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Falanas

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Posts posted by Falanas

  1. You never try to use $_GET['id'] as far as I can see.

     

     

    Jessica I believe I have used it (it is saved in '$correct_id').

     

    What worries me is that this code is unable to execuite:

     

     

    [$placesql = "SELECT tourism_sites.*, towns.*, place.*, photos.* FROM tourism_sites, towns, place, photos

    WHERE tourism_sites.cat_id = towns.id AND towns.id = place.cat_id AND place.id = photos.id AND photos.image_id = " . $correct_id;]

     

    No any number e.g 1, 2, 3 etc is seen on the URL instead, it redirects me to BASE_DIR i.e header("Location: " . BASE_DIR);

     

    Is it because the id sent from index.php to the url does not meet the below conditions?

     

    <?php

     

    if(isset($_GET['id']) == TRUE) {

    if(is_numeric($_GET['id']) == FALSE) {

    header("Location: " . BASE_DIR);

    }

    $correct_id = $_GET['id'];

     

    }

    else {

    header("Location: " . BASE_DIR);

    }

    ?>

     

    I expected to see something like - details.php?id=1 or details.php?id=2 depending on the link clicked. What do you say? Thanks.

    ===========================================================================================================

     

     

     

    <?php

     

    if(isset($_GET['id']) == TRUE) {

    if(is_numeric($_GET['id']) == FALSE) {

    header("Location: " . BASE_DIR);

    }

    $correct_id = $_GET['id'];

     

    }

    else {

    header("Location: " . BASE_DIR);

    }

    ?>

    <!-- start page -->

    <div id="page">

    <!-- start content -->

     

    <div id="content">

     

    <div class="post">

     

     

    <h2>Reservation</h2>

     

    <?php

    /*

    *this query links to variable . $sitenumrows['id'] of index.php to display further information about

    * $site2numrows['hotel_name']

    */

     

    $placesql = "SELECT tourism_sites.*, towns.*, place.*, photos.* FROM tourism_sites, towns, place, photos

    WHERE tourism_sites.cat_id = towns.id AND towns.id = place.cat_id AND place.id = photos.id AND photos.image_id = " . $correct_id;

    $placeresult = mysql_query($placesql);

    $placenumrow = mysql_num_rows($placeresult);

    if($placenumrow == 0){

    echo "<h2 class = 'results'>There was a problem querrying database</h2>";

    }

    else{

    while ($placenumrow == mysql_fetch_assoc($placeresult)){

    echo "<h2 class = 'title'>" . $placenumrow['town_name'] . "</p>";

    }

    }

     

    ?>

  2. Forgive me, I am only but a beginner in PHP & Mysql. I am trying to do a simple website but unfortunately I can see myself in a big mess:

     

    Getting 'id' from URL has been so hard for me and that's why I will appreciate if you could assist me. Here goes the code:

     

     

     

    index.php

     

     

    <?php

     

    //The first category (KENYA SAFARIS AS DEFAULT VALUE) of table - categoties

     

    $catsql = "SELECT * FROM categories WHERE categories.id = 1";

    $catresult = mysql_query($catsql);

    while($catrow = mysql_fetch_assoc($catresult)){

    echo "<h2> " .$catrow['cat_name'] . "</h2>";

     

    //list of available sites to be visited (OF - TABLE tourism sites)

    $sitesql = "SELECT * FROM tourism_sites WHERE cat_id = " .$catrow['id'] . ";";

    $siteresult = mysql_query($sitesql);

    $sitenumrows = mysql_num_rows($siteresult);

     

    if($sitenumrows==0){

    echo "<h2 class = 'results'>No listed safari site</h2>";

    }

    else{

    while($sitenumrows = mysql_fetch_assoc($siteresult)){

    echo "<ul>";

    echo "<li><a href='#.php?id =" . $sitenumrows['id'] . "'>" . $sitenumrows['hotel_name'] . "</a></li>";

    echo "</ul>";

    }

    }

     

    echo "</li> ";

    }

     

