Hi i wanted to know how i want to know how i would write the image tag and access the string to get the image out of the database. When i refresh my page it only shows the title that i have got out of the database and not the image i want to know how i can get the image out of the database in the heredoc.
In my database i have a string for example Images/pokemon.png
<?php
foreach ($result as $row) {
echo <<<_END
<h4 class="card-title mt-5"> $row->title</h4>
<img src="<?php echo base_url('application/' . $row->image)?>" alt=""/>
_END;
}
?>
The closing identifier must not be indented, so it should look like this.
<?php
foreach ($result as $row) {
echo <<<_END
<h4 class="card-title mt-5"> $row->title</h4>
<img src="<?php echo base_url('application/' . $row->image)?>" alt=""/>
_END;
}
?>
For reference
Related
trying to make this code render images in rows all to no avail. I initially tried to use boostrap's row and column classes and it didn't work. Then I tried the table element still no result. could you spot the problrm?
// Get images from the database
$query = $db->query("SELECT * FROM images ORDER BY id DESC LIMIT 5");
if($query->num_rows > 0){
while($row = $query->fetch_assoc()){
$imageURL = 'uploads/'.$row["file_name"];
?>
<!-- begin post -->
<div class="container recent-posts">
<table>
<tr>
<td class="card" style="width: 18rem;">
<img src="<?php echo $imageURL; ?>" alt="" />
<div class="card-body">
<h5 class="card-title">Card title</h5>
<p class="card-text">Some quick example text to build on the card title and make up the bulk of the card's content.</p>
Go somewhere
</div>
</td>
<?php }
}
else{ ?>
<p>No image(s) found...</p>
<?php } ?>
</tr>
</table>
</div>
I've swapped in your db code for an array and foreach, but it is in the same spirit.
Use indentation, to help you get your loops and tags in order.
<?php
$images = [
['file_name' => 'foo.jpg'],
['file_name' => 'bar.jpg'],
['file_name' => 'baz.jpg'],
]
?>
<html>
<?php if(!empty($images)) { ?>
<table>
<tr>
<?php foreach($images as $row) { $imageURL = 'uploads/'.$row["file_name"]; ?>
<td class="card" style="width: 18rem;">
<img src="<?php echo $imageURL; ?>" alt="" />
<div class="card-body">
<h5 class="card-title">Card title</h5>
<p class="card-text">Some quick example text to build on the card title and make up the bulk of the card's content.</p>
Go somewhere
</div>
</td>
<?php } ?>
</tr>
</table>
<?php } else { ?>
<p>No image(s) found...</p>
<?php } ?>
</html>
As others have said, there isn't anything really to be gained here by using a table.
Does it show you the rest of the html elements or does it show you the message that there is no picture?
Check with the inspect tool what location the src attribute of the img element is.
You can add a backslash in the $imageURL variable
$imageURL = '/uploads/' . $row["file_name"];
Show us the structure and a record of the images table.
Give us more details about what happens when the page is loaded. Like what it loading in your page.
I suggest you ditch the tables and use boostrap instead.
Example here: https://getbootstrap.com/docs/4.0/components/card/
In my MySQL database, the image links are saved in the format like http://www.old.com/image1.jpg. But I had to change the domains of the images and the new links appear like images.new.com/image1.jpg. I have been changing the images links with jQuery with the following function:
$(document).ready(function() {
$('img').each(function() {
var src = $(this).attr('src');;
$(this).attr('src', src.replace('www.old', 'images.new'));
});
});
But I am wondering is there any way to change part of the URL strings with PHP. I am getting the image URLs with the following PHP function.
