CODEWITHSUNDEEP
phpjson
Indian Thinking
Indian Thinking

Reputation: 103

convert large data in json format

i have a table named mobile in which all the mobile data is contained which have more then 7000 records now i want to display in json format but some how records are not showing here is my code kindly check ..

<?php
$conn = mysqli_connect("localhost","root","","test") or die ("Error ".mysqli_error($conn));
$sql = "select * from mobile";

$result = mysqli_query($conn, $sql) or die ("error" . mysqli_error($conn));
var_dump($result);

$myArray = array();


while($row = mysqli_fetch_assoc($result)){

    $myArray[] = $row;


}


mysqli_close($conn);

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

//$json = file_get_contents('json.json');

/*

$myArray = array("user1" => array("firstName" => "Mike2", "lastName" => "Smith" , "age" => 34),"user2" => array("firstName" => "Mike2", "lastName" => "Smith" , "age" => 34));

*/

$json = json_encode($myArray);

echo $json;


?>

table

enter image description here

Upvotes: 1

Views: 582

Answers (3)

Satish Saini
Satish Saini

Reputation: 2968

The problem with JSON_ENCODE in PHP is, it tends to add double quotes and escaping sequences which would increase the actual size of the JSON being imported. So, please try this. This worked for me.

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

$conn = mysqli_connect("localhost","root","","test") or die ("Error ".mysqli_error($conn));
$sql = "select * from mobile";
$result = mysqli_query($conn, $sql) or die ("error" . mysqli_error($conn));

$myArray = array();

while($row = mysqli_fetch_assoc($result)){
    $myArray[] = $row;
}

mysqli_close($conn);

$prefix = '';
echo '[';
foreach($myArray as $row) {
  echo $prefix, json_encode($row);
  $prefix = ',';
}
echo ']';
?>

Upvotes: 1

anis programmer
anis programmer

Reputation: 999

You can do this by following way:

var obj = JSON.parse($myArray);
echo obj;

Upvotes: 0

Shakti Phartiyal
Shakti Phartiyal

Reputation: 6254

I assume you are probably using ajax to get your data in the browser you have used a var_dump() in your php code followed by the json headers followed by the json result which will not work.

In order to set headers , the header() function should be called first before printing out any other thing, also you do no need to put in the var dump. Try the following code:

<?php
$conn = mysqli_connect("localhost","root","","test") or die ("Error ".mysqli_error($conn));
$sql = "select * from mobile";
$result = mysqli_query($conn, $sql) or die ("error" . mysqli_error($conn));
$myArray = array();

while($row = mysqli_fetch_assoc($result)){
    $myArray[] = $row;
}
mysqli_close($conn);
header('Content-Type: application/json');
$json = json_encode($myArray);
echo $json;
?>

Upvotes: 0

Related Questions