另一个mysqli_query（）期望参数1为mysqli，给定字符串

Question

i have read all the reference about this topic and follow as instruction. My other application using this step is running smoothly, but this simple query in part of my program keep showing this error :

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in D:\\xampp183-5\\htdocs\\Development\\Pajak-Penjualan_Auto_2-0.php on line 258

Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\\xampp183-5\\htdocs\\Development\\Pajak-Penjualan_Auto_2-0.php on line 258

Query Ref Penjualan Failed

Here the code :

$link = mysqli_connect('localhost', 'Username', 'Password', $_SESSION['Database']);
if (!$link) {
   die('Not connected : ' . mysqli_error());
}

$db_selected = mysqli_select_db($link, $_SESSION['Database']);
if (!$db_selected) {
  die ('Can\'t use foo : ' . mysqli_error());
}    

.
.
.//Some SqlCommand
$com = "
select h.Kode_FP, Tanggal, Ceil(
(
    SUM((Qty * Harga_Satuan)) - Potongan - Uang_Muka
)
) as Total, h.Staff, h.NPWP, h.Nama, k.Nama_Karyawan
from fp_penjualan_h h
LEFT JOIN fp_penjualan_d d ON h.Kode_FP = d.Kode_FP
LEFT JOIN karyawan k ON k.Kd_Karyawan = h.Staff
where Tanggal Between '".@$_REQUEST['Mulai']."' and '".@$_REQUEST['Sampai']."'
group by Tanggal, h.Kode_FP
".@$_REQUEST['Sort']."
";
$query = mysqli_query($link, $com) or die(mysqli_error($link)."Query Failed");
while($hasil = mysqli_fetch_array($query))
{
     $comRefPenjualan = "
     select COUNT(DISTINCT(rj.Kd_Penjualan)) as Ref_Jual
     from fp_ref_jual rj
     where rj.Kode_FP = '".$hasil[0]."'
     group by rj.Kode_FP
     ";
     $queryRefPenjualan = mysqli_query($link, $comRefPenjualan) or die(mysqli_error()."Query Ref Penjualan Failed");
     $hasilRefPenjualan = mysqli_fetch_array($queryRefPenjualan);
.
.
.

But whenever i query manual, there's no error, the query went success. What's wrong with my code ? Anyone help me please.

Answer 1

Instead of mysqli_error() it should be mysqli_error($link)

mysqli_error() needs to pass the connection parameter to the database.

To make your code better replace

if (!$link) {
   die('Not connected : ' . mysqli_error());
}

$db_selected = mysqli_select_db($link, $_SESSION['Database']);
if (!$db_selected) {
  die ('Can\'t use foo : ' . mysqli_error());
}

To

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

As your db is already selected in $link above

Also change

$query = mysqli_query($link, $com) or die(mysqli_error($link)."Query Failed");

To

if (!mysqli_query($link, $com)) {
    printf("Query Failed: %s\n", mysqli_error($link));
}else{
   $query = mysqli_query($link, $com); 
}

And this

 $queryRefPenjualan = mysqli_query($link, $comRefPenjualan) or die(mysqli_error()."Query Ref Penjualan Failed");

TO

if (!mysqli_query($link, $comRefPenjualan)) {
    printf("Query Ref Penjualan Failed: %s\n", mysqli_error($link));
} else {
   $queryRefPenjualan = mysqli_query($link, $comRefPenjualan); 
}