如何在新查询中使用html表值（从查询生成）？

Question

i've currently got a php/html table which holds values generated from a query:

while($row = mysql_fetch_array($query))
{
echo "<tbody>";
echo "<tr>";
echo "<td>" . $row['disease'] . "</td>";
echo "<td>
        <a href='result1.php' class='button1'>View Details</a>
        <a href='#' class='button1'>Book Appointment</a>
      </td>";
echo "</tr>";
echo "</tbody>"

;

What i would like to do is enable the user to click on the 'View Details' button to initiate another query (which is processed in result1.php):

$query = mysql_query("SELECT definition FROM tbl_disease WHERE disease = '" . $_GET[' $row[disease'] . "' ;  ")
or die(mysql_error());

This query should get details (definition) based on table values from the previous php page. At the moment i get an error on '$_GET[' $row[disease'] .'. Im new to this so I'm unsure if this is the right way to go about it?

Any ideas would be greatly appreciated, thanks.

Answer 1

In result1.php you have $_GET[' $row[disease'] but in the code with

<a href='result1.php' class='button1'>View Details</a>

you don't send the GET value. Change the above to:

<a href='result1.php?disease=".$row['disease']."' class='button1'>View Details</a>

and in result1.php

$query = mysql_query("SELECT definition FROM tbl_disease WHERE disease = '".$_GET['disease']."' ;  ")

To make it the right way with security just as @Dragon mentioned you should never do operations on mysql with $GET/$POST and other without proper formating.

For more info: mysql_escape_string htmlentities filter_input.