<?php
$imageQuery = $mysqli->query("SELECT imageURL FROM images WHERE album = 'UK' ORDER BY date ASC");
$images = $imageQuery->fetch_all(MYSQLI_ASSOC);
$images = array_chunk($images, 2);
?>
<div class="row">
<div class="col-4" id="box1">
<?php foreach (array_column($images, 0) as $image): ?>
<img class="img-fluid" src="<?= $image['imageURL']; ?>">
<?php endforeach; ?>
</div>
<div class="col-4" id="box2">
<?php foreach (array_column($images, 1) as $image): ?>
<img class="img-fluid" src="<?= $image['imageURL']; ?>">
<?php endforeach; ?>
</div>
</div>
With PHP how can I echo the new links for the images directly here in the img src? <img class="img-fluid" src="<?= $image['imageURL']; \\modified link here ?>">
PHP has it's own method for replacing strings, str_replace. The equivalent to your jQuery in PHP is:
str_replace('www.old', 'images.new', $image['imageURL'])
A better idea would be to update the values in your database.
An even better idea would be to not store the duplicate root URLs anywhere, and stitch them together in the application if you need absolute URLs, for some reason.
You could alternatively convert the text in the sql select statement:
SELECT REPLACE(imageURL, 'www.old', 'images.new') AS imageURL FROM ...
By using AS imageURL none of the rest of the code would need to change.
Actually I am beginner programmer in HTML, CSS and PHP. I have simple website for add and register in courses. The user should be add course into website and the courses should be posted on the site.so users can browse and register.
Actually my problem is how to call the course name from database and how to format it with HTML code as I want.
This is the page of courses which is content the list of available courses in website ( please note it is only HTML code, I do that to see how the page will be )
Screenshot of page:
So as you see, the first page include many this HTML code to add course list into website with the following code:
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% "></a> <div class="container">
<h4 class="textstyle"><b>Operating System</b> </h4>
<p class="textstyle">Free Course</p>
</div>
</div>
what i want do with PHP?
I want to write a PHP code to replace the P and h4 with the course name, cost of courses from my database for each available course.
Please note: the pic for each course it will be from my pc, no need to call the pic from database.
I tried to write PHP code like below:
<div>
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% "></a> <div class="container">
<?php
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT Course_Name,cost FROM `course`");
//while($res = mysql_fetch_array($result)) { // mysql_fetch_array is deprecated, we need to use mysqli_fetch_array
while($res = mysqli_fetch_array($result)) {
echo "<p>".$res['Course_Name']."</p>";
echo "<p>".$res['cost']."</p>";
}
?>
</div>
</div>
</div>
This is my result:
It's okay but I want the style to be like the first screenshot. each course should have picture.
After that when the user click on course name. I want move to another page which is content the course details ( for the same course that user clicked ) also it's called for my database
like this:
I hope any one help my to solve this problem only, I should solve this problem within 2 days only. and sorry if my explanation is bad.
Thanks in advance for everyone.
Put the code in a PHP loop.....
So, this
<div class="card card-1">
<a href="http://127.0.0.1/project2/course details/course1.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle"><b>Operating System</b> </h4>
<p class="textstyle">Free Course</p>
</div>
</div>
Becomes (after cleaning up the code a bit - I think you didn't mean to use two <p> in there, but I left them so you can see it. Note that using different lines for the segments makes it a lot easier to see what you have.)
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT Course_Name,cost FROM `course`");
$count = 0;
while($res = mysqli_fetch_array($result)) {
$count ++;
// NOTE: Here is the LOOP! - not outside the query, but INSIDE it
// First you 'jump out' of PHP, going back to HTML
?> <!-- now you are in HTML (when you need PHP again, you 'jump in' and 'jump out' as needed - see the code below....) -->
<div class="card card-<?php echo $count;?>">
<a href="http://127.0.0.1/project2/course details/course<?php echo $count;?>.php">
<img src="http://127.0.0.1/project2/icons/coursepic.jpg" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle">
<b><p><?php echo $res['Course_Name'];?></p></b>
</h4>
<p class="textstyle">
<p><?php echo $res['cost'];?></p>
</p>
</div>
</div>
<?php // we are in PHP again....
