Postingan lainnya
syntax ?> untuk end of file apa ya di php
<?php
require_once 'config.php';
$dbConn = mysqli_connect ($dbHost, $dbUser, $dbPass) or die ('MySQL connect failed. ' . mysqli_error());
mysqli_select_db($dbConn, $dbName) or die('Cannot select database. ' . mysqli_error());
function dbQuery($sql)
{
$sql = ['SELECT uid, uname, utype FROM tbl_users '];
$result = mysqli_query ("SELECT * from tbl_users where username='$dbUser' and password='$dbPass') or die(mysqli_connect_error());
return $result;
}
function dbAffectedRows()
{
global $dbConn;
return mysqli_affected_rows($dbConn);
}
function dbFetchArray($result, $resultType = MYSQLi_NUM) {
return mysqli_fetch_array($result, $resultType);
}
function dbFetchAssoc($result)
{
return mysqli_fetch_assoc($result);
}
function dbFetchRow($result)
{
return mysqli_fetch_row($result);
}
function dbFreeResult($result)
{
return mysqli_free_result($result);
}
function dbNumRows($result)
{
return mysqli_num_rows($result);
}
function dbSelect($dbName)
{
return mysqli_select_db($dbName);
}
function dbInsertId()
{
return mysqli_insert_id();
}
?>
berikut error nya:
Parse error: syntax error, unexpected end of file in B:\xampp baru\htdocs\asset_management\library\database.php on line 55
mohon bantuan nya, apa yang harus ditambahkan pada syntax ?> agar bisa tutup?
Tanggapan
coba kirim potongan kode-nya sebagai "script", jangan sebagai teks biasa, untuk membaca potongna kodenya agak susah
1 Jawaban:
<div>Ini coba saya betulkan dan rapihkan:</div><pre><?php require_once 'config.php'; $dbConn = mysqli_connect ($dbHost, $dbUser, $dbPass) or die ('MySQL connect failed. ' . mysqli_error()); mysqli_select_db($dbConn, $dbName) or die('Cannot select database. ' . mysqli_error()); function dbQuery($sql){ $sql = ['SELECT uid, uname, utype FROM tbl_users ']; $result = mysqli_query("SELECT * from tbl_users where username='$dbUser' and password='$dbPass'") or die(mysqli_connect_error()); return $result; } function dbAffectedRows(){ global $dbConn; return mysqli_affected_rows($dbConn); } function dbFetchArray($result, $resultType = MYSQLi_NUM) { return mysqli_fetch_array($result, $resultType); } function dbFetchAssoc($result){ return mysqli_fetch_assoc($result); } function dbFetchRow($result) { return mysqli_fetch_row($result); } function dbFreeResult($result){ return mysqli_free_result($result); } function dbNumRows($result){ return mysqli_num_rows($result); } function dbSelect($dbName){ return mysqli_select_db($dbName); } function dbInsertId(){ return mysqli_insert_id(); } ?></pre><div><br>Penjelasan:<br>Yang jadi masalah adalah di variable $result.</div><pre>$result = mysqli_query ("SELECT * from tbl_users where username='$dbUser' and password='$dbPass') or die(mysqli_connect_error());</pre><div><br>Di function mysqli_query() anda kurang teliti dengan menambahkan tanda petik ganda, yang benar adalah seperti ini:</div><pre>$result = mysqli_query("SELECT * from tbl_users where username='$dbUser' and password='$dbPass'") or die(mysqli_connect_error());</pre><div><br>Semoga cepat paham dan membantu.</div>
