Falanas
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Posts posted by Falanas
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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
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If you areusing wampserver, go to wampserver icon on the taskbar - down there. Left-click on it. Click on config files then you will see php.ini. All the best.
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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']}";
$i++;
Getting Variable From Url In Php & Mysql
in PHP Coding Help
Posted
Jessica I believe I have used it (it is saved in '$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>";
}
}
?>