如何在2个不同的php文件中共享php变量

Question

here is the code for first.php

<?php

mysql_connect("localhost","root",""); // host, username, password...

mysql_select_db("testdb");

$v1=$_REQUEST['usn'];

$q=mysql_query("select * from users where usn='$v1'");

while($row=mysql_fetch_assoc($q))

 $json_output[]=$row;

$json_output["re"]="success";

    print(json_encode($json_output));

    mysql_close();

?>

code for Second.php

<?php 

 mysql_connect("localhost","root",""); // host, username, password...

mysql_select_db("testdb"); // db name...

$q=mysql_query("select * from users where usn='$v1'");

   while($row=mysql_fetch_assoc($q))

            $json_output[]=$row;

    print(json_encode($json_output));

mysql_close();

?>

Now how to use the value of variable $v1 ,declared in first.php, in second.php.The value of $v1 is obtained through user's input via EditText widget in android.

I tried even with session variables but still not getting O/P:

first.php

   <?php
    session_start();
    mysql_connect("localhost","root",""); // host, username, password...
   mysql_select_db("testdb");
     $_SESSION["v1"]=$_REQUEST['usn'];

        $q=mysql_query("select * from users where usn='".$_SESSION["v1"]."'");
      while($row=mysql_fetch_assoc($q))
       $json_output[]=$row;
      $json_output["re"]="success";

      print(json_encode($json_output));

      mysql_close();

       ?>

second.php

 <?php 
 session_start();
 $v3 = $_SESSION["v1"];

   mysql_connect("localhost","root",""); // host, username, password...
   mysql_select_db("testdb"); // db name...

    $q=mysql_query("select * from users where usn='$v3'");
     while($row=mysql_fetch_assoc($q))
    $json_output[]=$row;

  print(json_encode($json_output));

    mysql_close();

    ?>

Answer 1

Well you can use sessions for more see here PHP Sessions

here is the fixed code for first.php

<?php
//Start session
session_start();

mysql_connect("localhost","root",""); // host, username, password...

mysql_select_db("testdb");

$v1=$_REQUEST['usn'];

//Set Session data
$_SESSION['v1']=$v1;

$q=mysql_query("select * from users where usn='$v1'");

while($row=mysql_fetch_assoc($q))

$json_output[]=$row;

$json_output["re"]="success";

print(json_encode($json_output));

mysql_close();

?>

code for Second.php

<?php 
 session_start();
mysql_connect("localhost","root",""); // host, username, password...

mysql_select_db("testdb"); // db name...

//Read from session 
$v1 = $_SESSION['v1'];

$q=mysql_query("select * from users where usn='$v1'");

while($row=mysql_fetch_assoc($q))

$json_output[]=$row;

print(json_encode($json_output));

mysql_close();

?>

Answer 2

in first.php you hav to save $v1 in a session variable .. like this

<?php
session_start(); // should be on top of your script
$v1 = $_REQUEST['usn'];
$_SESSION['v1'] = $v1;
?>

in Second.php you can use $_SESSION['usn']

<?php
session_start(); 
$v1 = $_SESSION['v1'];
$q=mysql_query("select * from users where usn='$v1'");

?>

NOTE : use of mysql_* is deprecated .. use mysqli

Answer 3

make a config.php file and place all common variables and values there

then require_once('config.php'); on the other php file

eg

config.php
<?
$v =1;
?>

second.php
<? 
require_once('config.php');
echo $v;
?>