Hi all I need Help.
$b is a variable vaue which I have populated from the databse and showing in field.
but when I use POST method its not working.

In Form.php

<form action="update.php" method="post">
<input type="text" name="siteid" value="<?= $b ?>" />
<input type="submit" name="subjoin" value="Update">
</form>

in update page how I show the value of site id. I am using post method. but field value is variable($b) so how I can I post the value to next page?
In update.php

$c = $_post["$b"]; 
echo $c; //its not working.

Thanks in advance

hello

you can get this value like this in update.php file

$c=$_post['siteid'];
echo "value".$c;

I hope your problem may be solve

Thanks.

$c= $_POST['siteid']
echo "value".$c

gives this error

Parse error: parse error in C:\xampp\htdocs\update.php on line 24

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<?
$con = mysql_connect("localhost","****","****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("onm", $con);

$result = mysql_query("SELECT * FROM sitepinfo
WHERE SiteId='C-SIL-4441' "); // I just get one value at this time


while($row = mysql_fetch_array($result))
  {
  echo "<table cellpadding=2 cellspacing=2 width=100%>
<tr>


</tr>";
  echo "<tr>";
  echo "<th bgcolor=#5D9BCC >SiteID</th>";
  echo "<td bgcolor=#FEE9A9>" . $row['SiteId'] . "</td>"; // show value in cell

 
 $b =$row['SiteId']; // store value in $b
 


  
 
  

  echo "</tr>";
 
  }
echo "</table>";


mysql_close($con);

?>
<form action="update.php" method="post">

<input type="text" name="userdate" value="<?= $b ?>" /> 

<input type="submit" name="subjoin" value="Update">

</form>

</body>
</html>

and update.php page is

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Update Record</title>
</head>

<body>



<?php




$con = mysql_connect("localhost","XXXXX","XXXXXXX");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 $c= $_POST['siteid']
echo "value".$c
print $c; //ITS NOT WORKING
mysql_select_db("onm", $con);

/*mysql_query("UPDATE SitePInfo SET Age = '36'
WHERE FirstName = 'Peter' AND LastName = 'Griffin'");
*/
mysql_close($con); 
?> 
</body>
</html>

$c= $_POST
echo "value".$c
gives this error

Parse error: parse error in C:\xampp\htdocs\update.php on line 24

hi
you have no pu ';' semicolon after the
this code

$c= $_POST['siteid'];
echo "value".$c;

parse error means you miss ; in script ok.

hi
this is your code but you can also do easy way in php
so i just ediit it like

<?php
while($row = mysql_fetch_array($result))
{ ?>
<table cellpadding=2 cellspacing=2 width=100%>
<tr>
</tr>
<tr>
<th bgcolor=#5D9BCC >SiteID</th>
<td bgcolor=#FEE9A9><?= $row['SiteId'] ?>/td>
</tr>
</table>
<?
$b =$row['SiteId']; // store value in $b
}

this code is not generate complexity.......
you save your time of solving your parse error right

Thanks

now its shwoing in result value

I am confused please give me updation example.
Ihave attache pic.

Attachments update.jpg 82.66 KB

hi

if you want to fetch only one row so that time no need of while loop ok

$row=mysql_fetch_array($result);

this is enough code.
now first of all you need to put name address and city value in text box ok

<form action="update.php" method="post">
<table>
<tr>
<td>name</td>
<td>
<input type="text" name="username" id="username" value="<?=$row['name'] ?>" /> </td></tr>
<tr>
<td>Address</td>
<td>
<input type="text" name="address" id="address" value="<?=$row['address']?>"></td></tr>
<tr>
<td>city</td>
<td>
<input type="text" id="city" name="city" value="<?=$row['city']?>">
</td>
</tr>
<tr>
<td colspan="2">
<input type="submit" value="Update"/>
</td>
</tr>
<input type="hidden" value="<?=$row["siteid"]?>" name="siteid" nmae="siteid">
</table>
</form>

now this in update.php file

$name=$_post["username"];
$address=$_post["address"];
$city==$_post["city"];
$id=$_post['siteid'];
$query="update table_name set name='$name',address='$address',city='$city' where siteid='$id'"'
mysql_query($query);

i posted complete code for update profile now you need to change design and table fields as per your requirment

Thanks

please check this hidden field is not showing any value now. and not working. how i change while loop to just statement. when i delete it gives error so please make this correct.

