Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 7 years ago.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Improve this question
I am getting an image from mysql database and trying to show it. I am using the following code. But its not working .What's the mistake i am making?
while ($row=mysql_fetch_array($result,MYSQL_ASSOC)) {
$photo=$row['photo'];
$name=$row['firstname'].' '.$row['lastname'];
$email=$row['email'];
echo "<tr><td>".'<img src="data:image/jpeg;base64,<?php echo base64_encode( $photo ); ?>" />'.'</td><td>'.$name.'</td><td>'.$email.'</td></tr>';
}
Strip the double quotes after img src because you have already used a single quote before and it is being taken a string so remove double quotes
Ok thanks. I could solve the problem. I just changed the echo line as following
echo "<tr><td>".'<img src="data:image/jpeg;base64,'.base64_encode($photo).'" alt="photo">'."</td><td>".$name."</td><td>".$email."</td></tr>";
Related
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 4 years ago.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Improve this question
i want to add a password to this line :
define('FTP_PASS' '#Aj\Zx5YJG')
but wp-confing will not recognize # and \ as characters and for that reason my website goes down.
How can i add those characters as text there?
thanks!
Typo? you're missing a comma between the two parameters and the quotes on the screenshot do not look regular.
http://php.net/manual/en/function.define.php
define('FTP_PASS','#Aj\Zx5YJG');
define('FTP_PASS' '\#Aj\\Zx5YJG')
Escape the characters or use different brackets
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 6 years ago.
Improve this question
I am trying to create an image that changes dependent on the genre grabbed from an icecast server, I am pretty sure I have the base code correct I think I've just incorrectly inputted the PHP variable.
<?php
$stats = $core->radioInfo( "http://http://sc.onlyhabbo.net:8124/status-json.xsl" );
?>
<img src=http://www.habbo.com/habbo-imaging/avatarimage?user=<?php
echo $stats['genre'];
?>&action=std&direction=2&head_direction=2&gesture=sml&size=m&img_format=gif/>
is the full code. Have I inputted the PHP variable incorrectly
Where are the quotes in your Html?
<img src="http://www.habbo.com/habbo-imaging/avatarimage?user=<?php
echo $stats['genre'];
?>&action=std&direction=2&head_direction=2&gesture=sml&size=m&img_format=gif"/>
UPDATE EVERYBODY
This is now resolved, I decided to go down the CURL route for this, and at first it didn't work until my host raised our CloudLinux Process Limit. I am unsure what the actual issue with this code was, but the CURL route works fine. Thank you for any answers
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 6 years ago.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Improve this question
I'm trying to include a file to current php page but the name of the file depens on the lang.
This line does not work:
include_once PATH.'lang/'.$_SESSION['lang'].'.php';
How can I achieve that?
Your problem should be the "PATH". It must be a string to concatenate with the other strings. For example:
include_once '/sites/all/files/'.'lang/'.$_SESSION['lang'].'.php';
Also be careful with the slashes that make your path to the file.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 8 years ago.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Improve this question
I hava a problem at my PHP script!
I want to do something like
echo "hello $var";
And the output should be "hello $var". So it should not take the value of $var, but a string "$var".
I want to use this to create a php file with a php script...
Is there any way to solve this?
Thanks in advance!
Use single quotes:
echo 'hello $var';
Or escape the $:
echo "hello \$var";
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 8 years ago.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Questions concerning problems with code you've written must describe the specific problem — and include valid code to reproduce it — in the question itself. See SSCCE.org for guidance.
Improve this question
I am following video tutorials on the Internet, and this is one of them. However, I am getting a syntax error and can't find where it is.
<?php
function hesapla($ilktarih,$sontarih){
$yil=$son-$ilk;
$ay=$yil*12;
$gun=$ay*30;
return array("$ilktarih","$sontarih","$yil","$ay","$gun");
}
$ilktarih=$_POST["ilk"];
$sontarih=$_POST["ikinci"];
hesapla($ilktarih,$sontarih);
list ("$ilktarih","$sontarih",$yil,$ay,$gun)=hesapla($ilktarih,$sontarih);
?>
Remove double quotes from list variables like this :
<?php
function hesapla($ilktarih,$sontarih){
$yil=$son-$ilk;
$ay=$yil*12;
$gun=$ay*30;
return array("$ilktarih","$sontarih","$yil","$ay","$gun");
}
$ilktarih=$_POST["ilk"];
$sontarih=$_POST["ikinci"];
hesapla($ilktarih,$sontarih);
list ($ilktarih,$sontarih,$yil,$ay,$gun)=hesapla($ilktarih,$sontarih);
?>