<?php
$host="localhost"; // Host name
$username="xxxx"; // Mysql username
$password="xxxx"; // Mysql password
$db_name="xxxx"; // Database name
$tbl_name="xxxx"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM xxxx WHERE branch = 'xx' AND xxxx.semester=x ORDER BY xxxx.xxxx";
$result=mysql_query($sql);
// Count table rows
$count=mysql_num_rows($result);
?>
<table width="500" border="0" cellspacing="1" cellpadding="0">
<form name="form1" method="post" action="">
<tr>
<td>
<table width="500" border="0" cellspacing="1" cellpadding="0">
<tr>
<td align="center"><strong>Id</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>may_tc_s1</strong></td>
<td align="center"><strong>may_ac_s1</strong></td>
</tr>
<?php
// Check if button name "submit" is active, do this
if (isset($_POST['Submit'])) {
$name = $_POST['name'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
for($i=0;$i<$count;$i++){
$sql1="UPDATE $tbl_name SET xxx='$name[$i]', xxx='$lastname[$i]', xxx='$email[$i]' WHERE xxx='$id[$i]'";
$result1=mysql_query($sql1);
}
}
if(isset($result1)){
header("location:try.php");
}
?>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center"><?php $id[]=$rows['idatten']; ?><?php echo $rows['idatten']; ?></td>
<td align="center"><input name="name[]" type="text" id="name" value="<?php echo $rows['username']; ?>"></td>
<td align="center"><input name="lastname[]" type="text" id="lastname" value="<?php echo $rows['may_tc_s1']; ?>"></td>
<td align="center"><input name="email[]" type="text" id="email" value="<?php echo $rows['may_ac_s1']; ?>"></td>
</tr>
<?php
}
?>
<tr>
<td colspan="4" align="center"><input type="submit" name="submit" value="Submit"></td>
</tr>
</table>
</td>
</tr>
</form>
</table>
我能够在文本字段中看到记录,但是输入了新记录 值没有得到更新????? 这个代码有什么问题????? 尝试了各种方法,无法找到任何正确的方法???
答案 0 :(得分:2)
参考此代码:
extract($_POST);
if (isset($Submit))
{
for($i=0;$i<$count;$i++)
{
$update=("UPDATE tbl_name SET name='$name[$i]', lname='$lastname[$i]',email='$email[$i]' WHERE id='$id[$i]'");
$res=mysql_query($update);
}
}
if(isset($res))
{
header("location:try.php");
}
?>
<?php
while($rows=mysql_fetch_array($result)){
?>
答案 1 :(得分:1)
您没有正确引用变量。
$sql1 = "UPDATE $tbl_name SET xxx='".$name[$i]."', xxx='".$lastname[$i]."', xxx='".$email[$i]."' WHERE xxx='".$id[$i]."'";