I am trying to display a image URL from mysql database into my website. It's saved as URL in my database.
http://www.upload.ee/image/5697422/prague-1168302_1920.jpg
this is the url which should appear as $rida['pilt']
<?php
$paring = 'SELECT * FROM postitus ORDER BY id';
$valjund = $yhendus->query($paring);
while($rida = mysqli_fetch_assoc($valjund))
?>
<img class="img-thumbnail" alt="city" src="<?php echo $rida['pilt']; ?>" style="width:250px;height:200px" />
I guess in your code you're getting error undefined index as you're accessing your variable out of while loop. Try below code:
<?php
$paring = 'SELECT * FROM postitus ORDER BY id';
$valjund = $yhendus->query($paring);
while($rida = mysqli_fetch_assoc($valjund)) : ?>
<img class='img-thumbnail' alt='city' src='<?php echo $rida['pilt']; ?>' style="width:250px;height:200px" />
<?php endwhile; ?>
You should recheck $rida. After you use "echo var_dump($rida);", you will know all parameters in array. You can choose trust array
Related
<?php foreach($works as $work) : ?>
<?php echo Asset::img('project-icons/icon/$work->cover_img', array('class'=>'img-responsive', 'alt'=>'...')); ?>
<?php endforeach; ?>
I am using FuelPHP to build a portfolio website. PHP v5.6
This is the "works" section where images and details of a "work" are fetched from database. Using foreach loop. I want to get the "cover_img" (which is the image name), inside the Asset::img(..)
How to do this? what to put in place of $work->cover_img?
Edit: I was able to get the result by using this:
<img src="<?php echo Uri::base(false); ?>/assets/img/project-icons/icon/<?php echo $work->cover_url; ?>" class="img-responsive" alt="..." />
Can this be achieved by using Asset::img() instead of Uri::base()?
Can't you just do:
<?php
echo Asset::img('project-icons/icon/'.$work->cover_img, array('class'=>'img-responsive', 'alt'=>'...'));
?>
...?
I'm fairly new to PHP and I've been trying to construct some code to print basic HTML, however the code causes an error 500 whenever used. I am guessing it is a syntax error since I've tried the code in a couple of forms and nothing seems to work (including removing the database lookup and just trying to compare to set values to each other). The script needs to get a variable from the db, compare it to a set value and print the HTML if true, here is the code I am trying:
<?php
$db = &JFactory::getDBO();
$id = JRequest::getString('id');
$db->setQuery('SELECT #__categories.title FROM #__content, #__categories WHERE #__content.catid = #__categories.id AND #__content.id = '.$id);
$category = $db->loadResult(); ?>
<?php if strcmp($category,"Blog")==0 : ?>
<div style="display: -webkit-inline-box" class="sharelogos">
<img src="/images/sharing-icons/facebook.png" width="30px" alt="Facebook" />
</div>
<?php endif; ?>
Any help will be appreciated, thanks!
You if is incorrect, try like this
<?php if (strcmp($category,"Blog")==0) { ?>
<div style="display: -webkit-inline-box" class="sharelogos">
<img src="/images/sharing-icons/facebook.png" width="30px" alt="Facebook" />
</div>
<?php } ?>
I am trying to display 4 images using PHP and MySQL database. I have to display the 4 images as rows.
I use the table cars with fields (id_car, car_image1, car_image2, car_image3, car_image4), with all the images being blob datatype.
$id = $_GET['id'];
$link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");
$sql = "SELECT car_image1, car_image2, car_image3, car_image4 FROM cars WHERE id_car='$id'";
$result = mysql_query("$sql");
mysql_close($link);
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
echo $row['car_image1'];
echo $row['car_image2'];
echo $row['car_image3'];
echo $row['car_image4'];
}
I can only display 1 image and not the other images. Since I am a newbie to this technology I need help.
That's how HTTP works (at least, current version): you cannot have a URL that points to more than one resource simultaneously. Your script needs to send one image.
Simply, call it four times:
<img src="/get-image.php?id=1">
<img src="/get-image.php?id=2">
<img src="/get-image.php?id=3">
<img src="/get-image.php?id=4">
You need to define path for images.
<?php
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
?>
<img src="path<?php echo $row['car_image1']; ?>" alt="" />
<img src="path<?php echo $row['car_image2']; ?>" alt="" />
<img src="path<?php echo $row['car_image3']; ?>" alt="" />
<img src="path<?php echo $row['car_image4']; ?>" alt="" />
<?php
}
?>
I'm trying to pass the path of an image ['guest'] from one page to another using a link. (I'm storing URLs of images in database)
Can't seem to get the image to display, which is a bigger image of 'url'. I'm doing it this way so that I can have a larger image displayed in the target page (does_this_work.php) plus adding some other bits on the page too.
