Okay, so this is my first project using PHP and MySQL and I thought a Facemash style site using HTML, PHP and MySQL would be good place to start.
Everything works correctly except calling the "updateHits" function as an image hyperlink doesn't behave as I'd expect.
I am confident that the MySQL database is functioning correctly and the pictures do display as expected. My research points towards the use of iFrames, jQuery or AJAX to update the "hits" field although I cannot understand how to apply them here.
I hope my code is readable and any advice would be greatly appreciated!
<html>
<body>
<?php
// Make a MySQL Connection
mysql_connect("localhost", "admin", "admin") or die(mysql_error());
mysql_select_db("facemash") or die(mysql_error());
// Select two random people
$personA = rand(1, 28);
$personB = rand(1, 28);
// Ensure that it is not the same person
if ($personB == $personA) {
$personB = rand(1, 28);
}
// Function to return path of photo
function photoPath ($person){
$query = mysql_query("SELECT photo FROM people WHERE id=$person");
$result = mysql_fetch_row($query);
$result = $result[0];
echo $result;
}
// Function to update the hits field
function updateHits($person){
$query = mysql_query("SELECT hits FROM people WHERE id=$person;");
$result = mysql_fetch_row($query);
$result = $result[0];
$result++;
mysql_query("UPDATE people SET hits = $result WHERE id=$person");
}
?>
<!--Image for personA-->
<img src="<?php photoPath($personA);?>"/>
<!--Image for personB-->
<a href="<?php updateHits($personB);?>"/><img src="<?php photoPath($personB);?>"/></a>
</body>
</html>
Thanks.
Erm, PHP does not work that way. ;)
PHP is serverside code, you're creating a link that points nowhere. To get the effect you want you'll need to make an AJAX call to the server to tell it to update the hits.
Related
// connects to database here
if(isset($_GET['id']))
{
$id = mysql_real_escape_string($_GET['id']);
$query = mysql_query("SELECT * FROM `blob` WHERE `id`='$id'");
$data = mysql_fetch_array($query);
header("content-type: image/jpeg");
echo $imageData;
}
else
{
echo "Error!";
}
<img src="showimage.php?id=1">
the images are then being shown through and image tag shown above but they just appear as small icons (the correct number for those listed in the database but not the actual image)
First, remove the <img src="showimage.php?id=1"> and put it on a different page; you can't keep it in the same page since it causes the web server to send headers before your script is even executed; your showimage.php should only contain PHP code, nothing else.
Then, change your code like this :
$data = mysql_fetch_array($query);
header("content-type: image/jpeg");
echo $data["image"];
I assume that your row is called "image" from your previous question.
Also, don't ask multiple questions for the same problem, you already got several answers on your previous question.
I want to use one single page with pre-defined divs, layout etc. as basis so that when a product is clicked on from elsewhere it loads that product info onto the page?
They way im doing it ill be sitting here till about 2020 still typing out product info onto pages.
EDIT*************
function product ()
{
$get = mysql_query ("SELECT id, name, description, price, imgcover FROM products WHERE size ='11'");
if (mysql_num_rows($get) == FALSE)
{
echo " There are no products to be displayed!";
}
else
{
while ($get_row = mysql_fetch_assoc($get))
{
echo "<div id='productbox'><a href=product1.php>".$get_row['name'].'<br />
'.$get_row['price']. '<br />
' .$get_row['description']. '<br />
' .$get_row['imgcover']. "</a></div>" ;
}
}
}
In addition one problem I have with that code is that the <a href> tag only goes to product1.php. Any ideas how I can make that link to blank product layout page that would be filled with the product info that the user has just clicked on, basically linking to itself on a blank layout page.
Thanks any help would be great!
Thanks Maxyy
Since you dont have code this is a general way of doing this. What you want is a template for the product page
Query the database
load the data into a variable
make a script that will print out the data from the variable into a product page
somescript.php
<?php
$productid = $_REQUEST['productid']; //Of course do sanitation
//before using get,post variables
//though you should be using mysqli_* functions as mysql_* are depreciated
$result = mysql_query("select * from sometable where id='{$productid}");
$product = mysql_fetch_object($result);
include("productpage.php");
productpage.php
<div class="Product">
<div class="picture"><img src="<?php echo $product->imghref;?>" /></div>
<div class="price"><?php echo $product->price;?></div>
</div>
so on and so fourth. Included scripts use whatever variables are currently in the scope of the calling function
If you are meaning to load the products into the same page without doing another page load you will need to use ajax. Which is javascript code that use XHR requests to return data from a server. You can either do pure javascript or a library like jQuery to simplify the process of doing a xhr request by using $.ajax calls.
