how to read json object from php

Question

I am newbie to json.I am trying to read values from MySQL database and encode the result to json data.for that i have written following coding

<?php
$conn=mysql_connect('localhost','root','');
$mydb=mysql_select_db("json",$conn) or die('failed');
$result=mysql_query("select * from register",$conn);
$rowsarray[]=array();
$rowsarra[]=null;
while($rows=mysql_fetch_assoc($result)){
$rowsarray[]=$rows;}
echo  json_encode($rowsarray); ?>

i am getting output as

[[],{"name":"$name","password":"$password","email":"$email","address":"$address"},{"name":"rushi","password":"dfsdf","email":"dsfgs","address":"ssfsfsdf"},{"name":"hmd","password":"123","email":"rdf","address":"sfdssf"},{"name":"sdfs","password":"sdfsf","email":"sdfsdf","address":"sdf"},{"name":"rushi","password":"sadsa","email":"xdfsaf","address":"sdfsdf"}]

when i view this in online jsonviewer i get the output as +JSON

+0

+1

+2

+3

+4

+5

how can i read the details inside the json object for '0','1',... in my android application

JsonArray jsonarray=new JsonArray(response.tostring);
JsonObject job=jsonarray.getstring("0");

is this the way to get the jsonobject? (or)

how can i give employee email in place of 0,1,2,... please help me out. if i am wrong anywhere please correct me. Thanks in advance

Answer 1

Use jsonlint for clear understanding about the format. Copy and paste your result in jsonlint and then just validate it like this.

[
[],
{
    "name": "$name",
    "password": "$password",
    "email": "$email",
    "address": "$address"
},
{
    "name": "rushi",
    "password": "dfsdf",
    "email": "dsfgs",
    "address": "ssfsfsdf"
},
{
    "name": "hmd",
    "password": "123",
    "email": "rdf",
    "address": "sfdssf"
},
{
    "name": "sdfs",
    "password": "sdfsf",
    "email": "sdfsdf",
    "address": "sdf"
},
{
    "name": "rushi",
    "password": "sadsa",
    "email": "xdfsaf",
    "address": "sdfsdf"
}
]

From that result of data you just write the below code.

JSONArray resultArray = new JSONArray(your result);
int length = resultArray.length();
for (int i = 0; i < length; i++) {
    JSONObject jsonItems = jsonFaovourites.getJSONObject(i);
    String name = jsonItems.getString("name");
    String email= jsonItems.getString("email");
    //password etc..
}

Answer 2

You should try this out,

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

$conn = new mysqli("localhost", "root", "myPassword", "json");


$result = $conn->query("SELECT * FROM register");

$outp = "";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    if ($outp != "") {$outp .= ",";}
    $outp .= '{"field1":"'  . $rs["field1"] . '",';
    $outp .= '"field2":"'   . $rs["field2"]        . '",';
    $outp .= '"fieldx":"'. $rs["fieldx"]     . '"}'; 

    //fieldx where x is the next column/filed if any and filed the name of the column the exact way it is on the database  
}
$outp ='{"records":['.$outp.']}';
$conn->close();

echo($outp);

//Note this returns json to the browser so no need to json_encode
?>

You can now read this as a normal JSON object, no need to encode. But to read a JSON string coming from your android application that would be a different approach

You will do something like

$data //data from android application 

$data = json_decode($data);

$data->field //to access the field/property in this object

Hope this solves your problem. Let me know if it doesn't