Passing data in a URL

[harkly]

I am new to php and cannot figure out how to move data from one page to another, I would like to do it in a URL. I don’t know the correct verbiage so it makes it harder for me to find this particular info.

 

I have a form, search.php, which passes the info to results.php which makes a list of all names in my database. I then want to click on the name(s) and have that pass the ‘artid’ to artist.php which will have detailed info on that particular name and some images which I want to be clickable, passing the ‘imgid’ to another page with detail info on the images.

 

So far my search.php works and I can get a list on the results.php but I can’t click on the name and pass that info on. All of my pages work independently.

 

In results.php I have:

 

echo “<a href=artist.php$id=artid>$fullName</a>\n”

 

I know a problem lies here (=artist.php$id=artid) but can’t figure it out. Do I need a form method on the search.php or am I completely off the mark here?

 

In artist.php:

 

$search=@_GET[“artid”];

 

$results = mysql_query(“SELECT * from image where artid like ‘AR3614’”);

My query works when I put in an actual ID (AR3614)

 

 

Could someone could point me in the right direction.

 

[Maq]

No, you don't need to put this in a form unless $id comes from an input field.  Change these lines and it should work.  If not echo them out and see what they output.

 

echo "".$fullName."
"

 

$search=$_GET['artid'];

[php-pendejo]

I know this is crazy and above a new php programmers understanding but...

 

if you want to get with more code and and make life easier place this in your code

 

$ArrayList = array("_SERVER", "_COOKIE", "_ENV", "_FILES", "_POST", "_GET", "_SESSION");

foreach($ArrayList as $gblArray) {

$keys = array_keys($$gblArray);

foreach($keys as $key){

@$$key = trim(${$gblArray}[$key]);

}

}

 

make you links as so

echo "<a href=artist.php?artid=".$id.">".$fullName."</a><br />"

 

then you can  do your mysql as so

$results = mysql_query(“SELECT * from image where artid like ‘$artid’”);

 

[harkly]

I got the second pass to work but I cannot get the variable to pass to my URL. Here's my code,

 

 

mysql_select_db("artdb");

 

 

$search=$_GET["search"];

 

$result = mysql_query("SELECT * FROM artist WHERE fullName LIKE '%$search%'");

 

while ($r=mysql_fetch_array($result))

 

{

 

$fullName=$r["fullName"];

 

$dob=$r["dob"];

 

$dod=$r["dod"];

 

$nationality =$r["nationality"];

 

$choosen_medium=$r["artForm"];

 

$artid=$r["artid"];

 

echo "<a href=artist.php?artid=".$artid.">".$fullName."</a><br />";

 

}

?>

[harkly]

 

Got it to work

 

this is what I used

 

echo "<a href='artist.php?artid=$r[artid]'>".$fullName."</a><br />";

[Maq]

Next time you post could you use code or php tags?  It's hard to read    Also please mark this as solved.