I asked a question earlier regarding retrieving a image from a database; as my query was not working properly. This is now working to some extent, as the query now works,correct image name appears and a placeholder box with a question mark in the middle where the image should be.
In my earlier post, i was directed to a number of posts, to fix the image not appearing image. I followed the posts by adding:
echo'<img src="data:image/jpeg;base64,' . base64_encode($img) . '" alt="Category_header"
width="100%" height="300px">';
To my code, however it is still now working, I know you can use a header, but i am really trying to avoid that method, unless it is the more efficient way.
This is what i am receiving at the present moment:
Code
mysqli_report(MYSQLI_REPORT_INDEX);
if (isset($_GET['rest_id'])) {
$Rest = $_GET['rest_id'];
$get_cat_img = "SELECT Cuisine_category,Category_img
FROM Rest_Category
INNER JOIN Rest_Details
ON Rest_Category.CategoryID = Rest_Details.Cat_ID
WHERE Rest_Details.Resturant_ID='$Rest'";
$results = mysqli_query($dbc, $get_cat_img) or die("query is not working");
$row=mysqli_fetch_array($results) or die ("q not working");
$img=$row['Category_img'];
echo $row['Category_img']; //shows the category_img name
// echo '<img src="'.$img.'" alt="background" style="width:100%;height:300px">';
echo'<img src="data:image/jpeg;base64,' . base64_encode($img) . '" alt="Category_header" width="100%" height="300px">';
}
mysqli_close($dbc);
I asked this question before but no one could help me unfort. I have images and headings coming from database now the problem is that only one image is displaying(the last image) i need both to display.
here is my revised code
$query = "SELECT page_title, page_image FROM pages WHERE id='$page'";
$result = mysqli_query($connection, $query);
confirm_query($result);
while ($page_fetch = mysqli_fetch_assoc($result)) {
$page_title = $page_fetch['page_title'];
$images = $page_fetch['page_image'];
echo "<div class=\"content \">";
echo "<h3 class=\"words\"> $page_title </h3>";
echo "<img src='pics/" . $images . "' width=\"340\" height=\"252\" alt=\"\" />";
echo "</div>"; //end box
} // close while loop
here is my database for pages
page_id id page_image page_title
1 1 ocean.jpg have a look at the ocean
2 1 house.jpg The house
just some extra info the images must display dynamically , as they coming in from a form to db to this page
Your are returning your images from your query.
Check the view source page, You will find your error their.
Your returing image name will not match with the image which you got in the folder. (For the second one)
I have the image URL saved in my mySQL database.($imageid).
How can I display images using their respective id? I want to display different images attached to different blog posts,as we see in regular blogs.
Thank you.
Unsure what stage you are, bit you need to query your 'table' for a single result, perhaps the ID? If the column was Imageid then you could display it like below in an array of results.
$con=mysqli_connect("localhost","user","pass","my_db");
$result = mysqli_query($con,"SELECT * FROM table");
while($row = mysqli_fetch_array($result))
{
echo '<img src="' . $row['Imageid'] . '">';
}
mysqli_close($con);
We have a small php script, that displays random adverts on our site . The banners are served from any location we designate.
What I really would like to know, or be pointed in the right direction is, can we somehow collate the number of impressions each banner gets and possibly the click count on that banner.
I am sure this can be done in php, and stored in a db. Just havent got a clue. I searched on Google, and seemingly everything I could find harks back to 2002 and 2003 lol.
Here is our script:
<?php
$Img1 = "path to image folder/banner.png";
$Alt1 = "alt text for banner";
$Url1 = "http://www.externaldomain.com";
$Img2 ="path to image folder/banner.png";
$Alt2 = "alt text for banner";
$Url2 = "http://www.externaldomain.com";
$Img3 ="path to image folder/banner.png";
$Alt3 = "alt text for banner";
$Url3 = "http://www.externaldomain.com";
$Img4 ="path to image folder/banner.png";
$Alt4 = "alt text for banner";
$Url4 = "http://www.externaldomain.com";
$Img5 ="path to image folder/banner.png";
$Alt5 = "alt text for banner";
$Url5 = "http://www.externaldomain.com";
$num = rand (1,5);
$Image = ${'Img'.$num};
$Alt = ${'Alt' .$num};
$URL = ${'Url'.$num};
Print "<img src=\"".$Image."\" alt=\"".$Alt."\" /"; ?>
To initiate the above code ( we fire an include request )
<?php include ("../adserver/thescriptabove.php"); ?>
I see that you already selected an answer, so I'm not sure if you figured it all out, but I was writing out a little tutorial for you. Finally got it done, hope it still helps you out.
