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cara mengatasi error You have an error in your SQL syntax;
master mohon bantuannya dah beberapa kali saya cek hasilnya error terus kira kira masalahnyaa dimanaa You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '' at line 1
<?php
include ('../conn.php');
if(isset($_GET['submit']) &&isset($_GET['nik']) && isset($_GET['tableid'])&& isset($_GET['ssf'])&& isset($_GET['userid'])){
$nik = $_GET['nik'];
$tableid = $_GET['tableid'];
$qty = $_GET['ssf'];
$userid = $_GET['userid'];
$sql=mysql_query('select * from data where nik like "%'.$nik.'%" order by id DESC limit 1')or die(mysql_error());
while($result=mysql_fetch_array($sql))
{
$id = ''.$result['id'].'';
$nik = ''.$result['nik'].'';
$nama = ''.$result['nama'].'';
$jml_bln = ''.$result['jml_bln'].'';
$harga = ''.$result['harga'].'';
$status = ''.$result['p_status'].'';
}
$sql=mysql_query('select * from startme')or die(mysql_error());
while($result=mysql_fetch_array($sql))
{
$id = ''.$result['id'].'';
$startme = ''.$result['startme'].'';
}
if($jml_bln <= '0')
{
echo'<script>';
echo'alert("YOUR STOCK FOR THIS ITEM IS ZERO"); ';
echo'window.location.href="tablein.php?tableid='.$tableid.'&userid='.$userid.'";';
echo'</script>';
}
else
{
if($jml_bln >= $qty)
{
$sql=mysql_query('UPDATE data SET h_beggining = '.$jml_bln.' WHERE id = '.$id.' order by id DESC limit 1')or die(mysql_error());
$jml_bln = $jml_bln - $qty;
$harga = $harga * $qty;
$sql=mysql_query('select * from members where member_id = '.$userid.' ')or die(mysql_error());
while($result=mysql_fetch_array($sql))
{
$name = ''.$result['FirstName'].'';
}
$sql=mysql_query('UPDATE data SET jml_bln = '.$jml_bln.' WHERE nik = '.$nik.' order by id DESC limit 1')or die(mysql_error());
$sql=mysql_query('UPDATE data SET harga = '.$harga.' WHERE nik = '.$nik.' order by id DESC limit 1')or die(mysql_error());
$sql=mysql_query('UPDATE tablename SET status = "occupy" ,assist_by = "'.$name.'" WHERE id = '.$tableid.' order by id DESC limit 1')or die(mysql_error());
$sql1 = "INSERT INTO payment('idno','nik','nama','jml_bln','harga','status','date','user')
VALUES ('{$tableid}','{$nik}','{$nama}','{$qty}','{$harga}','{$status}',NOW(),'{$userid}')";
$retval1 = mysql_query( $sql1);
if(!$retval1)
{
die('Could not enter data: ' . mysql_error());
}
$sql1 = "INSERT INTO payment_report('idno','nik','nama','jml_bln','harga','status','date','user')
VALUES ('{$tableid}','{$nik}','{$nama}','{$qty}','{$jml_bln}','{$status}',NOW(),'{$userid}')";
$retval1 = mysql_query( $sql1);
if(!$retval1)
{
die('Could not enter data: ' . mysql_error());
}
$sql=mysql_query("SELECT sum(harga) as price FROM payment where idno = ".$tableid."")or die(mysql_error());
while($result=mysql_fetch_array($sql))
{
$total = ''.$result['price'].'';
}
$sql=mysql_query('UPDATE tablename SET total = '.$total.' WHERE id = '.$tableid.' order by id DESC limit 1')or die(mysql_error());
}
else
{
echo'<script>';
echo'alert("YOUR STOCK ITEM IS NOT EQUAL!"); ';
echo'window.location.href="tablein.php?tableid='.$tableid.'&userid='.$userid.'";';
echo'</script>';
}
}
echo'<script>';
echo'alert("Item Succefully Add!!"); ';
echo'window.location.href="tablein.php?tableid='.$tableid.'&userid='.$userid.'";';
echo'</script>';
}
?>
5 Jawaban:
Bisa jadi inputnya kosong (ga ada nilai) gan.
'.$myinput.'
nah kalau $myinput ga ada nilainya ya jadi ''
maksud nya gimannaa gan kurang pahaamm sayaa
Itu query kan lebih dari 1 jadi cari query mana yg menyebabkan error. (tutup/comment query yg lain dulu lalu trace satu persatu)
misalkan query anda
UPDATE tablename SET total = '.$total.' WHERE
untuk memastikan $total ada nilai, bukan kosong sebab kalau kosong akan menjadi
UPDATE tablename SET total = '' WHERE
sisipkan var_dump sebelum query tersebut
var_dump($total);
die();
UPDATE tablename SET total = '.$total.' WHERE
<div>cara buka perintah" yang box item itu dimana ya?<br><br></div>