5 * 5 - 10; lukt niet?
oke mijn probleem is als volgt
Link:
http://www.dontsurfhere.com/game/test.php
Dit word niet uitgerekend het bedrag komt op 0 te staan steeds
Code (php)
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$af =$bank - $comp * $price;
Code (php)
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<?php
$query = "SELECT * FROM users";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
$query2 = "SELECT * FROM stock";
$result2 = mysql_query($query2) or die(mysql_error());
$row2 = mysql_fetch_array($result2) or die(mysql_error());
$comp = $_POST['buy'];
$af =$bank - $comp * $price;
$bank = $row['bank'];
$shares = $row2['shares'] - $_POST['buy'];
$preshares = $row2['shares'];
$price = $row2['price'];
if(($_POST['buy'])){
$result = mysql_query("UPDATE users SET bank='$af' WHERE bank='$bank'")
or die(mysql_error());
$result2 = mysql_query("UPDATE stock SET shares='$shares' WHERE shares='$preshares'")
or die(mysql_error());
}
echo'
<table><td align="center" width="400"><center>
<table border="1" cellspacing="0" cellpadding="2" bordercolor="black" width="100%" "> ';
echo '<tr><td class="header" align=center colspan=2><font color="#000000">Available money banked</font></td></tr><tr> ';
echo '<td bgcolor="#FFFFFF"><font color="#000000">Money</font></td><td bgcolor="#FFFFFF">';
echo '<font color="#000000"></font>$'.$row['bank'];
echo '</td> </tr><tr>';
echo '
<table border="1" align="center" width="400">
<form method="post" action="" name="f">
<td>Company</td>
<td>Shares</td>
<td>Price</td>
<td>Order</td>
</tr>';
echo '<td>'.$row2['company'];
echo'</td><td>'.$row2['shares'];
echo'</td><td>$'.$row2['price'];
echo' <td><input name="buy" size="2" maxlength="5" type="text" /><br />
</td>
</tr>
<tr>
<td>Totaal</td>
<td></td>
<td></td>
<td><input name="SUBMIT" value="Order" type="submit" /> </form></td>
</tr>';
echo '</table></tr>';
?>
$query = "SELECT * FROM users";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
$query2 = "SELECT * FROM stock";
$result2 = mysql_query($query2) or die(mysql_error());
$row2 = mysql_fetch_array($result2) or die(mysql_error());
$comp = $_POST['buy'];
$af =$bank - $comp * $price;
$bank = $row['bank'];
$shares = $row2['shares'] - $_POST['buy'];
$preshares = $row2['shares'];
$price = $row2['price'];
if(($_POST['buy'])){
$result = mysql_query("UPDATE users SET bank='$af' WHERE bank='$bank'")
or die(mysql_error());
$result2 = mysql_query("UPDATE stock SET shares='$shares' WHERE shares='$preshares'")
or die(mysql_error());
}
echo'
<table><td align="center" width="400"><center>
<table border="1" cellspacing="0" cellpadding="2" bordercolor="black" width="100%" "> ';
echo '<tr><td class="header" align=center colspan=2><font color="#000000">Available money banked</font></td></tr><tr> ';
echo '<td bgcolor="#FFFFFF"><font color="#000000">Money</font></td><td bgcolor="#FFFFFF">';
echo '<font color="#000000"></font>$'.$row['bank'];
echo '</td> </tr><tr>';
echo '
<table border="1" align="center" width="400">
<form method="post" action="" name="f">
<td>Company</td>
<td>Shares</td>
<td>Price</td>
<td>Order</td>
</tr>';
echo '<td>'.$row2['company'];
echo'</td><td>'.$row2['shares'];
echo'</td><td>$'.$row2['price'];
echo' <td><input name="buy" size="2" maxlength="5" type="text" /><br />
</td>
</tr>
<tr>
<td>Totaal</td>
<td></td>
<td></td>
<td><input name="SUBMIT" value="Order" type="submit" /> </form></td>
</tr>';
echo '</table></tr>';
?>
Gewijzigd op 01/01/1970 01:00:00 door Damiaan
Het lijkt erop dat die twee variabelen niet bestaan, dus dan zal het inderdaad niet werken...
ik nijg altijd stomme fouten te maken :D
Damiaan schreef op 06.01.2009 16:24:
Leer jezelf dan aan om op een goede manier foutafhandeling toe te passen. Dan zal PHP een foutmelding naar je hoofd slingeren als je een (domme) fout maakt...ik nijg altijd stomme fouten te maken :D
Geen variabelen kopiëren.
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Je definieert $bank en $price pas ná de berekening. Dan krijg je dus:
$af = null - $comp * null;