CODEWITHSUNDEEP
phpjson
Matteo Mosconi
Matteo Mosconi

Reputation: 244

How to retrive php data from mysql and convert it to json?

My mind got stuck when i try to develope this: i have a table in my database called "article" whit two column, "name" and "price". How can i extract all rows from my table and echo all column in JSON? i really can't understand how to convert result in JSON. My mind it's stuck like never before. i need to echo something like this:

{"items": {
    "items":[
        {"name": "firstitemname", 
         "price": "5"
        },
        {"name": "secondone", 
         "years": "3"
        }],
}}

Please help me fixing my buggy code!

<?php

$query = mysql_query("SELECT * FROM itemlist");
$nameitem = array();
$itemprice = array();
while($row = mysql_fetch_array($query)){
array_push($nameitem , $row['nome']);
array_push($itemprice, $row['pix']);
}

?>

Upvotes: 1

Views: 6311

Answers (4)

Masoud Siahkali
Masoud Siahkali

Reputation: 5331

Convert Data table to json with following code :

echo(json_encode($array));

for example ( select data from mysql and convert to json ): 

 public function SELECT($tableName,$conditions){

      $connection = mysqli_connect($hostname, $userName, $password,$dbName);
      try {

        if (!$connection)
            die("Connection failed: " . $connection->connect_error);
        else
        {
            $qry = "";
            if(!$this->IsNullOrEmptyString($conditions))
               $qry = "SELECT * FROM `".$tableName."` WHERE ".$conditions;
            else
               $qry = "SELECT * FROM `".$tableName."`";

            $result = mysqli_query( $connection, $qry);
            if($result) {
                $emparray = array();
                while($row =mysqli_fetch_assoc($result))
                    $emparray[] = $row;

                echo(json_encode($emparray));           
            }
            else
                echo(mysqli_error($connection));       
            } 
            mysqli_close($connection); 
      } catch(Exception $ex) {
          mysqli_close($connection);
          echo($ex->getMessage());
      }  
 }

Upvotes: 0

Pal Murugan
Pal Murugan

Reputation: 21

<?php
$result = mysql_query("select * from item_list");
$rows = array();
while($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
}
print json_encode($rows);
?>

Upvotes: 2

iambriansreed
iambriansreed

Reputation: 22241

You would simply edit your PHP as follows.

<?php

$query = mysql_query("SELECT * FROM itemlist");
$items = array();
while($row = mysql_fetch_array($query)){
$items[] = array('name' => $row['nome'], 'price' => $row['pix']);
}

echo json_encode(array('items'=>$items));

?>

http://php.net/manual/en/function.json-encode.php

JSON is super easy to deal with in PHP.

Upvotes: 4

somomomo
somomomo

Reputation: 295

If you're using PHP 5.2 or greater, you can use the json_encode function to do exactly what you're trying to do: http://www.php.net/manual/en/function.json-encode.php

For your code, you should be able to do something like this:

$query = mysql_query("SELECT * FROM itemlist");
$json_output = array();
while($row = mysql_fetch_assoc($query)){
    $json_output[] = json_encode($row);
}

Here, $json_output will contain an array of strings with the json encoded string of each row as each array element. You can output these as you please.

Upvotes: 2

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