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, Feb 16 '16 at 15:27
1. It is few lines of code and it is separated, function dataQuery comes from other file 2. It works without var. — stevenhawkingsbiggestfan, Feb 16 '16 at 15:30
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, Feb 16 '16 at 15:34
Then just say that it is better to use the var, nobody mentioned that... — stevenhawkingsbiggestfan, Feb 16 '16 at 15:59

score 3 · Answer 1 · 2016-02-16T15:33:40.043

3

Just change it to:

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

edited Feb 16 '16 at 15:33

answered Feb 16 '16 at 15:26

1

Damn I was so close, I tried to do echo "jArray = JSON.parse(. JSON_encode($work) .);"; – stevenhawkingsbiggestfan Feb 16 '16 at 15:29
`Uncaught SyntaxError: Unexpected token o` pointing on `[object Object],[object Object]` – stevenhawkingsbiggestfan Feb 16 '16 at 15:32
What is your code now? I think now its a problem with your js: http://stackoverflow.com/questions/8081701/i-keep-getting-uncaught-syntaxerror-unexpected-token-o – Feb 16 '16 at 15:34
When I get rid of JSON_parse it throws `Notice: Array to string conversion jArray = . JSON_encode(Array) .; ` I just added this attribute to connection `$dbh->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);` and changed echo `echo "jArray = . JSON_encode($work) .;";`. – stevenhawkingsbiggestfan Feb 16 '16 at 15:40

CodeGodie · Accepted Answer · 2016-02-16T15:46:55.900

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 '16 at 15:46

answered Feb 16 '16 at 15:41

CodeGodie

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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");` – stevenhawkingsbiggestfan Feb 16 '16 at 15:53
I would create that function in PHP, and use AJAX to run it through JS. – CodeGodie Feb 16 '16 at 15:54
Any tips? I never did ajax, I don't know where to start. – stevenhawkingsbiggestfan Feb 16 '16 at 16:02
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. – CodeGodie Feb 16 '16 at 16:04
Ok but I have to wait 90 minutes to do that. – stevenhawkingsbiggestfan Feb 16 '16 at 16:11
I see. Well, in the meantime you can check out this post, it will show you how to make ajax requests using jQuery: http://stackoverflow.com/questions/2269307/using-jquery-ajax-to-call-a-php-function – CodeGodie Feb 16 '16 at 16:27
http://stackoverflow.com/questions/35440184/making-mysql-queries-using-javascript-function there you go. – stevenhawkingsbiggestfan Feb 16 '16 at 18:21

Can't use PHP variable in JS

2 Answers2