mysqli_error()期望恰好有1个参数,第63行给出0 [英] mysqli_error() expects exactly 1 parameter, 0 given in line 63
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问题描述
我在php中的代码有问题...
I'm having an issues with my code in php...
mysqli_error()期望恰好有1个参数,第63行给出0.我得到了8个这样的错误.
mysqli_error() expects exactly 1 parameter, 0 given in on line 63. Im getting 8 of these errors.
这是第63行:--die('Error:'.mysqli_error());-
Here is line 63: --die('Error: ' . mysqli_error());--
这是它的功能:
if($email_exists == false && $pass1 == $pass2)
{
$sql="INSERT INTO users(u_Fname, u_Lname, u_email, u_Password)
VALUES('$Fname', '$Lname', '$EmailAddress', '$Password')";
if(!mysqli_query($link, $sql))
{
#$query = mysqli_query($myConnection, $sqlCommand);
#die (mysqli_error($myConnection));
die('Error: ' . mysqli_error());
}
header("location: login.php");
}
echo "<hr>";
如何解决此问题?
编辑完整脚本
<?php
session_start();
require_once('connect.php');
require_once "utils.php";
if(isset($_POST['submit']))
{
//CHECK EMPTY FORM DATA and SANITIZE
if(!empty($_POST['Username'])&&
!empty($_POST['Fname'])&&
!empty($_POST['Lname'])&&
!empty($_POST['EmailAddress'])&&
!empty($_POST['Password'])&&
!empty($_POST['Password1']))
{
$Username = mysqli_real_escape_string($link,htmlentities($_POST['Username']));
$Fname = mysqli_real_escape_string($link,htmlentities($_POST['Fname']));
$Lname = mysqli_real_escape_string($link,htmlentities($_POST['Lname']));
$Password = mysqli_real_escape_string($link,htmlentities($_POST['Password']));
$Password1 = mysqli_real_escape_string($link,htmlentities($_POST['Password1']));
$EmailAddress = mysqli_real_escape_string($link,htmlentities($_POST['EmailAddress']));
}
else { header("Location: register.php");exit(); }
if(isset($EmailAddress) && !empty($EmailAddress) &&
isset($Password) && !empty($Password))
{
$email_exists = false;
$sql="SELECT * FROM users WHERE u_EmailAddress='$EmailAddress'";
if($result = mysqli_query($link, $sql))
{
while($record = mysqli_fetch_row($result))
{
if($record = $EmailAddress)
{
$email_exists = true;
}
if($email_exists = true)
{
echo "<div id='reg_error1'>* The Email Used Already Exists! <br></div>";
}
}
}
if($Password != $Password1)
{
echo "<div id='reg_error2'>* Your Password Does Not Match!<br></div>";
}
if($email_exists == false && $Password == $Password1)
{
$sql="INSERT INTO users(u_Username, u_Fname, u_Lname, u_EmailAddress, u_Password)
VALUES('$_POST[$Username]', '$_POST[$Fname]', '$_POST[$Lname]', '$_POST[$EmailAddress]', '$_POST[$Password]')";
echo "<pre>"; print_r($sql); exit;
if(!mysqli_query($link, $sql))
{
#$query = mysqli_query($myConnection, $sqlCommand);
#die (mysqli_error($myConnection));
die('Error: ' . mysqli_error($link));
}
header("location: login.php");
}
echo "<hr>";
}
}
推荐答案
尝试一下.
mysqli_error($ con);
mysqli_error($con);
或用变量保存连接句柄的$ con替换
or replace $con with which ever variable holds the connection handle
(将$ con替换为$ link)
(edit: replace $con with $link in your case)
"mysqli_error()函数返回最近一次函数调用的最后一个错误描述,如果有的话."
"The mysqli_error() function returns the last error description for the most recent function call, if any."
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