如何使用php在json中转换mysql数据库表数据

Question

How to convert MySQL data base table into JSON data using PHP. Is there any way to do this?

Below is the php code I am using:

<?php 
$host = "emriphone.db.6420177.hostedresource.com"; 
$user = "emriphone"; 
$pass = "Light12-"; 
$database = "emriphone"; 

$linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host."); 
mysql_select_db($database, $linkID) or die("Could not find database."); 

$sth = mysql_query("SELECT * FROM ProviderAppointmentListings");
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
   $rows[] = $r;
}
print json_encode($rows);
?>

Answer 1

Try like this:

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

If you don't have json_encode add this before the code above:

if (!function_exists('json_encode'))
{
  function json_encode($a=false)
  {
    if (is_null($a)) return 'null';
    if ($a === false) return 'false';
    if ($a === true) return 'true';
    if (is_scalar($a))
    {
      if (is_float($a))
      {
        // Always use "." for floats.
        return floatval(str_replace(",", ".", strval($a)));
      }

      if (is_string($a))
      {
        static $jsonReplaces = array(array("\\", "/", "\n", "\t", "\r", "\b", "\f", '"'), array('\\\\', '\\/', '\\n', '\\t', '\\r', '\\b', '\\f', '\"'));
        return '"' . str_replace($jsonReplaces[0], $jsonReplaces[1], $a) . '"';
      }
      else
        return $a;
    }
    $isList = true;
    for ($i = 0, reset($a); $i < count($a); $i++, next($a))
    {
      if (key($a) !== $i)
      {
        $isList = false;
        break;
      }
    }
    $result = array();
    if ($isList)
    {
      foreach ($a as $v) $result[] = json_encode($v);
      return '[' . join(',', $result) . ']';
    }
    else
    {
      foreach ($a as $k => $v) $result[] = json_encode($k).':'.json_encode($v);
      return '{' . join(',', $result) . '}';
    }
  }
}

Answer 2

<?php
    $username = "user_name"; 
    $password = "password";   
    $host = "url";
    $database="database";

    $server = mysql_connect($host, $username, $password);
    $connection = mysql_select_db($database, $server);

    $myquery = "SELECT date,close FROM data2";
    echo "hi";
    $query = mysql_query($myquery);

    if ( ! $query ) {
        echo mysql_error();
        die;
    }    
    $data = array();    

    for ($x = 0; $x < mysql_num_rows($query); $x++) {
        $data[] = mysql_fetch_assoc($query);
    }

    echo json_encode($data);     

mysql_close($server);

?>

This code converts your mysql data in phpmyadmin to json. Works perfect

Answer 3

using this code :

json_encode($array)

for example :

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());
      }  
 }

Answer 4

As long as you are using MySQL server 5.7 or later, you can produce JSON data by using just SQL and nothing more, that is, you need PHP just to pass the SQL and get the JSON result. For example:

SELECT JSON_OBJECT( 'key1', column1,
                    'key2', JSON_OBJECT('key3', column2)) as fοο;

There is more that JSON_OBJECT ! Please check this page: https://dev.mysql.com/doc/refman/5.7/en/json-function-reference.html

Answer 5

I guess you mean the data in a MySQL-database table right?

In that case, have a look at PHP's json_encode() .

You can fetch the data from the db into an array and then covert it to JSON.

Answer 6

put the resultset of the query into an array and then use json_encode

$sql="Select * from table";
$l= array();
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
  $l[] = $row;
}
$j = json_encode($l);
echo $j;

you may use id table as index of the array.

Answer 7

<?php
$username = "user_name"; 
$password = "password";   
$host = "url";
$database="database";

$server = mysql_connect($host, $username, $password);
$connection = mysql_select_db($database, $server);

$myquery = "SELECT date,close FROM data2";
echo "hi";
$query = mysql_query($myquery);

if ( ! $query ) {
    echo mysql_error();
    die;
}    
$data = array();    

for ($x = 0; $x < mysql_num_rows($query); $x++) {
    $data[] = mysql_fetch_assoc($query);
}

echo json_encode($data);     
mysql_close($server);

?>