[RESOLVED] posting a hyperlink with a table ???

Aido82

posting a hyperlink with a table ???

Is there any way that i can post the value in of betIDinto a session or onto another page using a hyperlink in a table

I cant do it any way
wonderin if any gurus can help me

$i=1;

while($row = mysql_fetch_array($result))

{
$color = ($i % 2) ? "#F0E68C " : "#E0FFFF ";

echo "<tr bgcolor =", $color,">";
echo "<td><a href = 'seeBet.php'>", $row['betID'],"</a></td><td>",$row['fixtureID'], "</td><td>", $row['userName'], "</td><td>", $row['amount'],
"</td><td>", $row['prediction'],"</td><td>", $row['time'],"</td><td>",$row['result'],"</td><td>","</td>";
echo "</tr>";
$i++;

}
thanks

mykg4orce

you want to pass the value of betid to the seeBet.php I am guessing

why can't you do seeBet.php?id=$row['betid']

and use $_GET['id'] to retrieve it?

If you dont want to show the value in the url then use a form around the link. then use a javascript function to submit the form using the onClick handler of the <a tag.

Aido82

thanks for the reply

sorry if this sounds stupid but
where do i put the -- seeBet.php?id=$row['betid'] section

im not the best at php as you can see

edit is it like this

"<a href='seeBet.php?id=$row['betid']'>"

mykg4orce

thats it!

mykg4orce

on the seeBet page use $_GET['id'] to get your id back where you can use it with PHP. Also you must think about people who will change:

this:
seeBet.php?id=67

to this:
seeBet.php?id=lskdflksdnflaksldfkalskfnhahahah

take this into account and have proper error messages or redirect back to a safe page if $_GET['id'] is not what you were expecting.

meaning

if ($_GET['id'] == "") {
//meaning if the id is empty
echo "invalid ID";
}

you must check for this before any DB queries are made, after everything is kosher, then you can load your table data.

Aido82

i see what you mean about the php?id=111

but when i use the $_GET['id']; what value is that using or where does that go??

do i then need to use $betID= $_GET['id'];

did the above and it works but now get a error each time as below

Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively. in Unknown on line 0

mykg4orce

php?id=111

using $var = $_GET['id']; will return $var = 111;

without knowing the full extent of your script I cannot say why you are getting that error.

Aido82

thanks for the help mate

works a treat now🙂