How to print json array in jQuery?

Question

I create query in laravel 5.6

I need print json array in Jquery.

Return query in laravel is:

[
    {
        "day": 19,
        "count": 1
    },
    {
        "day": 18,
        "count": 3
    }
]

Script code is:

var chart = AmCharts.makeChart( "chartdiv", {
        "type": "serial",
        "theme": "light",
        "dataProvider": [
            {
                "day": 19,
                "count": 1
            },
            {
                "day": 18,
                "count": 3
            }
        ].reverse(),
        .
        .
        .
        .

I place {{ json_encode($query) }} in Jquery code:

var chart = AmCharts.makeChart( "chartdiv", {
    "type": "serial",
    "theme": "light",
    "dataProvider": [ {{ json_encode($query) }} ].reverse(),
    .
    .
    .
    .

But this code is not work...

How to issue this problem?

Given that the properties in the array you return are different to those expected in the actual code, I'm not sure why you'd expect this to work. You need to change the day and count property names. Also note that you don't need the extra [] around {{ json_encode($query) }} — Rory McCrossan
– Rory McCrossan, Commented Apr 19, 2018 at 14:57
@RoryMcCrossan Hi, my question is update. I can't remove [] becouse use ].reverse(). — utkarsh2k2
– utkarsh2k2, Commented Apr 19, 2018 at 15:01

Ravi · Accepted Answer · 2018-04-19 15:02:56Z

2

You need to parse json and remove an addition an array like below:

var chart = AmCharts.makeChart( "chartdiv", {
    "type": "serial",
    "theme": "light",
    "dataProvider": JSON.parse('{{ json_encode($query) }}').reverse(),
})

Hope this will help you

answered Apr 19, 2018 at 15:02

Ravi

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3 Comments

utkarsh2k2 Over a year ago

Uncaught SyntaxError: Unexpected token & in JSON at position 2 at JSON.parse (<anonymous>)

utkarsh2k2 Over a year ago

"dataProvider": JSON.parse('[{"day":19,"count":1},{"day":18,"count":3}]').reverse(),

Ravi Over a year ago

You have two way to fix this problem: 1) JSON.parse('{!! json_encode($query) !!}').reverse() 2) JSON.parse('<?php echo json_encode($query); ?>').reverse()

Anfath Hifans · Accepted Answer · 2018-04-19 18:16:21Z

0

Replace dataProvider by the following code

"dataProvider": JSON.parse(<?=json_encode($query);?>).reverse()

edited Apr 19, 2018 at 18:16

user6655984

answered Apr 19, 2018 at 16:47

Anfath Hifans

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