Your method seems to be working fine for serving banners, but if you're going to get into databases and tracking clicks/impressions, I would suggest that you go all in. So store your banner properties in the database as well. I'm going to get ahead and assume that your server/web host allows for a few free MySql databases.
What you need to do is create a database, as well as a User/Admin for the database. Then you're going to access the database with a MySql manager, most web hosts provide phpMyAdmin. Once you're inside the database, you need to set up a table to record your data.
Here's how I want you to set it up:
|Table Name: Banners |
|-------------------------|
|Field: | Type: |
|-------------------------|
|ID | int(5) | The Primary Key and Autoincrement attributes should be set on the ID field as well
|Image | varchar(100) |
|Url | varchar(100) |
|Caption | varchar(100) |
|Views | int(10) |
|Clicks | int(10) |
Now that you've got the database done, here comes the hard part, the PHP. I've pretty much done it for you, but it's untested, so I'm sure there will be bugs, that you will have to work out. But it should point you in the right direction, and help you learn.
<?PHP
// First of all, we need to connect to the MySql server
// For more info, check out: http://php.net/manual/en/function.mysql-select-db.php
$conn = mysql_connect("localhost", "username", "password");
if(!$conn){
die('Could not connect to the MySql Server ' . mysql_error());
}
// Now that we've connected to the MySql sever, we need to select the database
// More info can be found on the same link as above
$db_selected = mysql_select_db('my_database', $conn);
if(!$db_selected) {
die ('Could not select the MySql Database: ' . mysql_error());
}
// Now we need to check the URL parameters to see, if we came to this page normally or because a banner was clicked
// If normally, we serve a random banner and increment the Views field in the database
// Otherwise, we increment the Clicks field and direct the user to the banner's URL
if(!isset($_GET['Clicked'])){
// if the Clicked parameter is not set, we came to the page normally
// Let's select a random banner from the database
$result = mysql_query("SELECT * FROM banners ORDER BY RAND() LIMIT 1") or die(mysql_error());
$row = mysql_fetch_array(result);
// Now let's increment the Views field for the banner we selected
mysql_query("UPDATE banners SET Views = Views + 1 WHERE ID = '" . $row['ID'] . "'") or die(mysql_error());
// let's set the URL to this same page but with the Clicked parameter set
$url = "banner_server.php?Clicked=" . $row['ID'];
// Last but not least, we have to output the banner HTML
// Notice, I set the caption on both the 'alt' and 'title' attributes of the 'img' tag,
// that's because IE shows the value of the 'alt' tag when an image is hovered,
// whereas Firefox shows the value of the 'title' tag, just thought you might want that!
echo "<img src=\"" . $row['Image'] . "\" alt=\"" . $row['Caption'] . "\" title=\"" . $row['Caption'] . "\" />";
}else{
// Otherwise, increment the Clicks field and redirect
// First, let's get the ID of the banner that was clicked from the Clicked parameter
$id = (int)$_GET['Clicked'];
// now let's increment the Clicks field for the banner
mysql_query("UPDATE banners SET Clicks = Clicks + 1 WHERE ID = '" . $id . "'") or die(mysql_error());
// now let's select the banner from the database so we can redirect to the URL that's associated with it
$result = mysql_query("SELECT * FROM banners WHERE ID = '" . $id . "'") or die(mysql_error());
$row = mysql_fetch_array(result);
// redirect to the URL
header("location: " . $row['Url']);
}
// Close the MySql connection
mysql_close($conn);
?>
Good luck
why dont you just let google analytics do it for you? Fire off an event when the link is clicked and let google capture it?
onclick="_gaq.push(['_trackEvent', 'Event Name', 'click', 'Button title/name']);"
You can store the $num in the database pretty easy to get your impression count. Clicks require client side action. The easiest way is to call a javascript function that counts when the banner is clicked via AJAX:
print "<img src=\"".$Image."\" alt=\"".$Alt."\" /";
Then have your javascript function (countClick()) execute the AJAX that will tell the server the banner has been clicked.
Another way is to have a passthru page: mysite.com/linkout.php?q=www.google.com and then have linkout.php count that link and update the database, and then redirect them to the variable passed in the URL.
