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?
$dbConn = mysqli_connect ($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?
4 Kontribusi 3 Poin
- coba kirim potongan kode-nya sebagai "script", jangan sebagai teks biasa, untuk membaca potongna kodenya agak susah - @ahanafi
atau respon:
Jawaban
Ini coba saya betulkan dan rapihkan:
Penjelasan:
Yang jadi masalah adalah di variable $result.
Di function mysqli_query() anda kurang teliti dengan menambahkan tanda petik ganda, yang benar adalah seperti ini:
Semoga cepat paham dan membantu.
<?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(); } ?>
Penjelasan:
Yang jadi masalah adalah di variable $result.
$result = mysqli_query ("SELECT * from tbl_users where username='$dbUser' and password='$dbPass') or die(mysqli_connect_error());
Di function mysqli_query() anda kurang teliti dengan menambahkan tanda petik ganda, yang benar adalah seperti ini:
$result = mysqli_query("SELECT * from tbl_users where username='$dbUser' and password='$dbPass'") or die(mysqli_connect_error());
Semoga cepat paham dan membantu.
70 Kontribusi 160 Poin
Login untuk gabung berdiskusi
Pertanyaan Lainnya
Top Kontributor
- @ahanafi
813 Kontribusi 551 Poin
- @Nandar
647 Kontribusi 204 Poin
- @dianarifr
642 Kontribusi 316 Poin
- @Saputroandhi
509 Kontribusi 162 Poin
- @dodipsitorus
412 Kontribusi 145 Poin