HELP! display SQL query using a href code

hi there! i'm a newbie on php and i would like to ask on how to display the sql output correctly. It involves 2 php files and the codes are posted below.

display.php

echo "<a href=view_post.php?xval=";
echo $row["incval"];
echo ">";
echo $row["item_title"];
echo "</a>";



view_post.php

$result = mysql_query("select * from classified_postings where category ='xval'") or
die (mysql_error());

while ($row = mysql_fetch_array($result))

{
echo $row["date"];
echo "<br>";
echo $row["category"];
echo "<br>";

echo "<a href=view_post.php?xval=";
echo $row["incval"];
echo ">";
echo $row["item_title"];
echo "</a>";

echo "<br>";
echo $row["item_description"];
echo "<br><br>";
}

mysql_free_result($result);


any help would greatly be appreciated

 

help! please anyone

 

Can you explain what you are trying to do and what the error is you are getting. From looking over your code I spotted this error:

$result = mysql_query("select * from classified_postings where category ='xval'") 

PHP:

should be relaced with

$xval = (int)$_POST['xval'];
$result = mysql_query("select * from classified_postings where category ='$xval'");

PHP:

This is assuming that xval is a numeric value.

Brew

 

geeez thanks for the help.. let me try the code

 

What exactly is your question?

Peace,

 

Are you trying to echo the query that you're sending to the SQL server?

If so it may be worth trying storing the query as text in a variable, echoing the contents of the variable and then running mysql_query.

$query = "select something from somewhere";
echo $query;
$query = mysql_query($query);