Php yardim.

peacehawk

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20 Ara 2006
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Merhaba arkadaslar php ile bir hesapmakinesi yapmaya calisiyorum fakat, asagida kirmizi ile belirttigim satirda bir hata var. Ne oldugunu bulamadim yardimlarinizi bekliyorum..

PHP:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Untitled Document</title>
</head>
<form action=”" method=”post” name=”form1?><label></label>
<table width=”369? height=”162? border=”0?>
<tr>
<td width=”198? rowspan=”2?>Sayi 1 <input name=”sayi1? type=”text” id=”sayi1? /></td>
<td width=”155?><label><input name="radiobutton" type="radio" value="topla" /> Topla</label></td>
</tr>
<tr>
<td height=”31?><label><input name="radiobutton" type="radio" value="cikar" />Çıkar</label></td>
</tr>
<tr>
<td rowspan=”3?>Sayi 2<input name=”sayi2? type=”text” id=”sayi2? /> </td>
<td><label><input name="radiobutton" type="radio" value="Çarp" /> Çarp</label></td>
</tr>
<tr>
<td><input name="radiobutton" type="radio" value="bol" />Böl</td>
</tr>
<tr>
<td><label><input type="submit" name="Submit" value="Hesapla" /></label></td>
</tr>
<tr>
[COLOR="Red"]<td>Sonuç :<?$kontrol=$_POST['radiobutton'];if($kontrol=="topla") $sonuc=$sayi1+$sayi2;if($kontrol=="cikar") $sonuc=$sayi1-$sayi2;if($kontrol=="carp") $sonuc=$sayi1*$sayi2;if($kontrol=="bol") $sonuc=$sayi1/$sayi2;echo $sonuc;?></td>
<td> </td>[/COLOR]
</tr>
</table>
</form>
<body>
</body>
</html>
 
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