php - MySQL Query 双重查询

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Reference - What does this error mean in PHP? (38 个回答)

What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such (1 个回答)

已关闭 5 年前。

我尝试在 tbl_ordetails 中插入数据，然后更新 tbl_cart 中的数据，但是当我尝试插入数据时，结果总是“失败”，我不知道出了什么问题。请帮助我

<?php
include ("connection.php");
if(isset($_POST['btnSubmit']))
{




$fullname=$_POST['fullname'];
$address=$_POST['address'];
$phone_number=$_POST['phone_number'];
$city=$_POST['city'];
$customer=$_POST['customer'];

$query = "INSERT INTO tbl_orderdetails (fullname, address, phone_number, city, customer) VALUES ('$fullname, $address, $phone_number, $city, $customer')";
$query1 = "UPDATE tbl_cart SET status ='Ordered' WHERE customer=['$customer']";

 $result = mysqli_query($conn, $query);

   if(($conn->query($query1) === TRUE) && ($result->num_rows > 0)){
            echo "success";
            exit;

    }
    else{
         echo "failed";
         exit;

    }


}
?>

提前致谢

Answer 1

您在...中缺少的引用负载

$query = "INSERT INTO tbl_orderdetails (fullname, address, phone_number, city, customer) 
      VALUES ('$fullname, $address, $phone_number, $city, $customer')";

它获取所有值并构建 1 个值。应该是

$query = "INSERT INTO tbl_orderdetails (fullname, address, phone_number, city, customer) 
      VALUES ('$fullname', '$address', '$phone_number', '$city', '$customer')";

您的更新应该是(减[])

$query1 = "UPDATE tbl_cart SET status ='Ordered' WHERE customer='$customer'";

更好的是使用准备好的语句并绑定(bind)变量。

更新:

   $result = $conn->query($query);
   if ( $result === false ) {
       echo "error:".$conn->error;
       exit;
   }

   if($conn->query($query1) === TRUE){
            echo "success";
            exit;

    }
    else{
         echo "failed:".$conn->error;
         exit;

    }