Whats wrong in this one ??

[Anon]

Im trying to get data from mysql. Thsi works just fine, but if it finds something, it shows it as a link. Problem is it doeas not show data behind the link ?????? Here is the script:

And btw, i have that hits column on that table, if someone knows easyly how to update those hits everytime someone reads the data behind that link.
I really eed help on this one too...

<form method=post action="<?php echo $PHP_SELF ?>">
<font size=3 face=arial color=#330066><b>Etsi hakusanalla: &nbsp </b></font>
<SELECT NAME="value">
<!--<option value="id">ID-->
<option value="nimi">Nimi
<option value="koreo">Koreografi
<option value="musiikki">Musiikki
</select>
<br><br>
<input type=text name="keyword" size=35>
<input type=submit value="Hae kannasta">
</form>
<?php
if($keyword){
$db=mysql_connect("localhost","my_user","mypass");
mysql_select_db("mydatabase",$db);
if($value==nimi){
$sql="SELECT FROM mytable WHERE nimi='$keyword'";
}
else if($value==koreo){
$sql="SELECT FROM mytable WHERE koreo='$keyword'";
}
else{
$sql="SELECT * FROM mytable WHERE musiikki='$keyword'";
}
$result=mysql_query($sql);
if($myrow=mysql_fetch_array($result))
{
do{

printf("<a href=\"%s?id=%s\">%s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["nimi"], $myrow["lakana"], $myrow["hits"]);
}while($myrow=mysql_fetch_array($result));

}
else{
printf("Sorry, No records were found in our database!\n");

}
}
?>

[[deleted]]

your printf() function has three %s, but you give 5 arguments. You are missing two %s in the string.

[Anon]

hmm.. call me a new with this, but where do i put those two %s:s ?

Thanks a lot for your help !!!

[Anon]

hmm.. here is how i want it to work, but with search and what i want too is if i look from that previus script is that, if i look with nimi and type just part of a name, it lists all where typed nimi is present.
well, here is it without that seach thing.

<?php

$db = mysql_connect("localhost", "root");

mysql_select_db("mydatabase",$db);

if ($id) {

$result = mysql_query("SELECT * FROM table WHERE id=$id",$db);

$myrow = mysql_fetch_array($result);

printf("%s\n<br>",$myrow["lakana"]);

} else {

// show list

$result = mysql_query("SELECT * FROM table",$db);

if ($myrow = mysql_fetch_array($result)) {

  // display list if there are records to display

printf("<h1><b>Tanssit kannassa</b></h1><br>");
do {
printf("<a href=\"%s?id=%s\">%s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["nimi"], $myrow["hits"]);

  } while ($myrow = mysql_fetch_array($result));


} else {

  // no records to display

  echo "Sorry, no records were found!";

}

}

?>

this will print all from the table and gives it as a link, and when you click it is shows lakana..

[[deleted]]

Put the %s in the first argument of printf, at the point where you want th corresponding string to appear.
(the first %s gets replaces by the second argument, the second %s by the third argument etc.

Check out the manual for more info.

[Anon]

I know now wht you mean, but it does not do how i want it to. It prints that "lakana" right away. Not give lakanas name as a link.

-T-