How to open a popup window having the source code of a page

Question

I have a file - filedata.php. The file shows only a CLICK ME link when opened in the browser. On clicking the CLICK ME link a popup window should appear which has the source code of filedata.php. I wrote the following code but it isn't working (no pop-up window appears). Please help me to figure out the error.

SCRIPT -

<!DOCTYPE html>
<html>
<script src="jquery-1.9.1.js"></script>  
<script>
$(document).ready(function() {
   $("a.xy").click(function (event) {
        //Prevent default behavior
        event.preventDefault();
        var js_array = <?php echo json_encode($php_array) ?>;
        var disp = window.open('','','width=400,height=400');
        $(disp.document.body).text( js_array.join("\n") );
    });
});
</script>

PHP CODE -

<?php
// Get a file into an array.
echo '<a href="" class="xy" onclick="openWin()">CLICK ME!</a><br>';
$path='C:\wamp\www\directory_listing\filedata.php';
$lines = file($path);

// Loop through our array, show HTML source as HTML source; and line numbers too.
foreach ($lines as $line_num => $line) {
    $php_array[$line_num]="Line <b>{$line_num}</b> : " . htmlspecialchars($line) . "<br />";
}

$html = implode('', file($path));
$trimmed = file($path, FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES);
?>
</html>

onclick="openWin()" this means the click call a function: function openwin(){ //code } is missing from js use firebug. you will see the error in console — Catalin Sterian, Dec 27 '13 at 08:00
@Catalin >> I removed it, the error is actually with json_encode. The console says - "Uncaught SyntaxError: Unexpected identifier" — Rajarshi Sarkar, Dec 27 '13 at 08:28
I need a popup window that contains source code of filedata.php — Rajarshi Sarkar, Dec 27 '13 at 09:03

score 0 · Answer 1 · edited May 23 '17 at 10:25

0

it is possible to communicate with new window but it is not possible to pass dom to new window.You would basically need to convert the generated DOM to a string, pass it across to the new window and then parse it as though it was a document.

Like This

edited May 23 '17 at 10:25

Community

1
1

answered Dec 27 '13 at 08:07

Milind Anantwar

81,290
25
94
125

score -1 · Accepted Answer · answered Dec 27 '13 at 08:35

Your hava used the wrong way to append html code, change $(disp.document.body).text( js_array.join("\n") ); to disp.document.body.innerHTML = js_array.join("\n");.

Below is all my code:

<!DOCTYPE html>
<html>
<?php
// Get a file into an array.
echo '<a href="" class="xy" onclick="openWin()">CLICK ME!</a><br>';
$path='C:\wamp\www\directory_listing\filedata.php';
$lines = file($path);

// Loop through our array, show HTML source as HTML source; and line numbers too.
foreach ($lines as $line_num => $line)
{
    $php_array[$line_num]="Line <b>{$line_num}</b> : " . htmlspecialchars($line) . "<br />";
}

$html = implode('', file($path));
$trimmed = file($path, FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES);
?>
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script>
$(document).ready(function() {
   $("a.xy").click(function (event) {
        //Prevent default behavior
        event.preventDefault();
        var js_array = <?php echo json_encode($php_array) ?>;
        var disp = window.open('','','width=400,height=400');
        disp.document.body.innerHTML = js_array.join("\n");
    });
});
</script>
</html>

This is helpful. Thanks @wander. Also, try and catch block can be included for json_encode and the onclick attribute has to be removed from tag. — Rajarshi Sarkar, Dec 27 '13 at 09:41

How to open a popup window having the source code of a page

2 Answers2