Concatenation of php variable with javascript

Question

var settings = {
  "async": true,
  "crossDomain": true,
  "url": 'https://example.com/something.aspx?i='<? echo urlencode($_GET['id']); ?>,
  "method": "GET",
  "headers": {
    "cache-control": "no-cache",
  }
}

It doesn't work this way, the concatenation is wrong I think. Tried few ways still doesn't work.

"crossDomain": true, — This almost certainly doesn't do what you think it does. Look it up and decide if you really need it. — Quentin
– Quentin, Commented Jun 26, 2016 at 16:09
If you looked at what the browser received, surely it would have been fairly clear what was wrong. — T.J. Crowder
– T.J. Crowder, Commented Jun 26, 2016 at 16:09
'https://example.com/something.aspx?i=' + <? echo urlencode($_GET['id']); ?> — Bogdan Kuštan
– Bogdan Kuštan, Commented Jun 26, 2016 at 16:22

Quentin · Accepted Answer · 2016-06-26 16:09:17Z

5

You need to put the data inside the JavaScript string literal. Move the ' to after the extra data you are outputting.

answered Jun 26, 2016 at 16:09

Quentin

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Comments

user128161 · Accepted Answer · 2016-06-26 16:09:23Z

3

You just had the single quote the wrong side.

Don't forget you're outputting to HTML, so you don't have to concatenate a PHP variable to a JavaScript variable.

var settings = {
  "async": true,
  "crossDomain": true,
  "url": 'https://example.com/something.aspx?i=<?php echo urlencode($_GET['id']); ?>',
  "method": "GET",
  "headers": {
    "cache-control": "no-cache",
  }
}

answered Jun 26, 2016 at 16:09

user128161

Comments

DBX12 · Accepted Answer · 2016-06-26 17:30:55Z

-1

Maybe you can try it the other way round. Like

<?php
  echo 'var settings = {
    "async": true,
    "crossDomain": true,
    "url": "https://example.com/something.aspx?i='.urlencode($_GET['id']).'",
    "method": "GET",
    "headers": {
      "cache-control": "no-cache",
    }
  }'; 
?>

Please note the changes I made with the ' and " in the url attribute.

edited Jun 26, 2016 at 17:30

answered Jun 26, 2016 at 16:12

DBX12

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1 Comment

Chris Evans Over a year ago

You're still closing the " .. " string here, so this won't work.

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