Passing php database result to javascript array

Question

I have to create a dropdown list and a button that will add another dropdownlist once click. i used this javascript code:

var array = ["Volvo","Saab","Mercades","Audi"];

        var cell2 = row.insertCell(1);
        var element2 = document.createElement("select");
        element2.name="activity_dropdown"
        element2.id="activity_dropdown"
        cell2.appendChild(element2);

        for (var i = 0; i < array.length; i++) {
            var option = document.createElement("option");
            option.setAttribute("value", array[i]);
            option.text = array[i];
            element2.appendChild(option);
        }

but i want that my var array will come from the database result query. what can i do to pass php array to javascript array?

by the way.. i have tried to used json_encode but it seems not working.. here is the my code:

for data.php

    <?php
    include('connection/dbConn.php');


// get data and store in a json array
$activity=$conn->query("SELECT activity_name FROM rehab_activities");


while ($row =mysqli_fetch_array($activity)) {
    $activities[] = array(
    'actvityName' => $row['activity_name']
    );
}

echo json_encode($activities);
 ?>

and try it here, but nothing comes out:

<script>
 jQuery(document).ready(function(){
    jQuery("#add").click(function(){
    var res = new Array(); 
        jQuery.getJSON("data.php", function(data) {
    var i= 0;
    while(i<data.myarray.length){
                res[i]=data.myarray[i];
             i++;
    }
  jQuery("#result").html(res[0]);
            });
        });

    });
</script>
<body>
    <input type="button" id="add">
    <div id="result"></div>
</body>
</html>

Dante · Accepted Answer · 2015-01-29 18:27:37Z

5

Make php write that part of your script while it loops through your query. Like so:

var array = [
    <?php
        $query = mysql_query("SELECT * FROM cars");
        while ($car = mysql_fetch_assoc($query)) {
            $car_name = $car["name"];
            echo "'$car_name',";
        }
    ?>
];

For this to work, your file must be a php file and not a javascript file, since inside a php file you can also write html and therefore, javascript.

I personally solve this by making a file called db-to-js.php that only contains javascript arrays created by php. Whenever the database is modified, one or more of these javascript arrays are also modified since they are created every time this file is executed.

Lastly, you have to include the file where you need it as a script:

<script type="text/javascript" src="db-to-js.php"></script>

I hope this helps.

answered Jan 29, 2015 at 18:27

Dante

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2 Comments

jaspher chloe Over a year ago

Thanks a lot.. it took me days before i realized what you've been instructing to put <?php include "db-to-js.php;" ?>

Dante Over a year ago

Actually I suggested to include it in the page as a script, like in the second-to-last line. As html requests that document, the server will execute the php in it and leave only the javascript it has created. But include might work too. You're welcome.

Felix Runye · Accepted Answer · 2017-02-20 14:19:34Z

-1

I had a similar problem in vue js, renamed my 'app.js' file to 'app.php', where i added all my php functions and changed the reference script tag to

    <script> type='text/javascript' src='app.php' </script>

Worked For me!

answered Feb 20, 2017 at 14:19

Felix Runye

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