I'm still learning and can;t seem to see what I'm doing wrong. Any help appreciated,
<?php
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name
// Connect to server and select databse.
mysql_connect($host, $username, $password)or die("cannot connect");
mysql_select_db($db_name) or die("cannot select DB");
$photo=mysql_query("SELECT * FROM `images` ORDER BY (ID = 11) DESC, RAND() LIMIT 7");
while($get_photo=mysql_fetch_array($photo)){ ?>
<div style="width:300px;">
<a href="does_this_work.php?big_image=$get_photo['guest']>" target=""><img src="<?
echo $get_photo['url']; ?>" title="">
</div>
<? } ?>
I then use the following code to try and display the array data in the target file
<?php
echo "this is the page where you get a larger picture of image on previous page, plus
further info";
$big_image = $_GET['guest'];
echo $big_image;
?>
You're missing an opening tag in here (And a closing semi colon, but that's not as problematic here):
<a href="does_this_work.php?big_image=$get_photo['guest'] ?>"
Change to:
<a href="does_this_work.php?big_image=<?= $get_photo['guest']; ?>"
There are several errors in your code. First of all, this is how you use $_GET and $_POST:
ex. site.php?argument=value
To retrieve the value of argument, you need this code in site.php:
//The variable must not necessarily be $value
$value = $_GET['argument'];
//Alt.
$value = $_POST['argument'];
Secondly (like the other answers tell you) you are missing a php-tag here:
<a href="does_this_work.php?big_image=$get_photo['guest']>" target=""><img src="<? echo $get_photo['url']; ?>" title="">
Instead it should be:
<a href="does_this_work.php?big_image=<?php echo $get_photo['guest']; ?>" target=""><img src="<?php echo $get_photo['url']; ?>" title="">
Now, in order to make that compatible with your second code, you need to change the argument sent to guest like this:
<a href="does_this_work.php?guest=<?php echo $get_photo['guest']; ?>" target=""><img src="<?php echo $get_photo['url']; ?>" title="">
OR change the $_GET['guest']; to $_GET['big_image'];
I think I got it all right..
<a href="does_this_work.php?big_image=$get_photo['guest'] ?>" target=""><img src="<?
echo $get_photo['url']; ?>" title="">
You are missing your php starting tags. This should read:
<a href="does_this_work.php?big_image=<? $get_photo['guest'] ?>" target=""><img src="<?
echo $get_photo['url']; ?>" title="">
I can't seem to get my image to display properly. Previously, I have used the following code snippet and it worked perfectly.
catalog.php (worked perfectly):
<p class="image">
<a href="synopsis.php?id=<?php echo $row['id']; ?>">
<img src="getImage.php?id=<?php echo $row['id']; ?>" alt="" width="175" height="200" />
</a>
</p>
synopsis.php (not displaying image at all):
<?php
$id = $_GET['id'];
...?>
<p class="image">
<img border="0" class="floatleft" src="getImage.php?id=<?php echo $row['id']; ?>" width="250" height="400" />
<?php echo $row['synopsis']; ?>
</p>
where getimage.php:
<?php
$id = $_GET['id'];
$link = mysql_connect("localhost", "root", "");
mysql_select_db("dvddb", $link);
$sql = "SELECT dvdimage_path FROM dvd WHERE id=$id";
$result = mysql_query($sql, $link);
$row = mysql_fetch_assoc($result);
mysql_close($link);
header("Content-type: image/jpeg");
echo file_get_contents($row['dvdimage_path']);
?>
Any idea why can't I display this image?
EDIT 1:
So after debugging, I got an error message:
Undefined index: id in C:\xampp\htdocs\synopsis.php on line 106
so i went to add the following code into the php code just before echo $row['id']:
<p>getImage.php?id=<?php error_reporting(0); echo $row['id']; ?></p>
However,
the paragraph i got was just getImage.php?id=.
Then, i went into synopsis.php -> <img border="0" class="floatleft" src="getImage.php?id=<?php echo $row['id']; ?>
and changed that into:
<img border="0" class="floatleft" src="getImage.php?id=2">
Again, same problem happens, where i can't get the specific image out.
I suspect something is wrong with my getimage.php file. However, this getimage.php file has been working fine for other pages when i use the snippet.
My requirements are very simple:
In catalog.php, i populate images and text from dvd database using a while loop. Then, each of these images has got their specific primary ID. when i click the the images, they will go to the link: synopsis.php?id="primaryid" Then, using this "primaryid" i should be able use getimage.php?"primaryid" to generate an image on synopsis.php page.
EDIT 2:
actually, i made a syntax error somewhere. So this line:
<img border="0" class="floatleft" src="getImage.php?id=2">
is working perfectly, this means the fault lies in somewhere that i cant echo 'id' out correctly.
EDIT 3:
I have included the links to the relevant source code:
catalog.php
synopsis.php
getimage.php
sortmenu.css
style.css
database in xml format
Questions to ask yourself:
Is it really required that you use a php script to mimic the image? If not, just use the image path.
Is there any output before the header(); function in the getimage.php file? Even just a space before the
Is the image actually a JPEG?
Are there any errors coming up when you go to getimage.php?id=ID in your browser?
In synopsis.php you are getting the id from the querystring but then trying to use a database value. I cant see all of your code so im posting solutions to cover two scenarios
src="getImage.php?id=<?php echo $id; ?>"
or
src="getImage.php?id=<?php echo $row[$id]; ?>"