Question

当我尝试更新数据库中的表时，使用此代码我收到mysqli_error（）正好需要1个参数，错误地给出了0。

我不确定自己哪里出错了，因为我已经查看过这些字段以确认其长度。

<?php
// Establish a connection to the database
$con =  mysqli_connect("localhost", "root", "password");

// Check if there were any issues connecting to the database
if ( mysqli_connect_errno() != 0)
{
    die("An error occurred trying to establish a connection: " .mysqli_connect_error());
}

// Select the  database to work with.
$db  =  mysqli_select_db ($con, "SalesDB");

// Specify the query to run
$sqlString = "UPDATE Customer SET Cust_Fname ='".$_POST["FirstName"] ."', Cust_Lname='" .$_POST["LastName"] ."', Cust_Pasword='" .$_POST["Password"] ."', Cust_email='" .$_POST["Email"] ."', Cust_address='" .$_POST["Address"]  ."', Cust_city='" .$_POST["City"]  ."', Cust_state='" .$_POST["State"] ."', Cust_zip='" .$_POST["Zip"] ."', Locked_flag='" .$_POST["Lflag"]."', Emp_ID='" .$_POST["EmpID"]."' WHERE Cust_ID = '".$_POST["ID"] ."'";

// Run the query using the connection established earlier and assign the results to a results object.
$rs = mysqli_query ($con, $sqlString);


// Check if update was successful by checking whether the result returned is True
if ( !$rs)
{
    $result =  "There Was an Error Updating the Record".mysqli_error();
}
else
{
    $result =  "Record Updated Successfully";
}

// Close the connection
mysqli_close($con);
?>

Answer 1

您必须发送连接到mysqli_error

试试这个

if ( !$rs)
{
    $result =  "There Was an Error Updating the Record".mysqli_error($con);
}
else
{
    $result =  "Record Updated Successfully";
}

Answer 2

阅读您的错误代码。

在

// Check if update was successful by checking whether the result returned is True
if ( !$rs)
{
    $result =  "There Was an Error Updating the Record".mysqli_error();
}
else
{
    $result =  "Record Updated Successfully";
}

mysqli_error()期望连接的参数。传递给$con

mysqli_error（）正好需要1个参数，给出0

2 个答案: