php variables in jquery script

Question

I have these php variables:

$n= $json['health'];
$n2= $json2['health'];

The output are numbers.

Here is the jquery script:

<script>
$(function() {
var valMap = [0, 25, 50, 100, 250, 500];
$("#slider-range").slider({
    min: 1,
    max: valMap.length - 1,
    value: 0,
    slide: function(event, ui) {                        
        $("#amount").val(valMap[ui.value]);                
    }       
});
//$("#amount").val(valMap[ui.value]);
})
</script>

I tried to get the vars in the script as my php variables like that:

var valMap = ['<?php echo($n); echo($n2);?>';];

How can I set the variables in the right way?

what do $n and $n2 look like? can you show a var_dump($n)? — Jeff
– Jeff, Commented May 12, 2018 at 15:51
then it would be var valMap = ['<?php echo $n .", " . $n2; ?>']; — Jeff
– Jeff, Commented May 12, 2018 at 15:54
apart from the missing , you also had a ; to much here: ;?>';]; — Jeff
– Jeff, Commented May 12, 2018 at 15:55
@Jeff Why adding the single quote inside the var valMap = [ ] — benjarobin
– benjarobin, Commented May 12, 2018 at 15:58

benjarobin · Accepted Answer · 2018-05-12 15:53:41Z

1

You should try something like that:

var valMap = [<?php echo "$n, $n2"; ?>];

answered May 12, 2018 at 15:53

benjarobin

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Comments

Elias Gomes · Accepted Answer · 2018-05-12 16:50:02Z

0

Declare an array with these variables in your PHP code.

$n_arr=array($json['health'],$json2['health']);

Then in your script use that array like this.

var valMap = <?php echo json_encode($n_arr);  ?>;

This should be the tidy way to do it!

answered May 12, 2018 at 16:50

Elias Gomes

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Comments

Iynga Iyngaran Iyathurai · Accepted Answer · 2018-05-12 16:58:12Z

0

You should try something like that:

var valMap = [<?php echo $n.",".$n2; ?>];

edited May 12, 2018 at 16:58

answered May 12, 2018 at 16:16

Iynga Iyngaran Iyathurai

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3 Comments

benjarobin Over a year ago

No, this is going to print: var valMap = [3..67];

Iynga Iyngaran Iyathurai Over a year ago

sorry there was an issue in echo. I have changed it now. Please try again with my changes

benjarobin Over a year ago

Yes, now it's right, but this answer is very similar with my own answer, and this answer is already (almost) in a comment...

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