json data with multiple arrays displaying in a table

Question

i have a json data like so

{
    "code": 1,
    "data": [
        {
            "apple": [
                {
                    "id": 127,
                    "type": 1,
                    "color": green,                   
                    "stage": 1,                    
                    "status": 1                   
                },
                {
                    "id": 128,
                    "type": 2,
                    "color": red,                   
                    "stage": 1,                    
                    "status": 1  
                }
            ]
        },
        {
            "oranges": [
                {
                    "id":133
                   "type": 3,
                    "color": rainbow,                   
                    "stage": 1,                    
                    "status": 1   
                },
                {
                   "id":134
                   "type": 3,
                    "color": black,                   
                    "stage": 1,                    
                    "status": 1
                }
            ]
        },
        {
            "berry": [
                {
                    "id":4
                   "type": 2,
                    "color": white,                   
                    "stage": 1,                    
                    "status": 1
                }
            ]
        },
        {
            "watermelon": [
                {
                    "id":5
                   "type": 2,
                    "color": red and blue,                   
                    "stage": 1,                    
                    "status": 1
                }
            ]
        }
    ],
    "bleh": "Succesfully queried database"
}

i would like to create a table in php that goes somthing like this

Fruit |  Type | color
apple    1     green
apple    2     red
oranges  3     rainbow
oranges  3     black

so basically what i want is when ever a object like apple has more then one array inside it the table to display apple and the corresponding data to it. this is what i have so far

 $output = json_decode(JsonData);
     $result =$output['data'][0]['apple'];

<table>
    <thead> 
            <tr>                
                <th>Fruits</th> 
                <th>Type</th> 
                <th>color</th>                 
            </tr> 
            </thead> 
            <tbody> 
        <?php if(!isset($result)){ ?>
        <tr> 
            <td>Empty</td> 
            <td>Empty</td> 
            <td>Empty</td> 
            <td>Empty</td>       
        </tr> 
        <?php else{
            foreach($result as $apples){ ?>              
            <tr> 
                <td></td> 
                <td><?php echo "$apples["type"]";?></td> 
                <td><?php echo "$apples["color"]";?></td>                    
            </tr> 
          <?php } ?>
        <?php } ?>
            </tbody> 
        </table>

whats this <?php echo "apples["type"]";?>

ArtisticPhoenix
– ArtisticPhoenix

2018-04-18 00:52:56 +00:00
Commented Apr 18, 2018 at 0:52 — ArtisticPhoenix
– ArtisticPhoenix, Commented Apr 18, 2018 at 0:52

Nerdi.org · Accepted Answer · 2018-04-18 01:07:52Z

1

$arr = json_decode($json, true); 
foreach($arr['data'] as $fruit => $types){
 foreach($types as $info){
 echo $fruit;
 echo $info['type']; 
 echo $info['color'];
}
}

This is how you can get your JSON data into PHP. I trust you can create an HTML table or something to display this information.

answered Apr 18, 2018 at 1:07

Nerdi.org

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2 Comments

c319113 Over a year ago

hello when i echo $fruit i get a output as 0123

c319113 Over a year ago

also $info['types']; says undefended index

ArtisticPhoenix · Accepted Answer · 2018-04-18 01:56:56Z

1

Well just at a glance this is completely wrong syntax

<?php echo "apples["type"]";?>

Probably you need something like this

<?php echo $apples["type"];?>

What you had will probably give you a syntax error because PHP will see it like this

<?php echo "apples["   type   "]";    ?>

Where "apples[" is a complete string, type is an undefined constant that is unexpected in the location it's in, and then an extra string "]";

you can test it in this sandbox

https://3v4l.org/nR8vF

Which gives us this

Parse error: syntax error, unexpected 'type' (T_STRING), expecting ',' or ';' in /in/nR8vF on line 5

UPDATE

Well there are a few other obvious problems, just with this

$output = json_decode(JsonData);
$result = $output['data'][0]['apple'];

I assume JsonData is just a placeholder or are you missing the $ on that too?

Then as you don't have the second argument set to true, then data will be object style. Personally I would do

 $output = json_decode(JsonData, true);

And just use it as an array.

Others have covered the double "foreach" deal (for looping on fruit), so I won't re-hash that, I would have covered it but it was "Dinner" time and my wife gets a bit "Irritated" when I mess with code instead of coming up from the office (I have an office in the basement).

edited Apr 18, 2018 at 1:56

answered Apr 18, 2018 at 0:53

ArtisticPhoenix

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2 Comments

c319113 Over a year ago

its the same thing it doesn't matter if its inside a " its just easier to concatenate stuffs like that

ArtisticPhoenix Over a year ago

It's not the same because "apples is a string and $apples is a variable. You may have an argument if you had, had "$apples['type']" but that's not what you have. Even then I prefer to wrap it in {} such as "{$apples['type']}" as it makes it much more cleaner.

Second2None · Accepted Answer · 2018-04-18 01:20:27Z

0

If you are looking to show all fruits then you will need something along the lines of:

$result =$output['data'];

Then update your foreach loop

foreach($result as $key=>$fruit){ 
    foreach($fruit as $k=>$dets){?>              
        <tr> 
            <td></td> 
            <td><?php echo $dets["type"];?></td> 
            <td><?php echo $dets["color"];?></td>                    
        </tr> 
<?php } 
} ?>

That's also not valid json

edited Apr 18, 2018 at 1:20

answered Apr 18, 2018 at 1:07

Second2None

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1 Comment

c319113 Over a year ago

why do u say its not valid?

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