Variable in PHP http link

Question

I'm trying to get the cureent url and send that through in a link and I'm stuck. The link displays but it's missing the url that I want to include (content in the code sample here)

'href' => ( "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl='.$content .'")

lanzz · Accepted Answer · 2011-11-09 13:19:27Z

3

You're mixing single and double quotes. Try this:

'href' => ( "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl=".rawurlencode($content))

answered Nov 9, 2011 at 13:19

lanzz

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Aurelio De Rosa · Accepted Answer · 2011-11-09 13:20:02Z

3

First: use urlencode:

$content = urlencode($content);

Second: replace your single quote with double quote.

answered Nov 9, 2011 at 13:20

Aurelio De Rosa

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Comments

fuel37 · Accepted Answer · 2011-11-09 13:48:20Z

1

Your quotes aren't correct:

'href' => ( "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl=".$content)

You should also use urlencode() and urldecode() to ensure the $content variable is properly formatted

Edit Try this then:

$currentUrl = rawurlencode($_SERVER['PATH_INFO']);
$newUrl = "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl=";

header("Location: $newUrl.$currentUrl");

//Not sure if path_info will always contain the full url but there are lots of functions on the web to grab the current url that you can google for.

edited Nov 9, 2011 at 13:48

answered Nov 9, 2011 at 13:22

fuel37

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2 Comments

fuel37 Over a year ago

Your original post looks like a fragment of an array. should be $arr = array( 'href' => ( "chart.apis.google.com/…) ); Then you would access that variable as $arr['href'] so if you wanted to redirect to that url it would be header('Location: '.$arr['href']);

Dave Over a year ago

Thanks - your first solution was fine, I had the $content set in the wrong place. Thanks for your efforts. D

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