I know this question has been asked over 4 years ago, but since there's been no answer marked as right, I thought I might chip in.
First, let's upgrade from mysql and use mysqli - my personal favorite, you can also use PDO. Have you tried using $_GET to pull the id of whatever product you want to see and then displaying them all together or one at a time?
It could look something like this:
<?php // start by creating $mysqli connection
$host = "localhost";
$user = "username";
$pass = "password";
$db_name = "database";
$mysqli = new mysqli($host, $user, $pass, $db_name);
if($mysqli->connect_error)
{
die("Having some trouble pulling data");
exit();
}
Assuming the connection was made successfully we move on to checking for an ID being set. In this case I check it via an URL param assumed to be id. You can make it more complex, or take a different approach here.
if(isset($_GET['id']))
{
$id = htmlentities($_GET['id']);
$query = "SELECT * FROM table WHERE id = " . $id;
}
else
{
// if no ID is set, just bring all the results down
// then you can modify how, and which table the results
// are being used.
$query = "SELECT * FROM table ORDER BY id"; // the query can be changed to whatever you would be prefer
}
Once we have decided on a query we go on to start querying the database for information. I have three steps:
Check query >
Check table for records >
Loop through roles and create an object for each.
if($result = $mysqli->query($query))
{
if($result->num_rows > 0)
{
while ($row = $result->fetch_object())
{
// you can set up your element here
// you can set it up in whatever way you want
// to see your product being displayed, by simply
// using $row->column_name
// each column becomes an object here. So your id
// column would be pulled using $row->id
echo "<h1>" . $row->name . "</h1>";
echo "<p>" . $row->description . "</p>";
echo "<img src=" . $row->image_path . ">";
// etc ...
}
}
else
{
// if no records match the selected ID
echo "Nothing to see here...";
}
}
else
{
// if there's a problem with the query
echo "A slight problem with your query.";
}
$mysqli->close(); // close connection for safety
?>
I hope this answers your question and can help you if you are still stuck on this problem. This is the bare skeleton of what you can do with MySQLi and PHP, you could always use some Ajax to make the page more interactive, and user-friendly.
Adding content to a page on click needs to be done in either Javascript or in JQuery.
You can use ajax call to retrive the needed data from php page, Syntax is here.
Or you can also load a php page to a div content with .load() function in JQuery, Syntax is here.
I'm currently trying to learn HTML, AJAX, PHP, and MYSQL. I'm currently building a webapp just for internal use. It al seemed to work but now I'm experiencing some lag in my code using the MYSQL function.
Here's what happens:
First off all, I create some "profile images" form an sql database. trough PHP like this:
<?php
$price = 30;
$con = mysql_connect('localhost', 'root', 'root');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("Og297", $con);
$result = mysql_query("SELECT * FROM Og297.Drinkers");
while($row = mysql_fetch_array($result))
{
$picture = $row['Picture'];
$name = $row['Name'];
$nameup= $name."up";
$namep= $name."p";
$Onbetaaldresult = mysql_query("SELECT COUNT(Betaald) AS ob FROM Og297.Bierlijst
WHERE Betaald='1' AND Name='$name'");
$Onbetaaldarray= mysql_fetch_array($Onbetaaldresult);
$Onbetaald= $Onbetaaldarray['ob'];
$Betaaldresult = mysql_query("SELECT COUNT(Betaald) AS b FROM Og297.Bierlijst
WHERE Betaald='0' AND Name='$name'");
$Betaaldarray= mysql_fetch_array($Betaaldresult);
$Betaald= $Betaaldarray['b'];
echo "<div onclick=\"addBeer('$name','$price');\" class= 'thumbnail'><img src= $picture height= '90' width= '90' alt= $name title= $name><div class= 'overlay'><span id=$nameup class='unpaid'>$Onbetaald</span><span id= $namep class='paid'>/$Betaald</span></div></div>";
}
mysql_close($con);
?>
So this shows for every user a picture div, on top of that it shows to span's with the ID's Paid and Unpaid(The app is for keeping track how much beer we've paid and how much still need to be paid). This all works :)
next I have some functions;
On is addBeer, wich ads a beer into a table, including the name of that person. This is done trough Ajax.
