23 replies to this topic

[Jubal]

<?php

// retrieve the user name

$name = $_GET['name'];

// generate output depending on the user name received from client

$userNames = array('JAMES', 'BUBBLES', 'SUDOBAAL');

header('Content-Type: text/javascript');

if (in_array(strtoupper($name), $userNames))

  echo json_encode("1");

else if (trim($name) == '')

   echo json_encode("2");

else

   echo json_encode("3");

?>

This script takes a name keyed in by the user, checks it against its list, and if it matches passes a variable to the main javascript program. However, I need to edit it to check aganst the user data in my SQL table, or to put it simply, I need to replace the userNames array with the 'name' column of an SQL table. How might I do that?

Edited by Jaan, 18 January 2009 - 12:42 PM.

[Jubal]

Well it's working, but seemingly there's still something wrong since when I type in a correct name it still comes up as if i'd keyed in a wrong or nonexistent name, so I guess there must be something wrong with my connection...

TO check;
- I put localhost where you put localhost
- 'World' I changed to 'test' as that is my db name
- I put the name of my table in where it said '<yourtable>'

<?php

// retrieve the user name

$name = $_GET['name'];

// generate output depending on the user name received from client

$mysqli = new mysqli("localhost", "", "", "test");


/* check connection */

if ($mysqli->connect_error) {

    printf("Connect failed: %s\n", mysqli_connect_error());

    exit();

}


/* Select queries return a resultset */

if ($result = $mysqli->query("SELECT name FROM raiders")) {

    while($obj = $result->fetch_object()){

            $userName[]=$obj->name;

        }

    /* free result set */

    $result->close();

}


/* close connection */

$mysqli->close();


//$userNames = array('JAMES', 'BUBBLES', 'SUDOBAAL');


header('Content-Type: text/javascript');

if (in_array(strtoupper($name), $userName))

  echo json_encode("1");

else if (trim($name) == '')

   echo json_encode("2");

else

   echo json_encode("3");

?>

In terms of checking the connection, this definitely does work (it just loads some data from the server);

<?php

$conn = mysql_connect("localhost", "", "");

$name = $_GET['name'];

mysql_select_db("test", $conn);

$sql = "SELECT * FROM raiders where name = '$name' ";

$result = mysql_query($sql, $conn) or die(mysql_error());

while ($newArray = mysql_fetch_array($result)){

$x = $newArray['sl'];

}

header('Content-Type: text/javascript');

echo json_encode($x);

?>

Many thanks,

Jubal

[Jubal]

I did the echo loop, and ran the PHP on its own, I got this;

Quote

jamesbubbles"2"

The "2" is because I forgot to comment out the json_encode section, but the names are the ones from my table.

So it does look like it's finding the data, but not matching it to the strings sent in by the javascript... how odd...

THe second test doesn't seem to be working, perhaps I'm using an old version of PHP or somthing.. I'll check.

EDIT; PHP 5.2.4...

EDIT#2; Done it.

Quote

Array ( [0] => james [1] => bubbles )

[Jubal]

It comes up with 3 still...

I'd guess there's still some problem with data/data types; for some reason, what we're getting from the SQL isn't matching the strings I had in the original array or the input variable.

[Jubal]

Still working on this... I just can't see what could possibly not be working.

[Jubal]

Ohh... so if I put the variables into the SQL table in uppercase it might work?

[Jubal]

OK... I'll try that tomorrow, I need to get some sleep now for once.

Thanks