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<?
$con = mysql_connect("localhost","xxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("new", $con);

$result = mysql_query("SELECT * FROM abc
WHERE SiteId='C-SIL-4441' "); // I just get one value at this time


while($row = mysql_fetch_array($result))
  {
  echo "<table cellpadding=2 cellspacing=2 width=100%>
<tr>


</tr>";
  echo "<tr>";
  echo "<th bgcolor=#5D9BCC >SiteID</th>";
  echo "<td bgcolor=#FEE9A9>" . $row['SiteId'] . "</td>"; // show value in cell

 
 $b =$row['SiteId']; // store value in $b
 


  
 
  

  echo "</tr>";
 
  }
echo "</table>";


mysql_close($con);

?>
<form action="update_tulsa.php" method="post">
<table>
<tr>
<td>name</td>
<td>
<input type="text" name="username" id="username" value="<?=$row['name'] ?>" /> </td></tr>
<tr>
<td>Address</td>
<td>
<input type="text" name="address" id="address" value="<?=$row['address']?>"></td></tr>
<tr>
<td>city</td>
<td>
<input type="text" id="city" name="city" value="<?=$row['city']?>">
</td>
</tr>
<tr>
<td colspan="2">
<input type="submit" value="Update"/>
</td>
</tr>
<input type="hidden" value="<?=$row["siteid"]?>" name="siteid">
</table>

</body>
</html>

and update.php page is

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Update Record</title>
</head>

<body>



<?php




$con = mysql_connect("localhost","xxxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("new", $con);

$name=$_post["username"];
$address=$_post["address"];
$city==$_post["city"];
$id=$_post['siteid'];

$query=mysql_query("update table_name set name='$name',address='$address',city='$city' where siteid='$id'");



mysql_close($con); 
?> 
</body>
</html>

I also change table name with abc that was my mistake but its not posting values to update_tulsa.php

hi
after seen your code i notice that you have not complete form tag
after the table of first page right.
so now your are not able to view any posted fields ok

Thanks

Now its working.
Can I ask another Question?
But please remove the while loop and also tell me how I can get value from result in a varaible from report.php to form.php

thank you so much............
form.php here

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<?
$con = mysql_connect("localhost","xxxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("new", $con);

$result = mysql_query("SELECT * FROM abc
WHERE SiteId='C-SIL-4441' "); // I just get one value at this time


while($row = mysql_fetch_array($result))
  {
  echo "<table cellpadding=2 cellspacing=2 width=100%>
<tr>


</tr>";
  echo "<tr>";
  echo "<th bgcolor=#5D9BCC >SiteID</th>";
  echo "<td bgcolor=#FEE9A9>" . $row['SiteId'] . "</td>";
  echo "<td bgcolor=#FEE9A9>" . $row['Name'] . "</td>";
  echo "<td bgcolor=#FEE9A9>" . $row['Address'] . "</td>";
  echo "<td bgcolor=#FEE9A9>" . $row['City'] . "</td>"; // show value in cell

 
 $b =$row['SiteId']; 
  $c =$row['Name'];
   $d =$row['Address'];
    $e =$row['City'];// store value in $b
 



  
 
  

  echo "</tr>";
 
  }
echo "</table>";


mysql_close($con);

?>
<form action="update2.php" method="post">
<table>
<tr>
<td>name</td>
<td>
<input type="text" name="username" id="username" value="<?php echo $c; ?>" /> </td></tr>
<tr>
<td>Address</td>
<td>
<input type="text" name="address" id="address" value="<?php echo $d; ?>"></td></tr>
<tr>
<td>city</td>
<td>
<input type="text" id="city" name="city" value="<?php echo $e; ?>">
</td>
</tr>
<tr>
<td colspan="2">
<input type="submit" value="Update"/>
</td>
</tr>
<input type="text" value="<?php echo $b; ?>" name="siteid">
</table>

</body>
</html>
<body>



<?php




$con = mysql_connect("localhost","xxxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("new", $con);

if(!empty($_POST["siteid"]))
{
$a= $_POST["siteid"];
echo $a; //its working
}

if(!empty($_POST["username"]))
{
$c= $_POST["username"];
echo $c; //its working
}

if(!empty($_POST["address"]))
{
$d= $_POST["address"];
echo $d; //its working
}

if(!empty($_POST["city"]))
{
$e= $_POST["city"];
echo $e; //its working
}



$query=mysql_query("update abc set name='$c',address='$d',city='$e' where siteid='$a'");



mysql_close($con); 
?> 
</body>
</html>

hi

i just edit form.php file here
and findout that how it works

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<?
$con = mysql_connect("localhost","xxxx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("new", $con);

$result = mysql_query("SELECT * FROM abc
WHERE SiteId='C-SIL-4441' "); // I just get one value at this time


$row = mysql_fetch_array($result);
?>
  <table cellpadding=2 cellspacing=2 width=100%>
<tr>
</tr>
<tr><th bgcolor=#5D9BCC >SiteID</th>
<td bgcolor=#FEE9A9><?= $row['SiteId']?></td>
<td bgcolor=#FEE9A9><?= $row['Name']?></td>
<td bgcolor=#FEE9A9><? $row['Address']?></td>
<td bgcolor=#FEE9A9><?= $row['City'] ?></td>
</tr>
</table>
 
<? $b =$row['SiteId']; 
   $c =$row['Name'];
   $d =$row['Address'];
   $e =$row['City'];// store value in $b
mysql_close($con);
?>

hope you understand above code ok
if your problem solved then you need to mark as solved thread ok

Thanks

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