function addBeer (n,p)
{
$.post("AddBeer.php",{naam: n, price: p});
updateBeer(n);
}
So after having inserted the beer into the datbase, I want it to update the Paid and Unpaid span's. I do this by using updateBeer.
function updateBeer (n)
{
$.post("UpdateBeer.php",{naam: n},function(data) {
changeInfo(n,data.betaald,data.onbetaald);
},"json");
}
This returns the amount of paid and unpaid in a Json object. Those values(including the name of the person are then given to changeInfo function, wich will update the span's like this:
function changeInfo(n,b,ob)
{
window.alert("ja");
document.getElementById(n + "up").innerHTML=ob;
document.getElementById(n + "p").innerHTML="/" + b;
}
However, the info is not updated every time I click an image.. It lags behind, the beers are inserted tough, but the updating of the span's is just not happening. How come?
Thanks very much for even reading this huge question!
The problem in my code was that the AJAX request where async. Thus causing problems when the rest of the script was executed before the AJAX request was completed. The trick was to set the AJAX request to set Async to false!
Thanks!
I am working on something it has 2 pages. One is index.php and another one is admin.php.I am making CMS page where you can edit information on the page yourself. Then it will go to the database, where the information is stored. I also have to have it where the user can update the information on the page. I am getting a little bit confused here.For instance here I am calling the database and I am starting a function called get_content:
<?php
function dbConnect(){
$hostname="localhost";
$database="blank";
$mysql_login="blank";
$mysql_password="blank";
if(!($db=mysql_connect($hostname, $mysql_login, $mysql_password))){
echo"error on connect";
}
else{
if(!(mysql_select_db($database,$db))){
echo mysql_error();
echo "<br />error on database connection. Check your settings.";
}
else{
return $db;
}
}
function get_content(){
$sql = "Select PageID,PageHeading,SubHeading,PageTitle,MetaDescription,MetaKeywords From tblContent ";
$query = mysql_query($sql) or die(mysql_error());
while ($row =mysql_fetch_assoc($query,MYSQL_ASSOC)){
$title =$row['PageID'[;
$PageHeading =$row['PageHeading'];
$SubHeading = $row['SubHeading'];
$PageTitle = $row['PageTitle'];
$MetaDescription =$row['MetaDescription'];
$MetaKeywords = $row['MetaKeywords'];
?>
And then on the index page and I am going to echo it out in the spot that someone can change:
<h2><?php echo mysql_result($row,0,"SubHeading");?>A Valid XHTML and CSS Web Design by WG.</h2>
I do know that the function is not finished I am still working on that part. What I am wondering is am I echoing it out right or I am way off. This is my first time messing with CMS in php and I am still learning it. I am working with navicat and text pad on this, yes I know it is old school but that is what I am being shown with. But my index is a form not a blog. I have seen many of CMS pages for blogs not to many to be used with forms. Any input will be considered thanks for reading my question.
Your question is a bit confusing and your code very incomplete. I'ts hard to say if you do it the right way since I don't see the rest of the script. You need to connect to the database there as well and get your data. The $row variable only exists in the while statement inside you function get_content() though.
You could complete the get_content() and use it in the index.php as well. Remember that the variables you define inside a function only is available there though. If you need the data outside that function you need to return the values you need and save them to some other variable there. Put if you do the same as you've started doing in the get_content() function in index.php, then you just have to echo the variables you define. Like this:
<h2><?php echo $SubHeading; ?></h2>
or you could also do it like this somewhere inside the php tags:
echo '<h2>{$SubHeading}</h2>';
I hope that answers your question.
