Inserting value into database - problem.

[TheKizz]

Hi,

 

Sorry guys i'm kind of new to PHP, and I need help inserting a value into the database, I have the following code:

 

if($_POST['option1'] == 1) {

// You have attempted to jack her.

 

$failed = "You have failed to jack her";

$success = "You have successfully jacked her, and recieved £".$random_number." from the jacking";

 

$successorfail = array($failed,$success);

shuffle($successorfail);

 

 

if($successorfail = $success){

 

srand ((double) microtime( )* 100000000);

$random_number = rand(0,400);

 

$connect = mysql_connect("localhost","root","") or die("<center>Did not connect to database.</center>");

 

mysql_select_db("thegame") or die("<center>Couldn't find the database.</center>");

 

$query = mysql_query('INSERT INTO users (money) VALUE($random_number) WHERE username = "'.$_SESSION['username'].'"');

mysql_query($query);

 

echo $successorfail[0].".";

}

else

die ($failed);

 

}

 

What the code is supposed to do is when the first radio button is checked and the submit button is clicked, it should randomly select if the action "jacking or robbing" has failed or succeeded, if it has succeeded it should give a random number between 0 and 400, and add it into the database to a field called 'money' and display a message in the browser saying 'You have successfully jacked her.... etc', and if it has failed then just display the message 'You have failed to jack her.. etc'.

 

But with the code I already have, this isn't working and I don't know why.

 

Can anybody help?

 

- Kiran.

[xangelo]

get rid of the mysql_query($query) line. Also, are you receiving any errors?

[TheKizz]

I just removed that line, and yes I am. It just displays:

 

Y.

 

When I select the radiobox and click submit.

[xangelo]

Your check isn't actually doing anything.

 

if($successorfail == $success)

doesn't make sense. it should be

$successorfail[0] == $success

The first way will ALWAYS evaluate to true because you are assigning $success to $successorfail as opposed to performing a boolean check.

 

Is the database being updated?

[TheKizz]

Okay, thanks i've done that.

 

But it still doesn't work, it says I have been successful but it doesn't display or give an amount.

 

You have successfully jacked her, and recieved £ from the jacking.

 

That's what gets displayed.

[xangelo]

That is happening because you are using $random_number before it is actually initialized. If you move your

$random_number = rand(0,400);

line above the $success line you should be good.

 

Also, is the database being updated with the data even if it is not being displayed?

[TheKizz]

Okay, i've done that. It displays the string correctly and amount.

 

No, but it doesn't update the database data.

[xangelo]

I've re-written your code to make things a little easier to follow. If this doesn't work, make sure $_SESSION['username'] is being set.

 

<?php 
if($_POST['option'] == 1) {
  $amount = rand(0,400);
  $check = array(
    'You have failed to jack her',
    'You have successfully jacked her and receieved £'.$amount.' from the jacking'
  );
  
  if(rand(0,1) == 1) {
    $connect = mysql_connect("localhost","root","") or die("<center>Did not connect to database.</center>");
    mysql_select_db("thegame") or die("<center>Couldn't find the database.</center>");
    
    $sql = 'update users set money = money+'.$amount.' where username = "'.$_SESSION['username'].'"';
    if(mysql_query($sql)) {
      echo $check[1];
    }
  }
}
?>

[TheKizz]

Ahh thanks, it works now. Although I don't get this one little bit, what does this mean?

 

if(rand(0,1) == 1) {

 

That part..?

[xangelo]

Based on the $check array, 0 is fail, 1 is success. So I generate either 0 or 1 randomly, and if it's 1 then we proceed with the script.

[TheKizz]

Alright, thanks so much man.  Great help.