Question

我一直收到这些错误：

警告：mysqli_query（）期望参数1为mysqli，null给定   在/home/u590953899/public_html/app/signup/index.php第95行

警告：mysqli_error（）正好需要1个参数，给出0   第95行/home/u590953899/public_html/app/signup/index.php

第95行是：

 $result1 = mysqli_query($mysqli,$sql1) or die(mysqli_error());

整个代码：

<?php
session_start();
$mysqli1=mysqli_connect('localhost','u590953899_staff','DCr102404$','u590953899_dev') or die("Database Error");
if(isset($_POST['submit']))
{
 //whether the username is blank
 if($_POST['username'] == '')
 {
  $_SESSION['error']['username'] = "User Name is required.";
 }
else
 {
   //if it has the correct format whether the email has already exist
   $username= $_POST['username'];
   $sql1 = "SELECT * FROM user WHERE username = '$username'";
   $result1 = mysqli_query($mysqli,$sql1) or die(mysqli_error());
   if (mysqli_num_rows($result1) > 0)
            {
    $_SESSION['error']['username'] = "This username is already taken.";
   }
}
 //whether the email is blank
 if($_POST['email'] == '')
 {
  $_SESSION['error']['email'] = "E-mail is required.";
 }
 else
 {
  if(preg_match("/^([a-zA-Z0-9])+([a-zA-Z0-9._-])*@([a-zA-Z0-9_-])+([a-zA-Z0-9._-]+)+$/", $_POST['email']))
  {
   //if it has the correct format whether the email has already exist
   $email= $_POST['email'];
   $sql1 = "SELECT * FROM user WHERE email = '$email'";
   $result1 = mysqli_query($mysqli,$sql1) or die(mysqli_error());
   if (mysqli_num_rows($result1) > 0)
            {
    $_SESSION['error']['email'] = "This Email is already used.";
   }
  }
  else
  {
   //this error will set if the email format is not correct
   $_SESSION['error']['email'] = "Your email is not valid.";
  }
 }
 //whether the password is blank
 if($_POST['password'] == '')
 {
  $_SESSION['error']['password'] = "Password is required.";
 }
 //if the error exist, we will go to registration form
 if(isset($_SESSION['error']))
 {
  header("Location: http://app.bithumor.co/signup/");
  exit;
 }
 else
 {
   $username = $_POST['username'];
  $email = $_POST['email'];
  $password = $_POST['password'];
  $com_code = md5(uniqid(rand()));
$ip = $_SERVER['REMOTE_ADDR'];

  $sql2 = "INSERT INTO user (username, email, password, com_code) VALUES ('$username', '$email', '$password', '$com_code')";
  $result2 = mysqli_query($mysqli1,$sql2) or die(mysqli_error());

我到处寻找解释/解决方案，但没有任何工作。

Answer 1

看起来您在提交的代码中第3行命名了MySQLi连接“$ mysqli1”，但是您使用名为“$ mysqli”的变量作为mysqli_query函数中的第一个参数。

Answer 2

我想通了......

在说：

的行上

if ([entry[@"isInactive"] boolValue]) { // Original dispatch_async(dispatch_get_main_queue(), ^{ self.profileIconShadow.layer.shadowColor = [UIColor redColor].CGColor; [self.profileIconShadow.layer setNeedsDisplay]; } }

应该是$result1 = mysqli_query($mysqli,$sql1) or die(mysqli_error());而不是$mysqli1，因为$mysqli连接不存在

双mysqli_query（）期望参数1为mysqli，null给出警告

2 个答案: