使用php从sql数据库创建json对象

Question

I am trying to create a json object from my mysql database for a android project. I need an output something like this:

{
    "feed": [
        {
            "id": 1,
            "name": "National Geographic Channel",
            "image": "http://api.androidhive.info/feed/img/cosmos.jpg",
            "status": "\"Science is a beautiful and emotional human endeavor,\" says Brannon Braga, executive producer and director. \"And Cosmos is all about making science an experience.\"",
            "profilePic": "http://api.androidhive.info/feed/img/nat.jpg",
            "timeStamp": "1403375851930",
            "url": null
        },
        {
            "id": 2,
            "name": "TIME",
            "image": "http://api.androidhive.info/feed/img/time_best.jpg",
            "status": "30 years of Cirque du Soleil's best photos",
            "profilePic": "http://api.androidhive.info/feed/img/time.png",
            "timeStamp": "1403375851930",
            "url": "http://ti.me/1qW8MLB"
        }
]
}

But I am getting ouput something like this:

{"feed":[{"id":"0","name":"punith","image":"","status":"ucfyfcyfffffffffffffffffffffffffffffffff","profilePic":"http:\/\/api.androidhive.info\/feed\/img\/nat.jpg","timestamp":"1403375851930","url":""}]}

Everything is on a single line and the id attribute should not be quotes. Is there anything I could do.This is my php file

<?php
define('HOST','');
define('USER','');
define('PASS','');
define('DB','');

$con = mysqli_connect(HOST,USER,PASS,DB);

$sql = "select * from timeline";

$res = mysqli_query($con,$sql);

$result = array();

while($row = mysqli_fetch_array($res)){
array_push($result,
array('id'=>$row[0],
'name'=>$row[1],
'image'=>$row[2],
'status'=>$row[3],
'profilePic'=>$row[4],
'timestamp'=>$row[5],
'url'=>$row[6]
));
}

echo json_encode(array("feed"=>$result));

mysqli_close($con);

?>

And will it affect if the output is on a single line. The database contains exactly the same columns used as attributes.

Thanks in advance

Answer 1

Try encoding with JSON_PRETTY_PRINT

$json_string = json_encode($data, JSON_PRETTY_PRINT);

where data is your result array.

in your case

echo json_encode(array("feed"=>$result),JSON_PRETTY_PRINT);

refer this Tutorial from the official PHP docs .

Answer 2

ID in quotes

ID is in quotes because it is a string, not an integer. You can change that by changing this:

array('id'=>$row[0]

to this:

array('id'=>intval($row[0])

---

"Pretty Printing"

Putting it on multiple lines will only affect readability but not how the data is computed - but you can prettify it: Pretty-Printing JSON with PHP

$output = json_encode(array("feed"=>$result), JSON_PRETTY_PRINT);
echo $output;

Answer 3

与其他答案一样，包含json标头会更具有语义，我发现仅包含此标头也有助于json本身的外观，因此它的格式正确，而不是所有连续的字符串。

header('Content-Type: application/json');

Answer 4

For php>5.4

$json=json_encode(array("feed"=>$result),JSON_PRETTY_PRINT);
header('Content-Type: application/json');
print_r($json);