I echo out an image like that:
$newString = $thumbPre.'profilemain'.$thumbPost;
echo "<img src='http://render-api-us.worldofwarcraft.com/static-render/us/" . $newString. "' alt='error'>";
Now i want the image as a background-image, i tried it like that, but it doesn´t work:
echo '<div style="background-image:url('http://render-api-us.worldofwarcraft.com/static-render/us/" . $newString. "' alt='error');"></div>';
you need to use backslashes for nested apostrophes
echo '<div style="background-image:url(\'http://render-api-us.worldofwarcraft.com/static-render/us/' . $newString.'\' alt=\'error\');></div>';
Firstly, remove alt='error' because background-image does not have an alt parameter, img does (you probably thought you could use that from your original code). In trying to use that, your background will not show up.
And your background won't show unless you have content inside that div. I've added Content as an example.
echo '<div style="background-image:url(\'http://render-api-us.worldofwarcraft.com/static-render/us/' . $newString.'\');">Content</div>';
You either have to escape the encapsulating quotes, or remove them altogether.
echo '<div style="background-image:url(http://render-api-us.worldofwarcraft.com/static-render/us/' . $newString.');">Content</div>';
Error reporting would have also thrown you a parse error such as:
Parse error: syntax error, unexpected '' alt='' (T_CONSTANT_ENCAPSED_STRING), expecting ',' or ';'
http://php.net/manual/en/function.error-reporting.php
This is going to look super confusing but you need to 1, escape the quotes and 2, concatenate your path within that,
let me give you an example. I misread a bit of the question but this will serve you well moving forward especially for cleanliness sake
here is an example:
$imagePath ='PATH TO IMAGE HERE';
echo '<div style="background-image:url(\'' .$imagePath. '\')" >STUFF HERE </div>';
Related
This line of code is perfect for placing the image before the post's title
$title = ‘<img class=”icon_title” src=”‘. $img_source .'” />’ . $title;
But we want to place the image after the post's title. I think my code below is right
$title = $title . ‘ <img class=”icon_title” src=”‘. $img_source .'” />’;
But I am getting this error as shown in the screen shot:
Please can anyone help me to find the solution
It's happending due to your quote problem and you have not wrapped class attribute inside a string. Please follow the below code. It will help you. Convert to your keyboard to standard mode(US) and then type.
$title = $title.'<img class="icon-title" src="'.$img_source.'">';
Your " and ' were not correct type - perhaps you were editing in a word document/application. Consider writing code in applications designed for code to prevent this?
Below is the same line of code with the correct type of ' and ".
$title = $title . '<img class="icon_title" src="'. $img_source .'"/>';
I am using php and conditional code to give a dynamic url to a photo. The result should read as http://example.com/biophotos/1.jpg. But instead I am getting
http://example.com/%22http://example.com/biophotos/1.jpg%22
How can I force it to just give the one url and without the %22 space on the end?
if ($emresult[0]['photo'] = "y") {
echo '<img class=\"alignright\" src=\"http://example.com/biophotos/' .
$theID . '.jpg" width=\"150\" height=\"150\">';
}
else {
echo 'There is no author photo.';
}
There is no reason in escaping double quotes when your using single quotes.
echo '<img class="alignright" src="http://example.com/biophotos/' . $theID . '.jpg" width="150" height="150" />';
As much as I agree with #slik, you also might want to look into %22 (double quotes) added to url out of nowhere
Check if magic quotes is on in your php.ini file. You can look into html_entity_encode() to encode the %22 as a slash.
I am having trouble with my PHP code. I've been changing everything for 6 hours and I still get Parse errors no matter what I do. This is the code:
$slider3 = '<img src="'templates/' . $this->template . '/images/slider/slider3.jpg'">' . '" alt="' . $sitename . '" />';
The only way I can figure to not get it to throw an error is by writing it this way:
$slider3 = '<img src="templates/" . $this->template . "/images/slider/slider3.jpg" . "/>"';
but I don't think that's right.
I want $slider3 = "templates/MYTEMPLATE/images/slider/slider3.jpg" then later I will echo $slider3;
I get so confused with all the single and double quotation marks. I think the first one is right - I look at it and study it and it looks right to me. But it throws a parse error.
$slider3 = '<img src="templates/'.$this->template.'/images/slider/slider3.jpg"/>';
should work.
Explanation:
'<img src="templates/'
is a single-quoted string, which happens to contain a double-quote (which is needed for the html src attribute, or any other html attribute value really)
.
(dot) is the string concatenation operator. It concatenates ("glues") the first string together with...
$this->template
which is presumably a string containing the name of the template (not clear from your code example). Note that if $this->template comes from user input, or an otherwise unvalidated source, it could be used for cross-site scripting, eg. if it contains "><script>alert("XSS!")<script>, javascript is executed in the browser!
.
another concatenation with...
'/images/slider/slider3.jpg"/>'
which is another single-quoted string which happens to contain a double-quote, ending the src attribute value.
Try this:
$slider3 = '<img src="templates/"' . $this->template . '"/images/slider/slider3.jpg"/>';
$template = "MYTEMPLATE";
$slider3 = '<img src="templates/'.$template.'/images/slider/slider3.jpg"/>';
echo $slider3;
Will echo - >
<img src="templates/MYTEMPLATE/images/slider/slider3.jpg"/>
Just write:
<?php
$templates = "var";
echo "<img src='templates/${templates}/images/slider/slider3.jpg'/>";
it will result in
<img src='templates/var/images/slider/slider3.jpg'/>
Am echoing php variables which works fine but when i tried to output image, nothing seems to work
working.php
echo ("addMarker($lat, $lon,'<b>$name</b>$address<br><br>$desc');\n");
not_working.php
for image display, i added
<img src='http://localhost/services/status/" .$pic. "'>
hence
echo ("addMarker($lat, $lon,<img src='http://localhost/services/status/" .$pic. "'>,'<b>$name</b>$pic<br><br>$desc');\n");
Any Help
The php documentation about strings should clarify your issue, i hope. In simple words, variables are not expanded (parsed) in single quotes.
Best solution is to use sprintf:
sprintf('<img src="http://localhost/services/status/%s">', $pic);
OK solution:
echo '<img src="http://localhost/services/status/' . $pic . '">'
Not so ok solution:
echo "<img src=\"http://localhost/services/status/$pic\">"
I have this php line which works fine:
echo "<p>" . $post['message']. "</p>";
But I want to change it so it will link to my page (not to a single post). So it should look like that.
echo "<p>" . $post['message']. "</p>";
I have tried a lot many proposition gathered on different website, but each time I am getting an error.
Any idea ?
Thanks a lot!
Using single and double quotes, you avoid escaping issues. Try this:
echo '<p>'. $post['message']. '</p>';
i see that you didn't escaped from double quote that closes href attribute:
echo "<p><a href=\"https://www.facebook.com/rscmovement\" target=\"_blank\">"
I guess You have missed the back slash () before " after www.facebook.com/rscmovement.
"https://www.facebook.com/rscmovement\" "\"target=\"_blank\">" will