    ?>

     

    <li>

    <?php

    //The second category (UGANDA AS DEFAULT VALUE) of categories table

    $cat2sql = "SELECT * FROM categories WHERE categories.id = 2";

    $cat2result = mysql_query($cat2sql);

    while($cat2row = mysql_fetch_assoc($cat2result)){

     

    echo "<h2>" .$cat2row['cat_name'] . "</h2>";

     

    //list sites in Uganda to be visited

    $site2sql = "SELECT * FROM tourism_sites WHERE cat_id =" . $cat2row['id'] .";";

    $site2result = mysql_query($site2sql);

    $site2numrows = mysql_num_rows($site2result);

    if($site2numrows==0){

    echo "<h2 class = 'results'> No saved locations..! </h2>";

    }

    else{

    while($site2numrows = mysql_fetch_assoc($site2result)){

    echo "<ul>";

    echo "<li><a href='#.php?id =" . $site2numrows['id'] . "'>" . $site2numrows['hotel_name'] . "</a></li>";

    echo "</ul>";

    }

    }

     

    echo "</li>";

    }

     

    ?>

     

     

     

     

     

     

    <?php

     

     

    if(isset($_GET['id']) == TRUE) {

    if(is_numeric($_GET['id']) == FALSE) {

    header("Location: " . BASE_DIR);

    }

    $correct_id = $_GET['id'];

     

    }

    else {

    header("Location: " . BASE_DIR);

    }

    ?>

     

    <!-- start page -->

    <div id="page">

    <!-- start content -->

     

    <div id="content">

     

    <div class="post">

     

     

    <h2>Reservation</h2>

     

    <?php

    /*

    *this query links to variable . $sitenumrows['id'] of index.php to display further information about

    * $site2numrows['hotel_name']

    */

     

    $placesql = "SELECT tourism_sites.*, towns.*, place.*, photos.* FROM tourism_sites, towns, place, photos

    WHERE tourism_sites.cat_id = towns.id AND towns.id = place.cat_id AND place.id = photos.id AND photos.image_id = " . $correct_id;

    $placeresult = mysql_query($placesql);

    $placenumrow = mysql_num_rows($placeresult);

    if($placenumrow == 0){

    echo "<h2 class = 'results'>There was a problem querrying database</h2>";

    }

    else{

    while ($placenumrow == mysql_fetch_assoc($placeresult)){

    echo "<h2 class = 'title'>" . $placenumrow['town_name'] . "</p>";

    }

    }

     

    ?>

     

     

    please help

  3. I need help on how I can implement a drop down menu which queries mysql database and output the available data based on price range. This feature has been used here http://www.vebra.com - I will appreciate your help.

     

    $searchSQL = "SELECT  * FROM simple_search WHERE";

         

                 

          // grab the search types.

          $types = array();

          $types[] = isset($_GET['price'])?"`price` LIKE '%{$searchTermDB}%'":'';

          $types[] = isset($_GET['location'])?"`location` LIKE '%{$searchTermDB}%'":'';

     

    $types = array_filter($types, "removeEmpty"); // removes any item that was empty (not checked)

         

          if (count($types) < 1)

            $types[] = "`id` LIKE '%{$searchTermDB}%'"; // use the estate as a default search if none are checked

     

     

    $andOr = isset($_GET['matchall'])?'AND':'OR';

          $searchSQL .= implode(" {$andOr} ", $types) . " ORDER BY `price`"; // order by price.

     

          $searchResult = mysql_query($searchSQL) or die("There was an error.<br/>" . mysql_error() . "<br />SQL Was: {$searchSQL}");

         

          if (mysql_num_rows($searchResult) < 1) {

            $error[] = "The search term provided <i>{$searchTerms}</i> yielded no results.";

          }else {

     

              $results = array(); // the result array

            $i = 1;

            while ($row = mysql_fetch_assoc($searchResult)) {

                $results[] = "{$row['location']} <br/>{$row['image']}<br/>{$row['price']}";

     

    $i++;

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