在数据库中找到数据并插入数据库后，PHP将成功返回给Java

Question

<?php

include("db_config.php");

$myusername = $_GET['username'];
$mypassword = $_GET['password'];



$sql="SELECT id FROM tablename WHERE username='$myusername' and password='$mypassword'";

echo $sql;

$result=mysql_query($sql);
$row=mysql_fetch_array($result);
$active=$row['active'];

$count=mysql_num_rows($result);


 // If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
 echo "<strong> <font size='18'>Login Success</font></strong>";
 }
 else 
 {
echo "<strong> <font size='18'>Login Failed</font></strong>";
}

?>

I tried response array on success.. But in java file wen I Receive it .. is showing Null.. I am using this line to receive the success

  success=json.getInt(TAG_SUCCESS);

Also I used this in my JSON parser class to check if server returns success.

  if (response.getStatusLine().getStatusCode() == 200) 

And also The login is giving Login failed.. though I enter the username present in the database

Can any1 help me??

Answer 1

Not sure what library you are using BUT using a method "getInt()" on an HTML text "Login Success/Failed" looks very wrong.

Using a "json." method on HTML data also looks very wrong.

So get your php to emit valid JSON, and, get your Java to extract the Login Success or Failed text from hte JSON.

Answer 2

Correct me if I am wrong, but your php doesn't return a valid JSON, instead it returns HTML code. Naturally when you try to parse it as JSON, you get null . If your PHP is changed to this:

if($count==1) {
    $result['login'] == 'success';
}
else {
    $result['login'] == 'failed';
}
echo json_encode($result);

then you may be able to parse this in your java and use something like:

if("success".equals(json.getString("login")) {
    //login successful
}

This is not the only way to achieve this, but it should do the trick.