Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
I am echoing an image which works fine, but then i want to set how big it needs to be, and I just can't seem to get the syntax right.
This is what I have:
echo "<img src='" . $row['imglink'] . "' height="130" width="150"> ";
What am I doing wrong? If I remove the height and width it works fine.
You are mixing single and double quotes. So you get a syntax error.
This works:
echo "<img src='" . $row['imglink'] . "' height='130' width='150'> ";
If you like to output double quotes you have to escape:
echo "<img src=\"" . $row['imglink'] . "\" height=\"130\" width=\"150\"> ";
Or you switch to cover the string in single quotes:
echo '<img src="' . $row['imglink'] . '" height="130" width="150"> ';
try changing double quotes into single quotes for these attributes
height='130' width='150'
since double quotes are actually delimiting the PHP string. Optionally you may instead escape the double quotes like so
height=\"130\" width=\"150\"
but this syntax is less readable and more error prone (not recommended)
ok its here
echo '<img src="'$row['imglink']'" height="130" width="150">';
You can uses variables inside double quotes, like this:
<?
$img_url = $row['imglink'];
echo "<img src='$img_url' height='130' width='150'>";
?>
Sometimes the way that you're using causes to errors if you don't do it properly.Therefore you can try this way also.
<img src="<?php echo $row['imglink']?>" height="130" width="150" />
if you're using this inside a loop,you can do like this.
<?php foreach($result as $rows):?>
<img src="<?php echo $row['imglink']?>" height="130" width="150" />
<?php endforeach; ?>
Hope this helped to you.
Related
echo '<a href="LP_Teacher_Response_G4.php Lesson_id=$Lesson_id&user_id=$user_id&Lesson_Class_id=$Lesson_Class_id&class_id=$class_id">
<img height="62" src="../Images/Lock.png" width="76" /></a>';
Hi,
I am not sure why my parameters do not show in the link, instead it outputs the code statement, played around with it but not sure what needs to be done.
Try this way
echo "<img height=\"62\" src=\"../Images/Lock.png\" width=\"76\" />";
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.
This question already has answers here:
How to extract img src, title and alt from html using php? [duplicate]
(10 answers)
Closed 9 years ago.
How can I use img tag in php code?:
This is the code I have so far:
$message = $message."...<br />".$lang->messagemore."";
And I wish to use an image instead of ".$lang->messagemore." but I don't know how to use img tag in php :|
I have tried googling but I only find some tag info from html but not php code.
<img src="/res/gif/bullet_info_sq.gif" alt="" />
But I wish to use that in php so I could use your help.
Thank you very much!
tl;dr
Replace: $lang->messagemore
With: <img src="/res/gif/bullet_info_sq.gif" alt="" />
When you replace your PHP variable with the img tag, the HTML must still remain within the quotes of the PHP string, and thus the actual quotes for your img tag must be escaped, just as you've done for your a tag. You may also opt to use single quotes, to avoid having to escape. This should result any one of the following equivalent code snippets (I've added whitespace for readability):
Double quotes (with escaping):
$message = $message . "...<br />
<a href=\"" . $mybb->settings['bburl'] . "/" . $announcement['threadlink'] . "\">
<img src=\"/res/gif/bullet_info_sq.gif\" alt=\"\" />
</a>";
Single quotes for HTML attributes:
$message = $message . "...<br />
<a href='" . $mybb->settings['bburl'] . "/" . $announcement['threadlink'] . "'>
<img src='/res/gif/bullet_info_sq.gif' alt='' />
</a>";
Single quotes for PHP string:
$message = $message . '...<br />
<a href="' . $mybb->settings['bburl'] . '/' . $announcement['threadlink'] . '">
<img src="/res/gif/bullet_info_sq.gif" alt="" />
</a>';
Just echo the HTML code ( in your case , the <img> tag ) in PHP , just like you did for <a> tag.
You can use like this,
$message = $message."...<br /><img src='/res/gif/bullet_info_sq.gif' alt='' />";
use this:
$message = $message."...<br />".'<img src="/res/gif/bullet_info_sq.gif" alt="" />'."";
Use this
$message = $message."...<br /><img src=\"/res/gif/bullet_info_sq.gif\" alt=\"\" />";
It works like that:
$message = $message."...<br />"."<img src='".$aImageURL."' alt="" />";
Simply replace de img src with a PHP variable. ($aImageURL will be the URI of the image)
Take a look at the quotes (I put simply quotes in the src attribute, you can do that)
So simple, do not create mess with double quotes " PHP optimally used single quotes '. Use standard html within string!
echo ' <img src="php.gif" /> ';
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>";