Here are my 2 cents, assuming you have analytics on our site:
Use the following code when outputting a link:
<a class="ad" href="http://thecatisginger.com/" target="_blank" onclick="ga('send', 'event', 'Block-3-Ads', 'Click', 'The-Cat-Is-Ginger-Ad');"><img src="http://citybymouth.com/wp-content/uploads/2015/02/TCIG-Advert.jpg" onload="ga('send', 'event', 'Block-3-Ads', 'Impression', 'The-Cat-Is-Ginger-Ad', {'nonInteraction': 1});" /></a>
To explain:
<a class="ad" href="http://thecatisginger.com/" target="_blank"
Classic link a href link with class 'ad', links to a site, target opens in new tab. Easy.
onclick="ga('send', 'event', 'Block-3-Ads', 'Click', 'The-Cat-Is-Ginger-Ad');">
This is the newer 'analytics.js' google event tracking, onclick event code, that bascially says, hey, you've clicked this link, so 'send' this 'event' to my analytics 'Events' (which can be checked under "Realtime > Events" or "Behaviour > Events"). "Block-3-Ads" is the area on my website I personally know as the area I put ads, specifically its a right hand sidebar area, but it can be anything you want, so make yours a broad category type thing, like a box, inside which you will have different links you want to track. "Click" is simply the type of event you want to track, can be anything. "The-Cat-Is-Ginger-Ad" is this specific ad I want to track and have the information about.
<img src="http://citybymouth.com/wp-content/uploads/2015/02/TCIG-Advert.jpg"
Then you have an img with a src. Easy.
onload="ga('send', 'event', 'Block-3-Ads', 'Impression', 'The-Cat-Is-Ginger-Ad', {'nonInteraction': 1});" />
Then you have an onload event that fires when the image is loaded, it says, 'send' this 'event', categorise it as a 'Block-3-Ads' event, basically the image loading is counted as an 'Impression', before it was click, but not since this little 'onload' is called when it loads, its not a click, but a load / impression, and again, specifically, the ad loaded is 'The-Cat-Is-Ginger-Ad', and finally, there is passing the 'nonInteraction' parameter as 1, which is just telling google you are tracking a non interaction event.
Its pretty self explanatory, tho I may have typed too much.
WARNING: This is not perfect in that the page may load, and the ad may be below the viewport, out of sight of the user, and still get an impression, which is not accurate, so next I'll be working on firing the impression code just once when the user actually scrolls to the ad itself, and views it, instead of firing it every time a page loads that image.
Thanks to #Saad Imran for the code and also big thanks to the question poster
Bellow the update of the code in php 7 if someone else need it for later use
Note : The same Database table and then store this code in banner.php file
<?php
// to only count views
// select a random banner from the database to show
$rows = $db->QueryFetchArrayAll("SELECT * FROM banners ORDER BY RAND() LIMIT 1");
foreach ($rows as $row) {
// Now let's increment the Views field for the banner we selected
$url = "/banner.php?clicked=" . $row['ID'];
echo "<img src=\"" . $row['Image'] . "\" alt=\"" . $row['Caption'] . "\" title=\"" . $row['Caption'] . "\" />";
}
$db->Query("UPDATE banners SET Views = '".round($row['Views'] + 1)."' WHERE id = '" . $row['ID'] . "'");
// to count clicks need an if statement
if(isset($_GET['clicked'])){
$db->Query("UPDATE banners SET Clicks = '".round($row['Clicks'] + 1)."' WHERE id = '" . $row['ID'] . "'");
header("location: " . $row['Url']);
}
?>
Good luck everyone :)
How can i use php to get an image url from database (based on user id) and display it out as an image.
http://mysite.com/img.php?id=338576
The code below shows a php script goes to a database , check whether the specific user(using the id in the link above) exist then get the image url out of that user's row.
<?php
//connect to database
include ("connect.php");
//Get the values
$id = mysql_real_escape_string($_GET['id']);
//get the image url
$result = mysql_query("SELECT * FROM users WHERE id='$id'")
or die(mysql_error());
$row = mysql_fetch_array($result);
if($row)
{
$img_url = row['imgurl'];
mysql_close($connect);
}
else{
}
?>
The $Img_url is an actual image link www.asite.com/image.jpg
The problem is how can i use php to make an image from the image link($img_url)? By which http://mysite.com/img.php?id=338576 will turn into an image.
I hope someone could guide me
Thanks!
The easiest way:
header('Location: url/to/image');
You could also proxy the request, which uses your bandwidth:
echo(file_get_contents('url/to/image'));
This is pretty basic stuff. Am I understanding you correctly? I would just do this:
<?php
$img_html = '<img src="' . $img_url . '" />';
echo $img_html;
?>
Or check this answer:
How do I script a php file to display an image like <img src="/img.php?imageID=32" />?