Retriving Url ID for DB Query

vexer007

Hello, Greetings to you,

I am writting a code which basically its a url like below

<a href="http://cms.gre.uk/project/singleproduct.php?ID=<?php echo "".$id->value.""; ?>">View</a>

The above seems to work as the new URL now has the Value of the ID something like

http://cms.gre.uk/project/singleproduct.php?ID=3

But I was told that I can now perform a DB query with the value on the next page, this is giving errors I dont think the variable is been passed through to the next page. can somebody help me please?

Query is:

$rs = $conn->Execute("SELECT * FROM products WHERE id = '".$id->value."'");

Error: Fatal error: Call to a member function on a non-object in e:\project\singleproduct.php on line 8

Help Please, Thank you for your time!

🙁

alfoxy

Hey,

Try it without the ->value

i.e. before the $rs = $conn->Execute("SEL ...... just make sure the variable was passed successfully. Just

echo "Variable :".$id."<br>";

If this prints Variable : 3 then substitute $id->value with just $id

Hope this helps

vexer007

That didnt work neither.
🙁

cheese

Well if it is any help, here is the sort of thing I do.

This would be the list of links on the first page;

$query = mysql_query("SELECT * FROM mytable ORDER BY ID DESC LIMIT 10"); // this selects the last 10 entries in the table 


while ($row = mysql_fetch_array($query, MYSQL_NUM))
{

print "<a href='adminbirdview.php?id=". $row[0] ."'>". $row[1] ."</a>"; // $row[0] is the ID and $row[1] is the description that appears as the text for the hyperlink
}

Now to get that specific row from the query on the next page use something like this;

$id= $_GET[id]; // this is the information that was passed by your previous page.

$query = mysql_query("SELECT * FROM mytable WHERE ID= '". $id ."'"); // this query will only return the row for the ID passed from the previous page.


while ($row = mysql_fetch_array($query, MYSQL_NUM))
{

print "Record Number - ". $row[0] ."<br>";
print "Description - ". $row[1] ."<br>";
print "Another Field - ". $row[2] ."<br>";
//and add as many field as you want
}

By adding HTML tags you can create the output in to a table like this

$id= $_GET[id]; // this is the information that was passed by your previous page.

$query = mysql_query("SELECT * FROM mytable WHERE ID= '". $id ."'"); // this query will only return the row for the ID passed from the previous page.

print "<table>";
while ($row = mysql_fetch_array($query, MYSQL_NUM))
{

print "<tr><td>Record Number</td><td>". $row[0] ."</td></tr>";
print "<tr><td>Description </td><td>". $row[1] ."</td></tr>";
print "<tr><td>Another Field</td><td>". $row[2] ."</td></tr>";
//and add as many field as you want
}
print "</table>";

Well that should do something for you anyway.....

Cheers,

Neil.

vexer007

Thank to you all for your supports, I really appreciated it. Neil, your code was excellent, I actually just got it working about an hour before you posted the code, never the less that was very nicely and well explained set of code. Thank you and to the rest who have been posting suggestions since the begining.

All the best