Ben sorun nasıl html verilerin geri kalanını yüklerim sergiyi. İşte benim kod:
<?php
$con = mysql_connect("localhost","root","nitoryolai123$%^");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("school", $con);
$result = mysql_query("SELECT * FROM student WHERE IDNO='$_GET['id']'");
?>
<?php while ( $row = mysql_fetch_array($result) ) { ?>
<table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3">
<tr>
<form name="formcheck" method="post" action="updateact.php" onsubmit="return formCheck(this);">
<td>
<table border="0" cellpadding="3" cellspacing="1" bgcolor="">
<tr>
<td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td>
<tr>
<td width="30" height="35"><font size="2">*I D Number:</td>
<td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td>
</tr>
<tr>
<td width="30" height="35"><font size="2">*Year:</td>
<td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td>
Bu ben hata alıyorum parçasıdır:
mysql_select_db("school", $con);
$result = mysql_query("SELECT * FROM student WHERE IDNO='$_GET['id']'");
?>
Eğer ben de kimliği Numaralar diğer formlarına karşılık gelen verileri yükleyebilirsiniz böylece ben yapabileceğini ne tavsiye edersiniz?