[pr0slyte..]

I have

Code

$images = json_decode('{"small": "path-to-small", "large": "path-to-large"}');

When I want to print the path of image small or image large I'm using this:

Code

print $images->{'small'};
print $images->{'large'};

But how to print the path in echo, for example:

Code

echo '<a href="LARGE HERE" data-lightbox="mygallery" title=""><img src="SMALL HERE" alt=""></a>';

I think you get the point!
Thank you!

[Don Vito Corleone]

In

Code

<?php ... ?>

tags. Also, shouldn't you be using a foreach?

[pr0slyte..]

In

Code

<?php ... ?>

tags. Also, shouldn't you be using a foreach?

Of course I have <?php ?> and foreach but that's not my question.
I'm asking for the syntax in <a> and <img> tags for outputting the paths from json_decode in PHP.

[Don Vito Corleone]

Of course I have <?php ?> and foreach but that's not my question.
I'm asking for the syntax in <a> and <img> tags for outputting the paths from json_decode in PHP.

Put the results in variables like you said you have done, and the input it in the tags like the code below. If this is also not what you meant than probably someone else needs to help us out.

Code

a href="<?php (php code) ?>"

For image src it is the same I believe.

[pr0slyte..]

Put the results in variables like you said you have done, and the input it in the tags like the code below. If this is also not what you meant than probably someone else needs to help us out.

Code

a href="<?php (php code) ?>"

For image src it is the same I believe.

I have tried this already. Doesn't working. 😶

[TyrantKingBen]

I have tried this already. Doesn't working. 😶

Why doesn't it work? Your server should execute any PHP before it responds with the page to the client. What's it doing?

[Charger]

PHP Code


<?php 
$j = json_decode('{"small": "path-to-small", "large": "path-to-large"}');
foreach($j as $item)
{
 ?>

<a href=<?=$item->large;?> data-lightbox="mygallery" title=""><img src=<?=$item->small;?> alt=""></a>; 
  <?php
}
?>

[Disease]

I have

Code

$images = json_decode('{"small": "path-to-small", "large": "path-to-large"}');

When I want to print the path of image small or image large I'm using this:

Code

print $images->{'small'};
print $images->{'large'};

But how to print the path in echo, for example:

Code

echo '<a href="LARGE HERE" data-lightbox="mygallery" title=""><img src="SMALL HERE" alt=""></a>';

I think you get the point!
Thank you!

You're decoding it correctly. What you're trying to do is string concatenation though. PHP supports this in a number of ways to do this.

Concatenation Operator

PHP Code


<?php

$images = json_decode('{"small": "path-to-small", "large": "path-to-large"}');
echo 'Path to small: ' . $images->small 


String Interpolation - Simple

PHP Code


<?php

$images = json_decode('{"small": "path-to-small", "large": "path-to-large"}');
echo "Path to small: $images->small"; 


String Interpolation - Complex

PHP Code


<?php

$images = json_decode('{"small": "path-to-small", "large": "path-to-large"}');
echo "Path to small: {$images->small}"; 


There are others, but these three will handle most use cases. Concatenation is what you're likely going to use in your specific example. At least as it's structured now. When you need to content with quotations in your output you need to decide between concatenation and interpolation. For interpolation to work you need to use double quotes; PHP won't interpret what is inside a single quote string literal. But if you switch to using double quote strings, you need to escape any occurrence of double quotes in the output.

Another alternative, though overkill for your example, is the use of functions such as printf() or sprintf(). They are basically identical to each other; the difference being that printf() will immediately output the formatted string. You can store the output of sprintf() to a variable.

[pr0slyte..]

It works! Thanks you all! :)