array to js function from sql row

Question

i have an sql with 2 columns, latitude and longitude.

I am able to print out the entries of these two columns like:

$result = mysqli_query($con,"SELECT * FROM events");

while ($row = mysqli_fetch_array($result)) {

    echo $row['latitude'];
    echo $row['longitude'];

}

I also have a JS function, where i need a while to add as many JS lines as the sql query returns. Here is the JS:

var myMarkers = {"markers": [
{"latitude": "47.47256859162068", "longitude":"19.055979251861572", "icon": "marker.png", "baloon_text": 'This is <strong>Budapest</strong>'},
]
};

So i'd need something like: "latitude": $latitude[i] and "longitude": $longitude[i] all these in a for or a while

Is there a way that i could pass the content of the two sql rows to a js function, to be able to get all the line with the correct values?

Where does the value for ballon_text come from? Is it in the same database as lat long? — Steve
– Steve, Commented Jul 31, 2014 at 12:00

user399666 · Accepted Answer · 2014-07-31 12:09:54Z

2

An example of what to do can be seen here.

Convert the data into JSON.

$markers = array();
while ($row = mysqli_fetch_array($result)) {
    $markers[] = array(
        'longitude' => $row['longitude'],
        'latitude' => $row['latitude']
    );
}
$markersJson = json_encode($markers);

Then pass it to your JS:

<script type="text/javascript">
    var obj = JSON.parse('<?= $markersJson ; ?>');
    console.log(obj);
</script>

Check your console to view the structure of the data.

edited Jul 31, 2014 at 12:09

answered Jul 31, 2014 at 12:04

user399666

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4 Comments

Vivek Pratap Singh Over a year ago

Why there is need to create an empty array $markers when things can be done without creating it ?

user399666 Over a year ago

@Vivek Readability. Also allows me to narrow down the amount of columns if I need to.

user399666 Over a year ago

@Vivek Updated my answer to show you what I meant. My first approach was a tad bit "lazy". :)

Steve Over a year ago

is JSON.parse needed when printing an object declaration? Either way, ballon_text data needs to be included, though OP has not said where this data comes from

Vivek Pratap Singh · Accepted Answer · 2014-07-31 12:16:23Z

0

Convert the result of mysql data array into JSON.

$result_json = json_encode($result);

And parse this variable in jQuery fashion to access all values.

<script type="text/javascript">
    var obj= $.parseJSON('<?= $result_json; ?>'); //Parse the above encoded Json.
    //Here is how you iterate
    var json = $.parseJSON(obj);
    $(json).each(function(i,val){
        console.log(val);
    });
</script>

edited Jul 31, 2014 at 12:16

answered Jul 31, 2014 at 12:07

Vivek Pratap Singh

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Comments

imaibou · Accepted Answer · 2014-07-31 12:01:53Z

0

you can use something like this:

<script>
var myMarkers = {"markers": [

<?php
$result = mysqli_query($con,"SELECT * FROM events");

while ($row = mysqli_fetch_array($result)) {?>

    {"latitude": "<?php echo $row['latitude'];?>", "longitude":"<?php echo $row['longitude'];?>", "icon": "marker.png", "baloon_text": 'This is <strong>Budapest</strong>'},
<?php
}
?>
]
};
</script>

PHP is a server side language so it gets executed before javascript which is a client side language, the trick is to use :

"latitude": <?php echo $latitude[i];?>

answered Jul 31, 2014 at 12:01

imaibou

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Comments

Satish Sharma · Accepted Answer · 2014-07-31 12:02:12Z

0

you can try

<script>
<?php
$result = mysqli_query($con,"SELECT * FROM events");
$i=0;
while ($row = mysqli_fetch_array($result)) 
{   
?>
var myMarkers[<?php echo $i;?>] = {"markers": [{"latitude": "<?php echo $row['latitude'];?>", "longitude":"<?php echo $row['longitude'];?>", "icon": "marker.png", "baloon_text": 'This is <strong>Budapest</strong>'},]};
<?php
    $i+=1;
}
?>
</script>

answered Jul 31, 2014 at 12:02

Satish Sharma

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Comments

gnuanu · Accepted Answer · 2014-07-31 12:17:59Z

0

Fetch your data through the SQL query and put the output in a single array via your loop. Use json_encode() to serialise the array into a transportable form, pass it to JavaScript, for example by an XMLHttpRequest and parse it in JS with the internal JSON.parse().

edited Jul 31, 2014 at 12:17

gnuanu

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answered Jul 31, 2014 at 12:01

Freytag

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