Connect php with server - php

Question

I am connecting to server with help of php for an android application.
Name of Database in phpmyadmin is "student" , name of table is "data" and fields are "Name" and "EmpId"
This is what I coded n php and getting the error on the "$output" part as undefined variable
Here is the code:
<?php
$connection = connectionserver ();
function connectionserver (){
$con = mysql_connect("localhost", "root", "") or die ("connection not found");
if($con)
echo "Connection Created" ,"<br>";
$database = mysql_select_db ("student1", $con);
if($database) echo "Database Connected" , "<br>";
return $con;
}
$result = mysql_query("select * from data");
while ($row = mysql_fetch_assoc($result))
{
$output [] = $row;
}
print json_encode($output);
mysql_close($connection);
?>

Answer 1

declare $output as array before the while
$output = array();

Answer 2

$undefined_array[] = 'something' will not trigger an E_NOTICE error. However it is good practice to initialize the variable.
The error comes from the line with json_encode, most likely because your query didn't return any result, didnt get into the while loop, thus $output[] was never executed.

Answer 3

You may try this:
<?php
$connection = connectionserver ();
function connectionserver (){
$con = #mysql_connect("localhost","root","");
if(!$con) die("Can't connect!!");
$var2 = #mysql_select_db("student1",$con);
if(!$var2)
die("<br>"."can't select dataBase");
$result = mysql_query("select * from data");
while ($row = mysql_fetch_assoc($result))
{
$output[] = $row;
}
print json_encode($output);
mysql_close($con);
}
?>

Answer 4

Try this,
echo connectionserver();
function connectionserver (){
$con = mysql_connect("localhost", "root", "") or die ("connection not found");
$database = mysql_select_db ("student1", $con);
$result = mysql_query("select * from data") or die(mysql_error());
$output = array();
while ($row = mysql_fetch_assoc($result))
{
$output[] = $row;
}
mysql_close($con);
return json_encode($output);
}