details page from search results

Anon

How would I set up a details page that would display more information on the selected search results? Basically, I want the user to be able to perform a search and then click on a field in the records obtained from the search results to get additional information on any of the records they desire.

Someone posted a similar question in the past but I wasn't able to decipher the responses (no doubt due a lack of programming knowledge on my end). The database is set up with an auto_increment primary key (ID) and the detail page is named details.php

I have included a section of code from the results page with the hope that it will help to tie it together with the details page.

$result = mysql_query($query);
$num_results = mysql_num_rows($result);

echo "<p>Number of matching results: ".$num_results."</p>";

for ($i=0; $i <$num_results; $i++)
{
$row = mysql_fetch_array($result);
print "<a href = \"details.php\">Report Number</a>";
echo htmlspecialchars ( stripslashes($row["rptid"]));
echo " Author: ";
echo htmlspecialchars ( stripslashes($row["author"]));
echo " Title: ";
echo htmlspecialchars ( stripslashes($row["title"]));
echo " Source: ";
echo htmlspecialchars ( stripslashes($row["source"]));
echo "";
}

Thanks in advance for the help!!

Jeff

Anon

Forgive me if I'm oversimplifying this but what I would do:

print "<a href = details.php?ID=$ID>Report Number</a>";

then in details.php:

$query="SELECT * FROM db_name WHERE ID=$ID";

and then go from there...

hope that answers your question!

mondragon

Thanks for the advice and I don't think it is possible to oversimplify in my case. (Which I'm sure my next question will prove).

Once I have put
$query="SELECT * FROM db_name WHERE ID=$ID";
on the details page, how do I display the appropriate fields on the details page.

I have been attempting to use an array but I am failing miserably. Any suggestions??

Thanks!!

Jeff

Anon

This is what I've done to display appropriate fields. Change things to suite your fields. This is just a bit of code for you to see how it's done.

$sql = "SELECT *
	FROM $table_name
	WHERE id = \"$id\"
	";

$result = @mysql_query($sql,$connection) or die("Couldn't execute query.");


while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$name = $row['name'];
$email = $row['email'];
$city = $row['city'];
$state = $row['state'];
}

}

<HEAD>
<TITLE>Member Details</TITLE>
</HEAD>
<BODY background="background1.jpg">

<center>Member Details 
Details for <? echo "$username";?> from <? echo "$city"; ?></center>

<table cellspacing=0 cellpadding=30>
<tr>
<th align=center valign=center width=100 height=30>Profile
<th align=center valign=center width=100 height=30>My Picture
</tr>
<tr>
<td valign=top>
<? echo "Name: $name"; ?> 
<?
if ($city != "") {
echo "City: $city ";
}
if ($state !=""){
echo "State: $state ";
}

Keep using the if($FIELD !=""){ . . . for all the fields you want displayed. If the field is blank it will not be displayed.

use this where you want a few lines to seperate displayed fields into specific groups

<?
continue with the if($FIELD) . . .

The way I have it setup is in 2 columns if you want to use columns, put this where you want the 1'st column to end to begin the next column
End your last if with } then continue with this:
?>
<td valign=top>
<?

Start your next if($FIELD) . . .

to end use:
?>
</td>
</tr>
</table>
ANY OTHER HTML CODE YOU WANT TO DISPLAY
</BODY>

Hope this helps. I'm new to programming myself, I've found the following books to be useful: PHP Fast & Easy Web Development and Professional PHP Programming to be great help.

mondragon

Thanks for the help but it still is giving problems. Now it the error message "Supplied argument is not a valid MySQL result resource" in reference to:
while ($row = mysql_fetch_array($result)) {

Any suggections?

Thanks!!