   
   //get the options for the question1,question2 and so on and so on. depende kung ilang questions ang mafetch nya sa dbase.so pde kht ilang questions lagay ng user
   while ($row = mysql_fetch_assoc($result))
   {
      $questions[$i] = $row['question'];
      $op1[$i] = $row['op1'];
      $op2[$i] = $row['op2'];
      $op3[$i] = $row['op3'];
      $op4[$i] = $row['op4'];
      $answers[$i]=$row['answer'];
      $i++;
   }

   //view the data's in the array for debugging purposes
   echo 'answers: ';
   print_r($questions);
   echo '<br>';
   print_r($op1);
   echo '<br>';
   print_r($op2);
   echo '<br>';
   print_r($op3);
   echo '<br>';
   print_r($op4);
   echo '<br>';
   print_r($answers);
   echo '<br>';
   echo '<br>';


   //pass the arrays to session arrays to be used in other php pages
    $_SESSION['question']=$questions;
    $_SESSION['op1']=$op1;
    $_SESSION['op2']=$op2;
    $_SESSION['op3']=$op3;
    $_SESSION['op4']=$op4;
    $_SESSION['answers']=$answers;
   
   //print the arrays. debugging purposes
   //1 is the starting index of the array because the $i is equal to 1
   print_r($_SESSION['question']);
   echo '<br>';
   print_r($_SESSION['op1']);
   echo '<br>';
   print_r($_SESSION['op2']);
   echo '<br>';
   print_r($_SESSION['op3']);
   echo '<br>';
   print_r($_SESSION['op4']);
   echo '<br>';
   print_r($_SESSION['answers']);
   
   
   header("location:exam.php");

?>

</body>
</html>


----------------------------------------------------------exam.php--------------------------------------------------------------------

<?php
   //start session
   session_start();
   
   
   
   echo '<br>';
   echo '<br>';
   echo '<tr>';
   echo '<b>';
   echo 'Question: ';
   echo '<td align = center>'.$_SESSION['question'][$_SESSION['a']].'</td>';
   echo '</tr>';
   
   echo '<form method = "post" action= "nxt_question.php">';
   echo '<table border = "0">';
   echo '<tr>';
   //echo '<td>';
   echo "<td><Input type =\"Radio\" Name =\"choice\" value='".$_SESSION['op1'][$_SESSION['a']]."'>";
   echo '<td>'.$_SESSION['op1'][$_SESSION['a']].'</td>';
   echo '</tr>';
   echo '<tr>';
   echo "<td><Input type =\"Radio\" Name =\"choice\" value='".$_SESSION['op2'][$_SESSION['a']]."'>";
   echo '<td>'.$_SESSION['op2'][$_SESSION['a']].'</td>';
   echo '</td>';
   echo '</tr>';
   echo "<td><Input type =\"Radio\" Name =\"choice\" value='".$_SESSION['op3'][$_SESSION['a']]."'>";   
   echo '<td>'.$_SESSION['op3'][$_SESSION['a']].'</td>';
   echo '</tr>';
   echo "<td><Input type =\"Radio\" Name =\"choice\" value='".$_SESSION['op4'][$_SESSION['a']]."'>";
   echo '<td>'.$_SESSION['op4'][$_SESSION['a']].'</td>';
   echo '</tr>';
   echo '</table>';
   echo '<tr>
   <td>
   <input type = "submit" value="Next Question">
   </td>
   </tr>';
   echo '</form>';
   
   ?>

----------------------------------------------nxt_question.php-------------------------------------

<?php

   //start session
   session_start();

   //Checking of answers kada mag pindot next question button
   if(isset($_POST['choice']))
   {
      $ans_of_user=$_POST['choice'];
      $_SESSION['answers'][$_SESSION['a']];
      
      if (strcmp($ans_of_user,$_SESSION['answers'][$_SESSION['a']])==0)
         {
            $_SESSION["iCorrect"]++;
         }
   }

   //add 1 to the session a. session a is the iterator. serves as indexes of the arrays that stores the questions,answers etc...
   $_SESSION['a']++;
   
   //ridirect to exam.php if there are still number of items remaining.
   while($_SESSION['a']<=$_SESSION['num_of_questions'])
   {
      header("location:exam.php");
      
      exit();  //kill the script
   }
   
   echo 'exam finished!';
   echo '<br>';
   echo 'Your score is: ';
   echo $_SESSION["iCorrect"];
   
   //unset the session variables needed to be reset
   unset($_SESSION["iCorrect"]);
   unset($_SESSION["exam_id"]);
   unset($_SESSION["invitation_code"]);
   unset($_SESSION["a"]);
   unset($_SESSION["num_of_questions"]);
   unset($_SESSION["question"]);
   unset($_SESSION["op1"]);
   unset($_SESSION["op2"]);
   unset($_SESSION["op3"]);
   unset($_SESSION["op4"]);
   unset($_SESSION["answers"]);
   
?>

I'm trying to figure out how to create a temporary table of a query join, so I can paginate the results. I kind of made half of this up as i was going along - which is dangerous for a beginner like me.
I'm getting this error, and no results on the page:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource on line 179

also, is there a way to combine the two queries (create temp and insert from/into)


Code: [Select]
$sql2 = "CREATE TEMPORARY TABLE temp_search
    (plant_id INT PRIMARY KEY AUTO_INCREMENT,
     scientific_name VARCHAR(75),
     common_name VARCHAR(75),
     leaf_shape VARCHAR(75))";

$sqlx = "SELECT
                        descriptors.leaf_shape
                        ,plantae.scientific_name, plantae.common_name
                    FROM
                        descriptors
                    INNER JOIN
                        plantae ON (descriptors.plant_id = plantae.plant_name)
                    WHERE
                        descriptors.leaf_shape LIKE 'auriculate'
                        AND descriptors.leaf_venation LIKE '%$select3%'
                        AND descriptors.leaf_margin LIKE '%$select4%'

                    INTO temp_search (leaf_shape, scientific_name, common_name)";

$sql = "SELECT * FROM temp_search ORDER BY scientific_name ASC LIMIT $start, $limit";

$result = mysql_query($sql);


while($row = mysql_fetch_array($result))
{
echo  $row['scientific_name'] . $row['common_name'] . $row['leaf_shape'] ;
}