EDIT:
What you need in the index.php page is exactly what you seem to be doing in the admin file. You need to connect to db using mysql_connect() and select db with mysql_select_db(). You then need to select the data from the db using the appropriate query with $query = mysql_query($sql). If it's more then one row you want to display you need to put it in a while loop otherwise (which seems to be the case here) you just need to do one $row = mysql_fetch_assoc($query). After that you can get the data using $row['column_name']. If you have more than one row you can just use $row['column_name'] in side the while loop to get each consecutive row's data.
Here is an example index.php:
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password') or
die('Could not connect: ' . mysql_error());
mysql_select_db('database_name')) or die('Could not select database: ' .
mysql_error());
$sql = "SELECT SubHeading FROM tblContent WHERE PageID='1' LIMIT 1;";
$query = mysql_query($sql);
$row = mysql_fetch_assoc($query);
echo '<h2>{$row[\'SubHeading\']}</h2>';
mysql_close();
?>
This is just what you need to display the SubHeading from you database. You probably also need to handle your form and save the submitted data to the database in your admin.php file.
Please could someone help im building my first website that pulls info from a MySQL table, so far ive successfully managed to connect to the database and pull the information i need.
my website is set up to display a single record from the table, which it is doing however i need some way of changing the URL for each record, so i can link pages to specific records. i have seen on websites like facebook everyones profile ends with a unique number. e.g. http://www.facebook.com/profile.php?id=793636552
Id like to base my ID on the primary key on my table e.g. location_id
ive included my php code so far,
<?php
require "connect.php";
$query = "select * from location limit 1";
$result = #mysql_query($query, $connection)
or die ("Unable to perform query<br>$query");
?>
<?php
while($row= mysql_fetch_array($result))
{
?>
<?php echo $row['image'] ?>
<?php
}
?>
Thanks
Use $_GET to retrieve things from the script's query (aka command line, in a way):
<?php
$id = (intval)$_GET['id']; // force this query parameter to be treated as an integer
$query = "SELECT * FROM location WHERE id={$id};";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) == 0) {
echo 'nothing found';
} else {
$row = mysql_fetch_assoc($result);
echo $row['image'];
}
There are many things to consider if this is your first foray into MsSQL development.
SQL Injection
Someone might INSERT / DELETE, etc things via using your id from your url (be careful!, clean your input)
Leaking data
Someone might request id = 1234924 and you expected id = 12134 (so some sensitive data could be shown, etc;).
Use a light framework
If you haven't looked before, I would suggest something like a framework (CodeIgniter, or CakePHP), mysql calls, connections, validations are all boilerplate code (always have to do them). Best to save time and get into making your app rather than re-inventing the wheel.
Once you have selected the record from the database, you can redirect the user to a different url using the header() function. Example:
header('Location: http://yoursite.com/page.php?id=123');
You would need to create a link to the same (or a new page) with the URL as you desire, and then logic to check for the parameter to pull a certain image...
if you're listing all of them, you could:
echo "" . $row['name'] . ""
This would make the link.. now when they click it, in samepage.php you would want to look for it:
if (isset($_GET['id']) && is_numeric($_GET['id'])) {
//query the db and pull that image..
}
What you are looking for is the query string or get variables. You can access a get variable through php with $_GET['name']. For example:
http://www.facebook.com/profile.php?id=793636552
everything after the ? is the query string. The name of the variable is id, so to access it through your php you would use $_GET['id']. You can build onto these this an & in between the variables. For example:
http://www.facebook.com/profile.php?id=793636552&photo=12345
And here we have $_GET['id'] and $_GET['photo'].
variables can be pulled out of URL's very easily:
www.site.com/index.php?id=12345
we can access the number after id with $_GET['id']
echo $_GET['id'];
outputs:
12345
so if you had a list of records (or images, in your case), you can link to them even easier:
$query = mysql_query(...);
$numrows = mysql_num_rows($query);
for ($num=0;$num<=$numrows;$num++) {
$array = mysql_fetch_array($query);
echo "<a href=\"./index.php?id=". $row['id'] ."\" />Image #". $row['id'] ."</a>";
}
that will display all of your records like so:
Image #1 (links to: http://www.site.com/index.php?id=1)
Image #2 (links to: http://www.site.com/index.php?id=2)
Image #3 (links to: http://www.site.com/index.php?id=3)
...