這個查詢工作不

Question

$qurym="SELECT * FROM referal_member WHERE `ref_cusid` = '$id' && `confirm` = '1'"; 
$resm=mysqli_query($con,$qurym); 
$rowm=mysqli_fetch_array($resm); 

Answer 1

剛剛嘗試這樣的：如果字段（ref_cusid和confirm）是INT類型，然後傳遞值$id和1沒有單引號（'），否則做如下變更]

$qurym = "SELECT * FROM referal_member 
      WHERE `ref_cusid` = ' " . $id . "' 
      AND `confirm` = '1'"; 

$resm = mysqli_query($con,$qurym); 
// check query returns something or not 
if(mysqli_num_rows($resm)){ 
// fetch data 
    $rowm=mysqli_fetch_array($resm); 
} 
else{ 
// no data found 
} 

Answer 2

如果是整型數據類型，請從確認中刪除引號。

$qurym="SELECT * FROM referal_member WHERE `ref_cusid` = '$id' && `confirm` = 1"; 
$resm=mysqli_query($con,$qurym); 
$rowm=mysqli_fetch_array($resm); 
print_r($rowm); // to check results. 

或者如果ref_cusid也是整型。

$qurym="SELECT * FROM referal_member WHERE `ref_cusid` = $id && `confirm` = 1"; 
$resm=mysqli_query($con,$qurym); 
$rowm=mysqli_fetch_array($resm); 
print_r($rowm); // to check results.