I would like to add a profile image for this follow and unfollow system. How would I do this in this line?
I am thinking before the <tr> I would have a <img src = ...>
$t_rows .= '<tr><td>'.htmlspecialchars($v['NameFirst'].' '.$v['NameLast']).'</td><td>';
If you would like to add a profile image based off of a PHP variable that has already been set, you could echo the HTML content.
<?php
$myVariable = "https://www.google.com/images/branding/googlelogo/1x/googlelogo_color_272x92dp.png";
echo '<img src="' . $myVariable . '" alt="Some Text..."/>';
?>
Related
Hello I'm trying to echo an image on a view page in CodeIgniter but nothing is displayed on the page.
Here I'm making the variable:
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
And here I'm trying to echo the image:
<img src="<?php echo $image;?>">
Intro
This is the most basic php, so what's the addition of this question? Please read the basics of echo here: http://php.net/manual/en/function.echo.php (Example 1).
Learn the basics first!
Solution
1. Assign full html tag to variable and echo full html:
<?php
$image = '<img src="../../images/stoel.jpg" alt="Foo">';
echo $image;
?>
2. Or assign image path to variable and echo concat string:
<?php
$path = '../../images/stoel.jpg';
echo '<img src="' . $path . '" alt="Foo">';
?>
3. Or assign image path to variable and echo only this with php:
<?php
$path = '../../images/stoel.jpg';
?>
<img src="<?= $path; ?>" alt="Foo">
You are inserting full Image tag into src attribute.
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
and
<?php echo $image; ?>
or
<?php $image = "../../images/stoel.jpg"; ?>
and
<img src="<?php echo $image;?>">
<?php $image = "<img src='../../images/stoel.jpg' alt='img' />"; ?>
you try like this
<?php echo $image; ?>
You are already assign full image code in variable so you just need to print your variable:
<img src="<?php echo $image;?>">
To
<?php echo $image;?>
Make sure your images out side of the application folder then you can do something like
application
images
images > stoel.jpg
system
index.php
Use Base url from the url helper
<img src="<?php echo base_url('images/stoel.jpg');?>" />
Make sure you have set the base_url in config.php
$config['base_url'] = 'http://localhost/yourproject/';
You can also use HTML Helper img();
I think this is what you mean. Since you are echoing inside the src attribute, you do not need to store the whole <image>, just the path will do.
<?php
$image = "../../images/stoel.jpg";
?>
<img src="<?php echo $image;?>">
I have created some code to make buttons clickable and return a value.
When I click on the image in the browser the querystring is showing the x-y coordinates of the image I am currently in, while it should put the name of the image in the querystring.
I also replaced the "value" value to "test" just to check if it works but I still get x-y coordinates.
<?php
$dirname = "images_fotoviewer/";
$images = glob($dirname."*.jpg");
foreach($images as $image){
echo "<form action='24_fotoviewer.php' method='GET'>";
echo "<input type='image' src='" . $image . "' name='foto' value='" . $image . "'/>";
echo "</form>";
}
?>
This is the querystring I get:
iwp1_basis_php/24_fotoviewer.php?foto.x=164&foto.y=48
Could someone help me to get the name of the image in the querystring?
Thanks in advance.
If you use input type="image" you will get a submitbutton as an image. It is default behaviour to send the x/y coordinates where you clicked it.
So to get the behaviour you want to try this:
Replace the form with an hyperlink (a href=...)
Create a hyperlink that contains the name of the image
Something like this:
<?php
foreach($images as $image){
?>
<a href="24_fotoviewer.php?imgName=<?php echo urlencode($image); ?>">
<img src=<?php echo $image; ?>
</a>
<?php
}
?>
Hey guys got a question on outputting a dynamic PHP block for a dynamically created PHP page. In my code I am looking for a string in an HTML page thats been uploaded. Once found I am replacing the string with a block of PHP code, the HTML page will be saved as a PHP page to be used on the project. So as I am looping through the HTML I am replacing the string with this ($i is replaced with the number in the loop so I can use them in my array.)
$phpCodeNoLink = '<span id="Title'.$i.'"><?php echo $sl_result['.$i.'][2]; ?></span>
<a href="editor.php?<?php echo "vfSID=" . $sl_result['.$i.'][0] . "&vfSection=2&vfSLink=" . $sl_result['.$i.'][4] . "&vfOrderID=" . $sl_result['.$i.'][5] . "&vfID=" . $vfID; ?>" target="_parent">
<img src="images/btn_edit.gif" border="0" id="SL_editButton'.$i.'" class="editButton" />
</a>';
The problem is it is not outputting what I need, example of what it should look like
<span id="Title1"><?php echo $sl_result[1][2]; ?></span>
<a href="editor.php?<?php echo "vfSID=" . $sl_result[1][0] . "&vfSection=2&vfSLink=" . $sl_result[1][4] . "&vfOrderID=" . $sl_result[1][5] . "&vfID=" . $vfID; ?>" target="_parent">
<img src="images/btn_edit.gif" border="0" id="SL_editButton1" class="editButton" />
</a>
This is what I get in the PHP page once it's generated
<span id="Title0"><?php echo $sl_result[0][2]; ?></span>
<a href="editor.php?<?php%20echo%20%20" vfsid=" . $sl_result[0][0] . " .>" target="_parent">
<img src="images/btn_edit.gif" border="0" class="editButton"></a>
The PHP tags are being replaced and I am missing a whole block of code. Am I missing something any help would be much appreciated.
