警告：mysqli_error（）只需要1个参数，给定0 [英] Warning: mysqli_error() expects exactly 1 parameter, 0 given

问题描述

请帮助我查看下面的代码

 <？php  
 include（  includes / dbconnet.php）; 
 $ name = $ _ POST ['  name']; 
 $ email = $ _ POST ['  email']; 
 $ telephone = $ _ POST ['  telephone']; 
 $ subject = $ _ POST ['  subject']; 
 $ mesg = $ _ POST ['  mesg']; 
 $ sql =   INSERT INTO APPL SET name ='$ name'，
 email =' $ email'，
电话='$电话'，
 subject ='$ subject'，
 mesg ='$ mesg'; 
 $ result = mysqli_query（$ conn，$ sql）或 die（mysqli_error（））; 
 if（$ result == true）
 {
 header（'  location： ？的index.php成功）; 
 
} 
  else  
 {
  echo   错误; 
} 
 ？>  
 
 <？ php 
  $ conn  = mysqli_connect（'  localhost'，'  root'，'  '）或 die（mysqli_error（））; 
  / *  选择数据库* /  
  $ select_db  = mysqli_select_db（$ conn，'  practice'）或 die（mysqli_error（））; 
 ？>  

我得到的错误警告：mysqli_error（）预计只有1个参数，



我尝试了什么：



我试图对其进行故障排除并查看我的数据库连接，但我无法得到它

请帮助解决错误

解决方案

name =

_POST [' name'];

电子邮件=

Please kindly help me to look into the code bellow

<?php 
include("includes/dbconnet.php");
$name=$_POST['name'];
$email=$_POST['email'];
$telephone=$_POST['telephone'];
$subject=$_POST['subject'];
$mesg=$_POST['mesg'];
$sql="INSERT INTO pract SET name='$name',
email='$email',
telephone='$telephone',
subject='$subject',
mesg='$mesg'";
$result=mysqli_query($conn,$sql) or die (mysqli_error());
if($result==true)
{
header('location:index.php?success');

}
else
{
echo "Error";
}
?>

<?php 
$conn = mysqli_connect('localhost','root','') or die(mysqli_error());
/*Select Database*/
$select_db = mysqli_select_db($conn,'practise') or die(mysqli_error());
?>

The error I get "Warning: mysqli_error() expects exactly 1 parameter,

What I have tried:

I tried to troubleshoot it and look into my database connection but I couldn't get it
Please kindly help to resolved the error

解决方案

name=

_POST['name'];

email=

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