php返回null传递变量值

Question

The problem with code below is variable sp1. When replacing sp1 into "where" clause with a city name already existing in database, everything works well. But when i send startPoli1 variable from my app, php returns nothing. Logcat shows that startPoli1 is being sent every time. Any suggestion;

<?php
$con=mysql_connect("......","......","......" );
$database = "......";
$ok = mysql_select_db($database, $con);
mysql_set_charset("UTF8",$con); 


$us1 = $_POST['username1'];
$sp1 = $_POST['startPoli1'];
$fp1 = $_POST['finalPoli1'];
$w1 = $_POST['weight1'];
$em1 = $_POST['eidosmetaf1'];
$dD1 = $_POST['depDate1'];
$dT1 = $_POST['depTime1'];


$sql = mysql_query( "  SELECT `username1`,`startPoli1`, `finalPoli1`, `eidosmetaf1`, `weight1` , `depDate1` , `depTime1`, `tilefono1` 
 FROM customer ,registration1 
 where   (customer.startPoli1 = 'sp1')  and   
 (customer.username1 = registration1.username )");

    if($sql === FALSE) 
    { 
    die(mysql_error()); 
    }
    $results = array();
    while($row = mysql_fetch_assoc($sql))
{
   $results[] = array(
        'username1' => $row['username1'],
        'startPoli1' => $row['startPoli1'],
        'finalPoli1' => $row['finalPoli1'],
        'eidosmetaf1' => $row['eidosmetaf1'],
        'weight1' => $row['weight1'],
        'depDate1' => $row['depDate1'],
        'depTime1' => $row['depTime1'],
        'tilefono1' => $row['tilefono1']
         );
         }
    echo json_encode(array('select_itin_results' =>$results));
    mysql_close($con); 
?>

Answer 1

You forgot to add $ before the variable name. ie where (customer.startPoli1 = '$sp1') so the final query is using the string "sp1" instead of the value of the variable $sp1

After you understand that, learn about the proper way of making sql queries to avoid people messing with your database with sql injection