Figured it out, the PHP code was being parsed and removed by my inline CSS converter moving it above all the other parsing resolved it issue...
I am trying to display image from a blob field of a MySQL table. Looks like I have some sort of error in the following line. As soon as I put "header("Content-type: image/jpeg")" things get messed up and instead of displaying webpage, all source code of the page is displayed.
Please let me know how to correct.
<div class="image" align="left">
<a href="<?php header("Content-type: image/jpeg"); echo $rec['image']; ?>">
<img src="<?php echo $rec['image']; ?>" width="150" border="0"/>
</a>
</div><!-- image -->
You normally don't put the actual image contents in the src= attribute of the image tag. Instead, you point to the URL of an image file.
(There are ways to include the image source directly in the HTML, but it doesn't work consistantly with all browsers, and you still won't have your <a> link working properly.
Instead, the best way to do this is to create a separate PHP file to serve the image.
Your HTML:
<div class="image" align="left">
<img src="myimage.php?key=<?php echo($key) ?>" width="150" border="0"/>
</div><!-- image -->
myimage.php:
<?php
header("Content-type: image/jpeg");
$key = $_GET['key'];
// todo: load image for $key from database
echo $rec['image'];
You're trying to put the image data inline inside the content. The only feasible way to do this is via a Data URI data URI. Something like:
<img src="data:image/jpeg;base64,<?= base64_encode($rec['image']) ?>" width="150" border="0" />
However, what you probably want to do is put it into a separate script. So your HTML would be:
<img src="showimage.php?id=XXX" width="150" border="0" />
And your showimage.php script would be:
<?php
// Get $rec from database based on the $_GET['id']
header('Content-Type: image/jpeg');
echo $rec['image'];
?>
I've done something like that retrieving blob from my database in another way that you may find useful, here is the code example.. see if it suits your needs and if you needed anymore help let me know.
while ($row = mysql_fetch_array($hc_query2)) {
$title = $row['title'];
$text = $row['text'];
$image = $row ['image'];
$output ='<div class="HCInstance"><img src="data:image/jpeg;base64,' . base64_encode($image) . '" alt="High Council" width="100px" height="100px"/>
<div class="HCHeader"><h2>'.$title.'</h2></div><br/><div class="HCDetails"><p>'.$text.'</p></div></div>';
echo $output;
}
I don't know exactly where to start to display files from the file set in order to show a fancybox gallery on click. I would like to have the gallery open from a link. On click show the gallery, aka images that have the same rel but are set to display:none (easily controlled by my css). I can do it from selecting one image but am not sure how to pass the images from the file set into the view (I'm assuming I need to create some kind of function in my controller to get the fsID, just not sure how). I just need the first image to display on the page (thumbnail image), then click the link and it shows more full size images.
Basically, if you know Concrete5, I would like it to be like the image block, except that the administrator can choose a fileset instead of one image.
here is my view.php
$picture = $this->controller->getPicture();
if ($picture) {
$bigPicture = $image->getThumbnail($picture,600,600)->src;
$smallPicture = $image->getThumbnail($picture,200,200)->src;
echo "<img src=" . $smallPicture . " alt=" . $imageTitle . " title=" . $imageTitle . "/>";//thumbnail picture
echo "<div id=\"image-modal\">";
echo "{$linkText}";//open fancybox from link
echo "<div class=\"hiddenGallery\" style=\"display:none;\">";//hidden images
echo "<a href=\"images/pattern/t-103-n.jpg\" class=\"fancybox-thumb\" rel=" . $title . " title=" . $imageTitle . ">";
echo "<img src=\"images/pattern/t-103-n.jpg\" class=\"fancybox-thumb\" />";
echo "</a>";
echo "</div>";
echo "</div>";
}
my controller.php
function getPicture() {
if ($this->fIDpicture > 0) {
return File::getByID($this->fIDpicture);
}
return null;
}
my add.php
$al = Loader::helper('concrete/asset_library');
echo $al->image('ccm-b-image', 'fIDpicture', t('Choose File'),
$this->controller->getPicture());
echo '</div>';
Any and all help is much appreciated.
Well, two things :
You have to set the class="fancybox-thumb" AND the rel attribute to the <a> tag!!, not to the <img /> tag.
If you are planing to hide the rest of the elements of the gallery, don't set a display: none; css property to each of them, but rather wrap them in a hidden <div> container like :
<div style="display: none;">
<a class="fancybox-thumb" rel="gallery" href="images/02.jpg"></a>
<a class="fancybox-thumb" rel="gallery" href="images/03.jpg"></a>
<a class="fancybox-thumb" rel="gallery" href="images/04.jpg"></a>
... etc
</div>
I am using the rendered html, which is what it matters.
I have some code which handles the entire back-end (add/edit/controller) part of this equation:
https://github.com/jordanlev/c5_designer_gallery
Here's a tutorial that explains how to use it (with the example of the FlexSlider, but if you know how Fancybox works then it shouldn't be hard to understand what's going on):
http://c5blog.jordanlev.com/blog/2011/12/build-a-slideshow-block/