coding searching yang tak jalan

miss_tlot · #1 (**permalink**) 27-07-2003, 11:02 PM

hai...

baru-baru nie saya ada tanya pasal serching...dan ada sorang brother minta saya letak saya punya coding untu ditengok-tengokkan, so nie coding saya. Coding saya nie tak display apa-apa pun, kosong terus...tak jumpa apa yang saya cari walaupun dalam database ada data tu..so, sesapa yang baik hati tu, harap dapat tolong bagitau apa yang kurang atau lebih yek! TQ

//cari.php
------------------------------------------------------------------------------------
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>searching</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<form action="cari_2.php." method="post">
<table width="414" border="0" cellpadding="0" cellspacing="0">

<tr>
<td width="153" height="24" valign="top">
<input name=key type=text>
</td>
<td width="144" valign="top">  </td>
<td width="117" valign="top">
<input name="submit" type="submit"></td>
</tr>
<tr>
<td height="19" valign="top">*katakunci</td>
<td></td>
</tr>
</table>
</form>
</body>
</html>
------------------------------------------------------------------------------------
//cari_2.php
<?

/* mysql connection */
$host = "lacalhost";
$user = "";
$pass = "";
$db = "sppjibw";
import_request_variables('p', 'form_');
$connect = mysql_connect("localhost","","");
$open = mysql_select_db("sppjibw",$connect);

$key = "$form_key";

if($key=="")
{$key='%';}

if($result=="")
{$result='%';}

/*
if($open) {
echo "Database opened";
} else {
echo mysql_error();
}
*/

if(isset($_POST[‘submit’])) { //if submit button has been pressed
//take variable of form
$key = $_POST[‘key’];
//searching
$selec = mysql_query("SELECT result FROM carian WHERE key LIKE ‘%key%’,$conn");

$sql = mysql_query($selec);
//show the results
echo "Results";
print
$i=1;
while($row = mysql_fetch_array($sql))
{
print $row["result"];
print $row["key"];
$katakunci = $row["key"];
$hasil = $row["result"];
echo "$i $result ";
$i++;
}
if(!$sql) {
?><script language=’JavaScript’>alert(‘No record found’); </script><?
}
}
?>

zaimlah · #2 (**permalink**) 28-07-2003, 12:17 AM

PHP Code:


			
....

if(isset($_POST'submit'])) { //if submit button has been pressed

//take variable of form

$key = $_POST['key'];

//searching

$selec = mysql_query("SELECT result FROM carian WHERE key LIKE '%$key%',$conn"); //<--- sini dia, '%$key%' bukan '%key%'

....

(i think so)

mysticmind · #3 (**permalink**) 28-07-2003, 12:18 AM

PHP Code:


			
/* mysql connection */

$host = "lacalhost";

lacalhost?

zaimlah · #4 (**permalink**) 28-07-2003, 12:20 AM

one more thing, dia tak keluar apa-apa, atau ada keluar MySQL error? sebab line yg ni agak pelik sikit:

Code:

$selec = mysql_query("SELECT result FROM carian WHERE key LIKE ‘%key%’,$conn");

apa guna, dan dari mana datangnya '$conn' ??

ditambah kemudian (edit post):

see next post

zaimlah · #5 (**permalink**) 28-07-2003, 12:31 AM

mungkin macam ni:

Code:

$selec = mysql_query("SELECT result FROM carian WHERE key LIKE '%$key%'",$connect");

variable yg sebenar ialah $connect, bukan $conn ... dan mesti di luar sql query sebagai argument ke-2....

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