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up vote -2 down vote favorite

I have searched a lot and can't seem to find what I am looking for.

I have a .js field, and I really need help with some php code in side it. I have no idea how to put them together and would really appreciate some help.

The Javascript code(This code is inside the js file):

<span class="gravatar"><img src="',params.gravatar, '" width="23" height="23" onload="this.style.visibility=\'visible\'" />

where the <img src="',params.gravatar,'">

I want to add:

  <?php
$con=mysqli_connect("localhost","root","","facebook");

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM login WHERE username='$username'");
$username = $_SESSION['username'];

while($row = mysqli_fetch_array($result))
  {


    echo $row['profilepic'];echo "<br>";
mysqli_close($con);
    }
  ?>

Insted of ,params.gravatar,

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<img src="<?php echo $row['profilepic']; ?>" /> –  Shomz Jun 21 '13 at 18:54
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3 Answers

up vote 3 down vote

I assume you mean that the HTML snippet you are posting is inside a separate JS file. If that is the case, you can assign the profile picture URL to a global JS variable in your PHP code:

Your PHP file:

<?php
// Your DB call code goes here...
?>
<script>
    var profilePic = '<?=$row['profilepic']?>';
</script>
<script src="/your/jsFile.js"></script>

And in your jsFile.js:

var img = '<span class="gravatar"><img src="'+profilePic+'" width="23" height="23" />';
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up vote 1 down vote

If what you want was even possible, an attacker could just change your queries and execute them directly from javascript. That would be funny, but not very good in the end for developers.

What you should look for is asking javascript (through JQuery) to query the php code and from PHP do your queries and return the results to the javascript in JSON format.

Look up $.post in JQuery.

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up vote 0 down vote

I really need help with some php code in side it

No web server is executing PHP code inside a .js file per default, you have to create a .php file so your web server is executing the file's content. So this won't work without extra server setup, you need a different setup. I'm sorry…

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