Trying to encode an array to json in PHP

Question

I saw many threads on stackoverflow where people converted arrays into json. And I am pretty sure that this should work, but it does not. MySQL is fine and the query is working as intended:

$json = array();
if($result = $mysqli->query($query)){
    while($row = $result->fetch_array()){
        $json[]["id"] = $row[0];
        $json[]["trivialname"] = $row[1];
        $json[]["chemischername"] = $row[2];
        $json[]["formel"] = $row[3];
    }
    echo $json[0]["id"]; //just for testing.
    echo json_encode($json);
}

Sorry for bad English.

My problem is mainly that echo json_encode($json); doesnt show anything

EDIT: My Code look like this atm.

if($result = $mysqli->query($query)){
            $json = array();
            while($row = $result->fetch_assoc()){
                $json[]=$row;
                printf("%s (%s) ID: %s, Formel:%s <br>", $row["Trvialname"], $row["Chemischername"], $row["ID"],$row["Formel"]);
            }
            echo json_encode($json);

        }

and it shows me this: Phthalimid (1,2-Benzoldicarboximid) ID: 2, Formel:C8H5NO2 Phthalsäureanhydrid (2-Benzofuran-1,3-dion) ID: 3, Formel:C8H4O3

nothing else! i can change:

echo json_encode($json);

to

echo json_encode($json[0]);

and it will add the line :

{"ID":"2","Trvialname":"Phthalimid","Chemischername":"1,2-Benzoldicarboximid","Formel":"C8H5NO2"}

foreach($json[1] as $value){
    echo $value."<br>";
}

will show everything but

echo json_encode($json[1]);

doesnt.

Solution

$mysqli->set_charset('utf8');

thanks to Your Common Sense

You are adding a new item to $json array for each field in your while loop. You should add them all at once when using [] — Preston S
– Preston S, Commented Mar 19, 2014 at 17:27

Your Common Sense · Accepted Answer · 2014-03-19 18:43:48Z

2

$json = array();
while($row = $result->fetch_assoc()){
    $json[] = $row;
}
echo json_encode($json);

For the whole my life I'll never understand why PHP users always write so much code for the every trifle operation

You may also try json_last_error() and then add this call to your mysqli connect code

$mysqli->set_charset('utf8');

edited Mar 19, 2014 at 18:43

answered Mar 19, 2014 at 17:29

Your Common Sense

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3 Comments

Your Common Sense Over a year ago

then your query doesn't return anything

user3438815 Over a year ago

it does i swear to god echo $row[0]." "; echo $row[1]." "; echo $row[2]." "; echo $row[3]." "; returns everything i want

Your Common Sense Over a year ago

Then you are running not the code I posted but something else. Because the code I posted won't return $row[0]

Hieu Le · Accepted Answer · 2014-03-19 17:27:32Z

2

Update your code to

$json = array();
if($result = $mysqli->query($query)){
    while($row = $result->fetch_array()){
        $r = array();
        $r["id"] = $row[0];
        $r["trivialname"] = $row[1];
        $r["chemischername"] = $row[2];
        $r["formel"] = $row[3];
        $json[] = $r;
    }
    echo $json[0]["id"]; //just for testing.
    echo json_encode($json);
}

answered Mar 19, 2014 at 17:27

Hieu Le

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1 Comment

Hieu Le Over a year ago

so $result is false. If your query return an empty array, at least you could see the string [] as an empty json array.

gen_Eric · Accepted Answer · 2014-03-19 17:28:58Z

0

You are not building your array the way you think you are. Each element you are adding is adding a new row to the array, that's what $json[] does. You will wind up with an array that looks like:

array(
    array('id' => 1),
    array('trivialname' => 'foo'),
    # etc.
)

What you want to do is:

$json[] = array(
    "id" => $row[0],
    "trivialname" => $row[1],
    "chemischername" => $row[2],
    "formel" => $row[3]
);

This will build the structure you desire.

answered Mar 19, 2014 at 17:28

gen_Eric

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