Can't use PHP variable in JS

Question

Just trying to get the result of query to the javascript array or something.

<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<title>Admin check</title>
<meta charset="UTF-8">
</head>

<body>
<script type='text/javascript'> 
<?php
include 'pdo_connect.php';
function pleaseWork() {
    return dataQuery("SELECT * FROM `grupy`")->fetchAll();
}

$work = pleaseWork();

echo "jArray = JSON.parse(<?php echo JSON_encode($work);?>);";
?>
</script>

Got code like that and there is the result:

<br />
<b>Notice</b>:  Array to string conversion in      <b>/virtual/chemioterapia2137.cba.pl/adminCheck.php</b> on line <b>20</b><br />
jArray = JSON.parse(<?php echo JSON_encode(Array);?>);  //jArray = JSON.parse('');

How can I get it working?

Sorry but this is a good example of "spaghetti code" You need to separate your code. A good rule is to separate PHP from JS as much as possible. — CodeGodie
– CodeGodie, Commented Feb 16, 2016 at 15:27
It never ceases to amaze me - people come here for help, but then complain about the comments. yes, it may work without the var, but it's not right without the var. Listen to the comments, people are trying to help! — random_user_name
– random_user_name, Commented Feb 16, 2016 at 15:34

CodeGodie · Accepted Answer · 2016-02-16 15:46:55Z

3

A couple of errors:

Your code is very messy. Its easy to separate this. Try to place the most PHP outside of your JS.
When using json_encode it is already turning your array into a readable JS object. So no need for parsing.

Rewrite your code in this manner, I created the array result manually to mimic your DB results:

<?php

function pleaseWork()
{
    return array(
        array(
            "name" => "John",
            "age" => 52
        ),
        array(
            "name" => "Jane",
            "age" => 48
        )
    );
}

$work = pleaseWork();
$json = json_encode($work);
?>
<html>
    <head>
        <script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
        <title>Admin check</title>
        <meta charset="UTF-8">
    </head>
    <body>
        <script type='text/javascript'>
            var jArray = <?= $json ?>;
            console.log(jArray);
        </script>
    </body>
</html>

your browser's console result:

0: Object
  age: 52
  name: "John"

1: Object
  age: 48
  name: "Jane"

edited Feb 16, 2016 at 15:46

answered Feb 16, 2016 at 15:41

CodeGodie

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7 Comments

stevenhawkingsbiggestfan Over a year ago

And what if I want to make php function that I can use in javascript for making querys? Like makeQuery("DELETE * FROM ``grupy`` WHERE 1");

CodeGodie Over a year ago

I would create that function in PHP, and use AJAX to run it through JS.

stevenhawkingsbiggestfan Over a year ago

Any tips? I never did ajax, I don't know where to start.

CodeGodie Over a year ago

Sure. I suggest opening up a new question here on SO and I can provide a thorough answer. Just let me know when you do so that I can check it out.

stevenhawkingsbiggestfan Over a year ago

Ok but I have to wait 90 minutes to do that.

|

score 3 · Accepted Answer · 2016-02-16 15:33:40Z

3

Just change it to:

echo "var jArray = JSON.parse(". JSON_encode($work) .");";

edited Feb 16, 2016 at 15:33

answered Feb 16, 2016 at 15:26

user5653854

4 Comments

stevenhawkingsbiggestfan Over a year ago

Damn I was so close, I tried to do echo "jArray = JSON.parse(. JSON_encode($work) .);";

stevenhawkingsbiggestfan Over a year ago

Uncaught SyntaxError: Unexpected token o pointing on [object Object],[object Object]

user5653854 Over a year ago

What is your code now? I think now its a problem with your js: stackoverflow.com/questions/8081701/…

stevenhawkingsbiggestfan Over a year ago

When I get rid of JSON_parse it throws <b>Notice</b>: Array to string conversion</b> jArray = . JSON_encode(Array) .; </script> I just added this attribute to connection $dbh->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC); and changed echo echo "jArray = . JSON_encode($work) .;";.

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