Pass php value to html [closed] - php

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I am trying to pass a simple php value to html but I cannot format it in the right way. How can I do this?
<?php
$html = '
<img src="'$val'.svg" alt="15"> //Does not work
';
$val = 5;
?>

You would try to format it this way:
<?php
$val = 5;
$html = "<img src='${val}.svg' alt='15'> ";
?>

Related

foreach loop in Multidimensional Arrays [closed]

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Closed 5 years ago.
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$marks=array(
array(a1,a2,a3),
array(b1,b2,b3),
array(c1,c2,c3)
);
so i want this in foreach loop
example
echo "text-start".$marks[0]."text11".$marks[1]."text22".$marks[2]."text-end";
and result to be:
text-starta1text11a2text22a3text-end
text-startb1text11b2text22b3text-end
text-startc1text11c2text22c3text-end
i tried this but it show a1a2a3 b1b2b3
and i cant add text between them
for($r=0;$r<count($marks);$r++)
{
for($c=0;$c<count($marks[$r]);$c++)
{
echo $marks[$r][$c]."test";
}
echo "111<br>";
}
Basics...
foreach($marks as $marky_marks){
echo "text-start".$marky_marks[0]."text11".$marky_marks[1]."text22".$marky_marks[2]."text-end";
}

Disable errors when getting postal code using ipapi.co [closed]

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Closed 6 years ago.
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I'm retrieving the postal code by using -
<?php $pstcde = file_get_contents('https://ipapi.co/'.$_SERVER['REMOTE_ADDR'] . '/postal/'); ?>
It works great but if the postal code is not found, it says, "None". I can't find a way to disable the message.
You can empty the variable if the respond is None. With a simple check.
<?php
$pstcde = file_get_contents('https://ipapi.co/'.$_SERVER['REMOTE_ADDR'] . '/postal/');
if($pstcde === "None")
$pstcde = ""; //Empty string. Or you can use unset($pstcde); if you want to unset the variable.
?>

Cannot put $url; on "define" PHP [closed]

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Closed 7 years ago.
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Im newbie in php im learning it since some weeks ago but i didnt understand one thing.
I have a
$url = "http://example.com/";
define('LOCATION', '<?php echo $url; ?>');
but it is not working. I didnt understand why "define" term cannot read
<?php
$url = "http://example.com/";
define('LOCATION', $url);
1.You are alreay in php tag,so no need of any extra php tag.
2.And when assigning value to a variable , we never use echo in php . Its used to display value in html only.

i want to remove %20 from the url in php [closed]

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Closed 8 years ago.
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This is my page url
http://myaliveidea.com/news/news/readmore/56/%20When%20Harry%20and%20Sally%20met%20New%20York%20City
but i want my page url like this
http://myaliveidea.com/news/news/readmore/59/When-Harry-and-Sally-met-New-York- City
the solution of this question is
<?php
$url = str_replace(' ','-',$sport['title']);
$url = str_replace(":",'',$url);
$url = str_replace("'",'',$url);
?>
<?php echo $sport_top['title'];?>
Use urldecode();
$url = urldecode($url);
then replace spaces(' ') by - using str_replace();

PHP: easy random picture code [closed]

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Closed 8 years ago.
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My code is random but its does not print out the picture just the text. When I add the IMAGE tag. Its go all wrong. Where do I go wrong?
$plaatje[] = 'afbeelding1.jpg';
$plaatje[] = 'afbeelding2.jpg';
$plaatje[] = 'afbeelding3.jpg';
$plaatje[] = 'afbeelding4.jpg';
$nummer2 = mt_rand(1,4);
echo "$plaatje[$nummer2]";
Use array_rand() - Picks one or more random entries out of an array
echo $plaatje[array_rand($plaatje)];
With an image tag, something like:
<img src="<?php echo $plaatje[array_rand($plaatje)]; ?>" />
or
echo "<img src='".$plaatje[array_rand($plaatje)]."' />";
Here is the documentation
Although you should add the image tag to show where it goes wrong, your code already has a problem as array indices are 0-based.
So you would need:
$nummer2 = mt_rand(0,3);

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