使用mysql json，ajax检索多个数据

Question

I am Posting my values from database here, code:

<?php 
$host = "localhost";
$user = "root";
$pass = "";

$databaseName = "ani";
$tableName = "balls";

//--------------------------------------------------------------------------
// 1) Connect to mysql database
//--------------------------------------------------------------------------
include 'DB.php';
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);

//--------------------------------------------------------------------------
// 2) Query database for data
//--------------------------------------------------------------------------
$result = mysql_query("SELECT * FROM $tableName");         
$array = mysql_fetch_row($result);                          

//echo json
echo json_encode($array);
?>

Then I am trying to write out all rows of my array but can't get to understand how. However, this is the code from my script. Hope you can help somehow. Thanks.

<script id="source" language="javascript" type="text/javascript">

setInterval("yourAjaxCall()",1000);
function yourAjaxCall() 
{
    $.ajax({                                      
        url: 'api.php',     //the script to call to get data          
        data:  "id=1",      //you can insert url argumnets here to pass to api.php
                            //for example "id=5&parent=6"
        dataType: 'json',           //data format
        success: function(data)     //on recieve of reply
        {   
            var id1 = data[0];          //get id of first row
            var vcolor1 = data[1];      //get color of first row            
            $("#square").css({"background-color" : vcolor1});   
            $color1 = vcolor1; //saving value
            var $color1;

            //HERE I WANT TO BE ABLE TO GET NEXT ROW OF MY JSON ARRAY. HOW?
        } 
    });
};

</script>

I can't find a solution so I am very greatful for good and simple answers!

Answer 1

In PHP file do it:

$result = mysql_query("SELECT * FROM $tableName");         
$dataArray = array();
while($array = mysql_fetch_assoc($result)){
    $dataArray[] = $array;
} 

echo json_encode($dataArray);

AND in JS:

success: function(data)
{ 
   var obj = JSON.parse(data);
   $.each(obj,function(index,value)){
      alert(index+" : "+value); //here you can get all data
   }
}