Pretty straight forward simple question, can you open a php code block to call image information in html? I don't think I phrased that right. Here is my code:
<img src="../inventory_images/' . <?php echo $item_number; ?> . '.jpg" width="150" height="150" border="2" />
This code is within the tags
I'm just trying to post a photo using the $item_number variable (which is also the name of the image file i.e. $item_number = T3144 and the image file is name T3144.jpg ). Also if there is a better way to accomplish this suggestions are happily accepted. Sorry to take up bandwidth with such a remedial question but for some reason I can't seem to answer this question in research. Thanks for taking the time everyone.
Your code is wrong, try:
<img src="../inventory_images/<?php echo $item_number;?>.jpg" width="150" height="150" border="2" />
with what you have it looks like the code you had would print
src="../inventory_images/' . whateveritem_numberis . '.jpg"
Yes that is perfectly fine, but make sure that this code is in a file that ends with .php or it will not get parsed by PHP. Also, you need to take out the single quotes and periods:
<img src="../inventory_images/<?php echo $item_number; ?>.jpg" width="150" height="150" border="2" />
Unless the above HTML is in an echo statement, you need to change it to this:
<img src="../inventory_images/<?php echo $item_number; ?>.jpg" width="150" height="150" border="2" />
That will in-turn look like this:
<img src="../inventory_images/T3144.jpg" width="150" height="150" border="2" />
Of course, that is going off of your example where $item_number = 'T3144';.
The single quotes and periods are used for concatenating variables inside of strings.
Related
I wanted to pull out html code Target URL and forgive with a variable
<img src="https://www.test.de/adtrack/3a10056f3.img" border="0" />
$url = https://www.test.de/adtrack/3a10056f3.html
the $url wants to echo that too in php
Just echo the whole thing with your variable added in it.
$url = 'https://www.test.de/adtrack/3a10056f3.html'
echo '<img src="https://www.test.de/adtrack/3a10056f3.img" border="0" />';
or, echo it inside your html like this:
<img src="https://www.test.de/adtrack/3a10056f3.img" border="0" />
If you do as above, remember that your file extension should be ".php" and not ".html" or else it won't work.
I have a string, an img url, in a php block called $str. And I want to set that string as the img src in an img src but its not working.
<img src="<?php $str ?>" height="50px" width="50px">
how else can i set the src to the $str string?
<img src="<?php echo $str;?>" height="50px" width="50px">
Well, I found correct answer for this problem. Nothing works well from above answers, that codes only print out source string to html page on site.
This works for me (I have my function that return source string of picture):
require_once("../../my_coded_php_functions.php");
<?php echo '<img src="' . getSourcePathOfImage() . '" />' ?>
This site(article) helped me to find and understand solution:
http://php.net/manual/en/faq.html.php
<?php echo '<img src="'.$str.'" height="50px" width="50px">' ?>
Other ways, although is not recommended. You may try :
<img src="<?=$str?>" height="50px" width="50px">
this will work, only if (in php.ini)
short_open_tag = On
I am trying to set size of an image in PHP but it's not working..
echo "<img src=".$media."width=200 height=200";
$media is the src link. Without the width and height attributes, it works perfectly well. I think that 200 must be enclosed in double quotations but I am unable to do that. Any ideas on how to solve this problem?
width is currently concatenated with your file name. Use:
echo '<img src="'.$media.'" width="200" height="200" />';
The /> closing tag is necessary to correctly render your image element.
The quotes around tag names are recommended. Omitting these quotes will only cause issues when:
The attribute value contain spaces, or
You forgot to add a space after the attribute value
Yet another alternative, just for kicks:
echo <<<EOL
<img src="{$media}" width="200" height="200" />
EOL;
aka a HEREDOC.
That is because that is not proper HTML.
In your code you'd get something like:
<img src=image.jpgwidth=200 height=200
And what you need is:
<img src="image.jpg" width="200" height="200" />
So do:
echo '<img src="' . $media . '" width="200" height="200" />';
echo '<img src="' . $media . '" width="200" height="200" />';
You should be able to make it work with the following code
echo "<img src=\"".$media."\" width=200 height=200>";
Here's my code:
<embed src="/sound/lowyourchicken.mp3"
width="140" height="40" autostart="true" loop="TRUE">
</embed>
I would like the src for the .mp3 to take in to account that there are many randomly named .mp3 files in the /sound/ directory, and to choose one at random each time the page is opened. Any clues for me?
My server is PHP enabled but I'd like to keep this as simple as possible.
This should do it:
$files = glob("/path/to/directory/*.mp3");
$random = array_rand($files)
Then do this:
<embed src="<?php echo $random ?>"
width="140" height="40" autostart="true" loop="TRUE">
</embed>
array_rand returns what random index it chose, so you'll need to do this:
<embed src="<?php $files[ $random ] ?>"
Try This:
It will Work, I used original code found in answers and did some tweaking by adding array($files) in the $random = array_rand(); variable statement
You will first need to put the PHP code in the body like this
<?php
$files = glob("assets/songs/SayYesToLove/*.mp3");
$random = array_rand(array($files));
?>
next add this just outside that php code in the body
<embed src="<?php echo $files[$random]; ?>" width="140" height="40" autostart="true" loop="TRUE">
</embed>
Please Notice the echo output in the src file. This will ensure it gets outputted to your page. Also don't forget to use the ; at the end of every php variable statement as this can through some errors.
how to insert this javascript code :
thejavascript
inside img src in php
<img src="http:'.$iv[$j].'.jpg" height="220" width="200" alt="Image '.ucwords($kw).'" "thejavascript">
Thanks
You can use this jQuery script to add and onClick event in your balise img but you must adding an attribute id into your image. An id must be unique in your html page.
$("img#idImage").click(function({
window.open('welcome.html','welcome');
}));
$kw = $ucwords($kw);
"<img id='idImage' src='http:{$iv[$j]}.jpg' height='220' width='200' alt='Image {$kw}' />";
But the best will be to separate attributes height and width into a CSS stylesheet.
Insert the following HTML outside PHP brackets and it should work according to the way you posted it. I'm making a few assumptions, one being that the link you posted wraps the image code and that the PHP variables turn the image into valid code.
<a href="javascript:void(0);" onclick="window.open('welcome.html','welcome')">
<img src="http:<?=$iv[$j]; ?>.jpg" height="220" width="200" alt="Image <?=ucwords($kw); ?>">
</a>
More simple, replace "thejavascript" (with the "") by :
onclick="window.open(\'welcome.html\',\'welcome\')"
A bit different approach, onclick event is on img tag.
$javascript="window.open('welcome.html','welcome')";
$img='<img src="http:'.$iv[$j].'.jpg" height="220" width="200" onclick="'.$javascript.'">';
echo $img;