php variable to javascript with json_encode

Question

I know this topic was already discussed a few times but I can't seem to find what I'm doing wrong.

What I'm trying to do: The user types in a number and by clicking on the button creates a table with that number of columns.

Heres the php:

<?php
$twig = require_once('bootstrap.php');
$hostname = 'localhost';
$username = 'root';
$password =  '';
$conn = new PDO("mysql:host=$hostname;dbname=mydb", $username, $password);

echo $twig->render('index.html', array());

$numOfRows = 1;

if(isset($_POST['button'])){
    $numOfRows = $_POST['num_input'];
}

html/javascript:

<html>
<head>
<script>
function insertRows(){
    var numOfRows = <?php echo json_encode($numOfRows) ?>;
    var out = "<table><tr><th>test</th>";

    for (i = 0; i < numOfRows; i++){
        out += "<th>test</th>";
    }

    out += "</tr></table>";
    document.getElementById("table").innerHTML = out;
}
</script>
</head>

<body>
<form action="index.php" method="post">
  <textarea id="num_input" name ="num_input"></textarea>
  <button type="button"  name="button" onclick="insertRows()"> Go </button>
</form>

<p id="table"></p>

</body>
</html>

Theres no error or anything since I'm not using a IDE, just doing it in vim but the error is that is just doesn't happen. If i change "numOfRows" in the for loop to a number it works, so I'm pretty sure the json_encode is the problem.

Thanks!

EDIT: Just to test it, I used a string variable $str = "test"; the php file, and instead of using the for loop, I just edited javascript to

var str = <?php echo json_encode($str); ?>;
alert(str);

and I also tried

var str = <?php echo $str; ?>;
alert(str);

but nothing works.

That's the js var numOfRows not the php one. Why would I have to use $ in javascript variable? — mcCat, Oct 12 '15 at 15:29
Why is `numOfRows` actually a count of the number of columns? — Phylogenesis, Oct 12 '15 at 15:32
devlin carnate: why not, what else would be good to use? I'm new to php and I'm just playing around, I'm happy to try out something else :) — mcCat, Oct 12 '15 at 15:37
@mcCat If you want to send a comment to someone, you have to put @ before their name. — Barmar, Oct 12 '15 at 15:54

Mikki Tiger · Answer 1 · 2015-10-16T09:15:22.633

4

json_encode is not necessary in this case.

Simply replace

var numOfRows = <?php echo json_encode($numOfRows); ?>;

with

var numOfRows = <?php echo (int)$numOfRows; ?>;

edited Oct 16 '15 at 09:15

answered Oct 12 '15 at 15:37

Mikki Tiger

41
3

Added semicolons inside PHP part. Maybe it will help. – Mikki Tiger Oct 16 '15 at 09:17
If that also doesn't work, then there's nothing wrong with `json_encode()`. The problem is that `$numOfRows` isn't defined when that line of code runs. `$numOfRows` is defined in your first code block, but how are the two code blocks connected to each other? Setting it in one place doesn't magically make it work elsewhere.# – Simba Oct 16 '15 at 09:39

mrhn · Answer 2 · 2015-10-12T15:49:24.387

0

Edit: You are missing a ; on the

<?php echo json_encode($numOfRows) ?>

Should be

<?php echo json_encode($numOfRows);?>

And in these cases, if would be good to check the server log, this will automaticly make you better at finding these mistakes yourself.

You are mixing up ints and strings. The database will in PHP always return strings and the way you are using the variable as an int in a for loop.

The following change i believe would achieve the right result.

$numOfRows = intval($_POST['num_input']);

Where you use PHP's conversion to integer function there is at a global level.

edited Oct 12 '15 at 15:49

answered Oct 12 '15 at 15:31

mrhn

17,961
4
27
46

But `i < numOfRows` will convert `numOfRows` to an integer before comparing, so the result should be the same. – Barmar Oct 12 '15 at 15:33
Tried it but unfortunately it didn't help :/ – mcCat Oct 12 '15 at 15:37
Updated my post, got that version working in phpfiddle without the javascript thou – mrhn Oct 12 '15 at 15:49
; at the end doesn't help either. – mcCat Oct 12 '15 at 16:09
@Martin Henriksen how do I do that? :) Do you mean debugging in developer tools? – mcCat Oct 12 '15 at 18:03
What php version do you have installed and what server are you running it on? – mrhn Oct 12 '15 at 18:03
@Martin Heriksen 5.6.12 and I'm using xampp – mcCat Oct 12 '15 at 18:27
http://stackoverflow.com/questions/3719549/where-does-phps-error-log-reside-in-xampp – mrhn Oct 12 '15 at 18:32

score -1 · Answer 3 · answered Oct 12 '15 at 15:27

-1

You did not forget any $. JS does not need $ for variables.

As far as your json_encode is concerned, if you are just passing an integer from PHP to JS, there is no need to json_encode. Just pass the variable to JS as <?=$numOfRows?> in the JS source.

answered Oct 12 '15 at 15:27

Andrew Coder

1,058
7
13

Don't assume they have short-tags enabled. – Barmar Oct 12 '15 at 15:30
1

WIth PHP 5.4+ the `=` *(echo shorthand)* is available without having short tags enabled ... granted you're then assuming they've either got short tags enabled *or* 5.4+ : http://php.net/manual/en/migration54.new-features.php – CD001 Oct 12 '15 at 15:43

php variable to javascript with json_encode

3 Answers3