Make PHP generated URL trigerable in Javascript function

Question

I have a php script that I'm trying to trigger with Javascript. I'm trying to set the URL dynamically with php using a variable but am not having any luck, does anyone know how I should approach this?

<a href="#" onclick="return getOutput();">Click here</a>

<?php $update_url = 'https://' . $_SERVER['SERVER_NAME'] . '/xyz.php'; ?>

function getOutput() {
  getRequest(
    var str = "<?php echo $update_url ?>", // URL for the PHP file
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}

It should be an string, but it looks to me that you're passing 3 arguments to getRequest() function, so you cannot have var str = there. You could define it some where else or directly pass it — Adam Azad
– Adam Azad, Commented Jul 5, 2016 at 7:49
var is a statement, not an expression. You can't use statements as arguments. — gcampbell
– gcampbell, Commented Jul 5, 2016 at 7:49

Adam Azad · Accepted Answer · 2016-07-05 07:53:18Z

1

Here's what you can do; pass the PHP value as an string.

<?php $update_url = 'https://' . $_SERVER['SERVER_NAME'] . '/xyz.php'; ?>

function getOutput() {
  getRequest(
       "<?php echo $update_url ?>", // URL for the PHP file
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}

OR

Define as JavaScript value in the global scope (this may not work)

var updateURL = "<?php $update_url = 'https://' . $_SERVER['SERVER_NAME'] . '/xyz.php'; ?>";

function getOutput() {
  getRequest(
       updateURL, // URL for the PHP file
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}

answered Jul 5, 2016 at 7:53

Adam Azad

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Comments

PaulH · Accepted Answer · 2016-07-05 07:57:26Z

0

You can't have var str = as a parameter to a function(). You can't write

functionName(var a='a', b); // this will not work

This is a way to solve that:

function getOutput() {
  var str = "<?php echo $update_url ?>"; // URL for the PHP file
  getRequest(
       str,
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}

and this way is shorter:

function getOutput() {
  getRequest(
       <?php echo $update_url ?>, // URL for the PHP file
       drawOutput,  // handle successful request
       drawError    // handle error
  );
  return false;
}

edited Jul 5, 2016 at 7:57

answered Jul 5, 2016 at 7:50

PaulH

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