PHP not outputting new lines from JSON data

Question

I'm getting a feed that contains json data that is then decoded by php (json_decode) and echoed back out to my html page.

The feed contains \n characters but when the PHP echos them out, the HTML doesn't recognize them as new lines or line breaks so the HTML text just comes out as one big wall of text.

Can someone give me a hint as to what I should be doing to make the php echo out a
or something similar when the json data has \n in it?

As an example, the json data might contain "FRIDAY THROUGH WEDNESDAY.\n\nMORE RAIN IS" but when the HTML is generated it just looks like "FRIDAY THROUGH WEDNESDAY. MORE RAIN IS" all on the same line.

Thanks!

i don't think this is a problem with your json - how are you displaying it? — Daniel A. White
– Daniel A. White, Commented Jun 21, 2013 at 14:45
probalby the duplicate of that stackoverflow.com/questions/42068/… — Ivan Kuckir
– Ivan Kuckir, Commented Jun 21, 2013 at 14:49

Valeh Hajiyev · Accepted Answer · 2013-06-21 14:49:05Z

2

Try:

echo nl2br($json_string);

answered Jun 21, 2013 at 14:49

Valeh Hajiyev

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Dany Caissy · Accepted Answer · 2013-06-21 15:05:33Z

1

You must replace your '\n' with the equivalent in HTML, which is the br tag.

You can output your string in the following way :

echo str_replace("\n", "<br />", $yourString);

edited Jun 21, 2013 at 15:05

answered Jun 21, 2013 at 14:46

Dany Caissy

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6 Comments

Question Asker Over a year ago

The solution you posted prior to editing worked but I can't get it to work with nl2br.

Dany Caissy Over a year ago

Wierd, maybe it's your version of PHP? I put it back to how it originally was, for the sake of diversity.

Question Asker Over a year ago

Yeah maybe... do you know if nl2br works only when echoing? Because I have a bunch of echo statements that I didn't want to have to update I just tried changing the string like this: $page = nl2br(file_get_contents('JSON_FEED_URL'));

Dany Caissy Over a year ago

It won't only work while echoing, but it won't modify the string you pass to it. It will return the resulting string, so you need to either echo or save the string that it returns.

Question Asker Over a year ago

Thanks, I edited my above response about the same time you replied to show that I am passing the string to nl2br. Would you expect that to work?

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Community · Accepted Answer · 2023-11-17 19:42:58Z

0

From php.net:

https://www.php.net/manual/fr/function.json-decode.php#112084

function json_decode_nice($json, $assoc = TRUE){
    $json = str_replace(array("\n","\r"),"\\n",$json);
    $json = preg_replace('/([{,]+)(\s*)([^"]+?)\s*:/','$1"$3":',$json);
    $json = preg_replace('/(,)\s*}$/','}',$json);
    return json_decode($json,$assoc);
}

edited Nov 17, 2023 at 19:42

CommunityBot

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answered Jun 21, 2013 at 14:48

Armel Larcier

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