Onay kutusunu kullanarak mysql tablo güncelleniyor

0 Cevap php

i birden fazla onay kutusu ile bir mysql tablosunu güncelleştirmek için çalışacağız, ama benim kod çalışmak için görünmüyor, bu yüzden herhangi bir yardıma açığız. Benim kod:

<strong>Update <strong class="highlight">multiple</strong> <strong class="highlight">rows</strong> <strong class="highlight">in</strong> <strong class="highlight">mysql</strong></strong><br> 

<?php 
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="root"; // Mysql password 
$db_name="db_test"; // Database name 
$tbl_name="test_table"; // 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 $tbl_name"; 
$result=mysql_query($sql); 

// Count table <strong class="highlight">rows</strong> 
$count=mysql_num_rows($result); 
?> 
<table width="500" border="0" cellspacing="1" cellpadding="0"> 
<form name="form1" method="post" action="<? echo $_SERVER['REQUEST_URI']; ?>"> 
<tr> 
<td> 
<table width="500" border="0" cellspacing="1" cellpadding="0"> 


<tr> 
<td align="center" bgcolor="#FFFFFF"><strong>Id</strong></td>
<td align="center" bgcolor="#FFFFFF"><strong>Name</strong></td>
<td align="center" bgcolor="#FFFFFF"><strong>Email</strong></td>
<td align="center" bgcolor="#FFFFFF"><strong>Description</strong></td>
<td align="center" bgcolor="#FFFFFF"><strong>Phone Number</strong></td>
<td align="center" bgcolor="#FFFFFF"><strong>Operation</strong></td>
</tr> 
<?php 
while($rows=mysql_fetch_array($result))
{ 
?> 
<tr> 
<td align="center"><input type="hidden" name="id[]" value="<? echo $rows['id']; ?>" /><? echo $rows['id']; ?></td> 
<td align="center"><input name="name<? echo $rows['id']; ?>" type="text" id="name" value="<? echo $rows['name']; ?>"></td> 
<td align="center"><input name="email<? echo $rows['id']; ?>" type="text" id="email" value="<? echo $rows['email']; ?>"></td> 
<td align="center"><input name="description<? echo $rows['id']; ?>" type="text" id="description" value="<? echo $rows['description']; ?>"></td> 
<td align="center"><input name="phone_number<? echo $rows['id']; ?>" type="text" id="phone_number" value="<? echo $rows['phone_number']; ?>"></td>
<td align="center"><input name="operation<? echo $rows['id']; ?>" type="text" id="operation" value="<? echo $rows['operation']; ?>"></td>
<td align="center"><input name="ONOFF<? echo $rows['id']; ?>" type="checkbox" id="ONOFF" value="1" 
<?php if ($rows['ONOFF'] ==1) { echo "checked";} else {} ?> 
</td> 
</tr> 
<?php 
} 
?> 
<tr> 
<td colspan="4" align="center"><input type="submit" name="Submit" value="Submit"></td> 
</tr> 
</table> 
</td> 
</tr> 
</form> 
</table> 
<?php 
// Check if button name "Submit" is active, do this 
if($Submit)
{ 
 foreach($_POST['id'] as $id)
 { 
  $onoff = 0;
    if (isset($_POST["ONOFF".$id]))
    {
        $onoff = 1;
    }

  $sql1="UPDATE ".$tbl_name." SET name='".$_POST["name".$id]."', email='".$_POST["email".$id]."', description='".$_POST["description".$id]."', phone_number='".$_POST["phone_number".$id]."', operation='".$_POST["operation".$id]."', ONOFF='".$onoff."' WHERE id='".$id."'";  
  $result1=mysql_query($sql1);
 } 
} 

if($result1){ 
header("location:test_update2.php");
} 
mysql_close(); 
?>

Senin için teşekkürler yardımcı olur.

Saygılar.

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