Hello Experts,

I am stuck in database query. Well i am making a page which fetch all my database value. For which below is my code:

PHP Code:
<?php
$con
=mysqli_connect("localhost","root","","test");
// Check connection
if (mysqli_connect_errno())
  {
  echo 
"Failed to connect to MySQL: " mysqli_connect_error();
  }

$result mysqli_query($con,"SELECT * FROM testimonials");

echo 
"<table border='1'>
<tr>
<th>ID</th>
<th>Name</th>
<th>Email_id</th>
<th>Review_Title</th>
<th>Review</th>
<th>Date</th>
<th>Post</th>
</tr>"
;

while(
$row mysqli_fetch_array($result))
  {
  echo 
"<tr>";
  echo 
"<td>" $row['id'] . "</td>";
  echo 
"<td>" $row['name'] . "</td>";
  echo 
"<td>" $row['email_id'] . "</td>";
  echo 
"<td>" $row['review_title'] . "</td>";
    echo 
"<td>" $row['review'] . "</td>";
  echo 
"<td>" $row['date'] . "</td>";
   echo 
"<td>" "<a href='row_test.php'>Post</a>""</td>";
  echo 
"</tr>";
  }
echo 
"</table>";

mysqli_close($con);
?>
This is done properly but in my last column "POST" der are links.What i want is wen click on one particular row link on the next page it must display my selected row contents. I tried the below code so far:

PHP Code:
<?php
$con
=mysqli_connect("localhost","root","","test");
if (
mysqli_connect_errno())
  {
  echo 
"Failed to connect to MySQL: " mysqli_connect_error();
  }
 
$txt$_POST[id];
$result mysqli_query($con,"SELECT * FROM testimonials where id=3");
while(
$row mysqli_fetch_array($result))
  {
  echo 
"<tr>";
  echo 
"<td>" $row['name'] . "</td>" ."<br>";
  echo 
"<td>" $row['review_title'] . "</td>";
  echo 
"<td>" $row['review'] . "</td>";
  echo 
"</tr>";
  }
echo 
"</table>";

mysqli_close($con);
?>

But this only output after mentioning the "id" ie: id=3, but i want it automatically after clicking on the link for the selected row.
Please some one can help me out with the code....
Any help is appreciated.