Passing JSON to Array in PHP

Question

I'm working on a filter in which results are filtered right away, I'm wondering if that may be the cause of the problem so I thought I would ask and see if anyone could give me a pointer on how to proceed.

      <script>
        var services = [
 <?php
//Variables for connecting to your database.
//These variable values come from your hosting account.
$hostname = "###";
$username = "###";
$dbname = "###";

//These variable values need to be changed by you before deploying
$password = "###";
$usertable = "###";
$url = "permalink";
$title = "Address";
$amount = "rent";
$id = "id";
$status = "Beds";
$nonprofit = "Address";

//Connecting to your database
mysql_connect($hostname, $username, $password) OR DIE ("He's dead Jim");
mysql_select_db($dbname);

//Fetching from your database table.
$query = "SELECT * FROM $usertable";
$result = mysql_query($query);

if ($result) {
    while($row = mysql_fetch_array($result)) {
        $url = $row["$url"];
        $title = $row["$title"];
        $amount = $row["$amount"];
        $id = $row["$id"];
        $status = $row["$status"];
        $nonprofit = $row["$nonprofit"];

        echo '{"permalink": "';
        echo "{$url}";
        echo '",';
        echo '"title": "';
        echo "{$title}";
        echo '",';
        echo '"amount":';
        echo "{$amount}";
        echo ',';
        echo '"id": "';
        echo "{$id}";
        echo '",';
        echo '"status": "';
        echo "{$status}";
        echo '",';
        echo '"address": "';
        echo "{$address}";
        echo '",';
        echo '},';

    }
}
?>       

]
                //]]>
      </script>


  <script id="template" type="text/html">
       <a title="{{title}}" href="{{permalink}}">
         <div class="fs_box hide-for-small-down">
           <div class="fs_left">
             <span class="fs_head">{{title}}</span>
             <span class="fs_id"><img src="images/{{id}}.jpg" width="75%" height="75%" onError="this.onerror=null;this.src='images/logo.png';"></span>
       <span class="fs_status">{{status}}</span>
             <span class="fs_disc">{{address}}</span>
           </div>
           <div class="fs_price">${{amount}}+</div>
           <div class="clear"></div>
        </div>
      </a>
    </script>

I'm expecting it to produce a bunch of results that then are filtered criteria which are elsewhere in the page. When I try it currently just as a php code it outputs fine. However, when I try it in the php file that this should go in it produces nothing. Or does it dislike being in a script? Thanks for any help!

Why are you adding [ before the php script? You can make an array in php the use json_encode to encode the array to json. — Abdessabour Mtk
– Abdessabour Mtk, Commented Oct 1, 2014 at 20:24
@user3980820 I was putting the JSON between that or could I put the SQL to create an array then put that array in between the [ ]? — obi1kenobi2
– obi1kenobi2, Commented Oct 1, 2014 at 20:26
mysql_query is an obsolete interface and should not be used in new applications and will be removed in future versions of PHP. A modern replacement like PDO is not hard to learn. If you're new to PHP, a guide like PHP The Right Way can help explain best practices. — tadman
– tadman, Commented Oct 1, 2014 at 20:27
Additionally, is there any reason you're not using the json_encode method to create JSON rather than doing this completely ineffective version? That doesn't escape anything correctly and will probably produce a lot of invalid JSON output. — tadman
– tadman, Commented Oct 1, 2014 at 20:28
I'll take a look at the json_encode thanks for the suggestions! — obi1kenobi2
– obi1kenobi2, Commented Oct 1, 2014 at 20:32

David Jones · Accepted Answer · 2014-10-01 20:40:30Z

2

You can use json_decode and json_encode to turn an array to json and json back to an array.

Also someone will probably mention that you should not be using the mysql_* functions in PHP as they are depreciated.

Something like this:

 <?php
//Variables for connecting to your database.
//These variable values come from your hosting account.
$hostname = "###";
$username = "###";
$dbname = "###";

//These variable values need to be changed by you before deploying
$password = "###";
$usertable = "###";
$url = "permalink";
$title = "Address";
$amount = "rent";
$id = "id";
$status = "Beds";
$nonprofit = "Address";

//Connecting to your database
mysql_connect($hostname, $username, $password) OR DIE ("He's dead Jim");
mysql_select_db($dbname);

//Fetching from your database table.
$query = "SELECT * FROM $usertable";
$result = mysql_query($query);

if ($result) {
    $results = array()
    while($row = mysql_fetch_array($result)) {
        $results[] = $row;
    }

    $json =  json_encode($results);
}
?>       

]

<script>
    var services = <?php echo $json; ?>;
</script>

This would give you a json object to use to render in your script.

edited Oct 1, 2014 at 20:40

answered Oct 1, 2014 at 20:23

David Jones

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2 Comments

tadman Over a year ago

It'd be worth converting the example given in the question to a version that uses the built-in JSON function. It's way cleaner when expressed that way and has the additional benefit of actually working.

David Jones Over a year ago

I have edited my answer to build a some json using json_encode

user2676655 · Accepted Answer · 2014-10-01 20:29:20Z

-1

What extension is the file you are saving? If it's not .php or an extension set to render php, then you'll just have the code show up as test. You might want to pull out the "die" statement after the db connect. This looks like you are running php in a .js file so you probably want the entire file to write out rather than stop because you couldn't connect to the database (or at least give 0 results, maybe a warning)

answered Oct 1, 2014 at 20:29

user2676655

441 bronze badge

1 Comment

tadman Over a year ago

It's not a JavaScript file. It's just running when rendering a <script> tag on a page.

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