}
That should do what you asked for - though I would go a step (well, more than one...) further and make as much of this dynamic as you can.
For this I will presume that:
your database table has a column called 'id' (if it doesn't, you should have) and it relates to the course number (you could make a course number column if they don't match up, but I'm keeping it simple)
you have all your pictures labeled 'coursepicX' where the X is the course number.
We'll use 'coursepic' as a default in case there isn't a picture yet...
Now, the code is more dynamic!
include_once("db.php");
$result = mysqli_query(OpenCon(), "SELECT id,Course_Name,cost FROM `course`");
while($res = mysqli_fetch_array($result)) {
// NOTE: Here is the LOOP! - not outside the query, but INSIDE it
// First you 'jump out' of PHP, going back to HTML
?> <!-- now you are in HTML (when you need PHP again, you 'jump in' and 'jump out' as needed - see the code below....) -->
<div class="card card-<?php echo $res['id']?>">
<a href="http://127.0.0.1/project2/course details/course<?php echo $res['id']?>.php">
<?php
$pic = "http://127.0.0.1/project2/icons/coursepic.jpg";
if(file_exists("http://127.0.0.1/project2/icons/course" . $res['id'] . ".jpg") {
$pic = "http://127.0.0.1/project2/icons/course" . $res['id'] . ".jpg";
}
<img src="<?php echo $pic; ?>" alt="Avatar" style="width:101% ">
</a>
<div class="container">
<h4 class="textstyle">
<b><p><?php echo $res['Course_Name'];?></p></b>
</h4>
<p class="textstyle">
<p><?php echo $res['cost'];?></p>
</p>
</div>
</div>
<?php // we are in PHP again....
}
Note that this is the basic 'shopping cart' sort of program - you will likely use it many (many) times in your career.
Happy Coding!
I wrote a code which is retrive image paths and display on the website in bootstrap grid style.But it does not showing the image. Code is working fine, Please help me. here is my code
<div class="row">
<?while ($row = mysql_fetch_assoc($query)) {?>
<div class = "col-md-3">
<?php echo $row['keywords'];?>
<?php $imagePath = $row['video_url'];?>
<?php echo $imagePath;?>
<div id="video_thumbnail">
<a href="#" class="thumbnail">
<?php echo '<img src="' . $imagePath . '">'; ?>
<img src="<?php echo file_dir . '/' . $imagePath; ?>" height="100" width="100"/>
</a>
</div>
</div>
<?php } ?>
</div>
unless file_dir is a constant using DEFINE, I suspect it's because you didn't put a $ in front of it $file_dir
You are also displaying two files. One with the file path and one without.
Chances are, the mysql query is returning a path which is not linked ....
ie <image src="myImage.jpg" /> is not the same as <image src="images/myImage.jpg" />
As #pmahomme said, right click the element and check the pat and if need be add the additional requirements
I am trying to load pictures from database to jquery slider
infact what is stored at database is the name of the picture
i implemented this code, it's working without errors but it showing me nothing at slider all is empty
<div id="slider">
<?php
while ($result = mysqli_fetch_object($banner)):
?>
<img src="images/banner/<? $banner['picture'];?>/" width="950" height="400"alt="" title="<strong><? echo $banner['title'];?></strong><span><? echo $banner['description'];?> </span>" />
<?php
endwhile
?>
</div>
name of the table at database is banner in which i have id(INT), picture(varchar 50) title(varchar(100), description(longblob)
query is working and returning number of selected rows
but nothing is shown
You need to echo the result rather than just use the variable...
<div id="slider">
<?php
while ($result = mysqli_fetch_object($banner)):
?>
<img src="images/banner/<?php echo $banner['picture'];?>" width="950" height="400" alt="" title="<strong><? echo $banner['title'];?></strong><span><?php echo $banner['description'];?></span>" />
<?php
endwhile;
?>
</div>