select record from mysql_fetch_assoc after putting into table

Question

not sure how feasable this is, but I have just rolled my own user search form, which simply queries my database and returns all the results with any given username, or similar using the LIKE 'some_username%' statement.

My search works great, and im really chuffed with myself as I am a php and mysql novice.

I used a mysql_fetch_assoc($result) statement, and then used a while loop to echo out each row from the database into an html table.

What I would then like to be able to do, is select a record from the table, and open a new page, which is populated with all the fields for that record, which I can then use to edit and update the user settings.

I thought perhaps one way to do it, is to perhaps echo out a form instead? that way I can have a button next to each row, to post the fields into some php code on my new page? I thought this may be a bit clunky though, and not sure how I would go about echoeing out a different form for each row.

Don;t know if anyone had any ideas on the best way to do this? If you need any code examples of what im working with, I can post them here.

Thanks very much!! Eds

Answer 1

not a form but a hyperlink.
I wonder why you aren't familiar with this way of opening new pages as it is used everywhere.

just create a hyperlink

<a href="edit.php?id=1">name</a>

here is a sketch example of such an application, editing only one field, but you can add any number as well:

a main script:

<?  
mysql_connect(); 
mysql_select_db("new"); 
$table = "test"; 
if($_SERVER['REQUEST_METHOD']=='POST') { //form handler part: 
  $name = mysql_real_escape_string($_POST['name']); 
  if ($id = intval($_POST['id'])) { 
    $query="UPDATE $table SET name='$name' WHERE id=$id"; 
  } else { 
    $query="INSERT INTO $table SET name='$name'"; 
  } 
  mysql_query($query) or trigger_error(mysql_error()." in ".$query); 
  header("Location: http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF']);  
  exit;  
}  
if (!isset($_GET['id'])) { //listing part: 
  $LIST=array(); 
  $query="SELECT * FROM $table";  
  $res=mysql_query($query); 
  while($row=mysql_fetch_assoc($res)) $LIST[]=$row; 
  include 'list.php'; 
} else { // form displaying part: 
  if ($id=intval($_GET['id'])) { 
    $query="SELECT * FROM $table WHERE id=$id";  
    $res=mysql_query($query); 
    $row=mysql_fetch_assoc($res); 
    foreach ($row as $k => $v) $row[$k]=htmlspecialchars($v); 
  } else { 
    $row['name']=''; 
    $row['id']=0; 
  } 
  include 'form.php'; 
}  
?>

and two simple templates responsible for output,
one for the displaying the form, form.php

<? include TPL_TOP ?>
<form method="POST">
<input type="text" name="name" value="<?=$row['name']?>"><br>
<input type="hidden" name="id" value="<?=$row['id']?>">
<input type="submit"><br>
<a href="?">Return to the list</a>
</form>
<? include TPL_BOTTOM ?>

and one to display the list, list.php

<? include TPL_TOP ?>
<a href="?id=0">Add item</a>
<? foreach ($LIST as $row): ?>
<li><a href="?id=<?=$row['id']?>"><?=$row['name']?></a>
<? endforeach ?>
<? include TPL_BOTTOM ?>

Answer 2

always start php like this <?php, usually php manual configuration do not support short tag like this <?. for hyperlink just use view record It is just query string and get id on next page like this $id = $_GET['id']; Hope u will understand..