PHP代码中的错误 [英] Errors In PHP code

问题描述

我是第一次进行PHP编码.我遇到以下错误:

I am doing PHP coding for first time. I got the following errors:

错误:

警告:mysqli_real_escape_string()期望参数1为mysqli， 第9行的C:\ xampp \ htdocs \ 331002.php中给出的布尔值

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 9

警告:mysqli_real_escape_string()期望参数1为mysqli， 在第10行的C:\ xampp \ htdocs \ 331002.php中给出的布尔值

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 10

警告:mysqli_real_escape_string()期望参数1为mysqli， 在第11行的C:\ xampp \ htdocs \ 331002.php中给出的布尔值

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 11

注意:未定义的索引:C:\ xampp \ htdocs \ 331002.php中的empID 12

Notice: Undefined index: empID in C:\xampp\htdocs\331002.php on line 12

警告:mysqli_real_escape_string()期望参数1为mysqli， 在第12行的C:\ xampp \ htdocs \ 331002.php中给出的布尔值

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 12

警告:mysqli_query()期望参数1为mysqli，布尔值 在第17行的C:\ xampp \ htdocs \ 331002.php中给出

Warning: mysqli_query() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 17

警告:mysqli_error()期望参数1为mysqli，布尔值 在第18行的C:\ xampp \ htdocs \ 331002.php中给出

Warning: mysqli_error() expects parameter 1 to be mysqli, boolean given in C:\xampp\htdocs\331002.php on line 18

这是我的代码:

<?php
$con=mysqli_connect("localhost","root","root","student");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
$FName = mysqli_real_escape_string($con, $_POST['firstName']);
$LName = mysqli_real_escape_string($con, $_POST['lastName']);
$Salary = mysqli_real_escape_string($con, $_POST['salary']);
$ID = mysqli_real_escape_string($con, $_POST['empID']);

$sql="INSERT INTO PersonInfo (FName, LName, empID, Salary)
VALUES ('$FName', '$LName','$ID','$Salary')";

if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}
echo "1 record added";

mysqli_close($con);
?>

推荐答案

通常的检查连接"模式如下:

The usual "check connection" pattern is the following:

<?php
$con=mysqli_connect("localhost","root","root","student");
// Check connection
if (false === $con) {
  // die will "finish" the script
  die("Failed to connect to MySQL: " . mysqli_connect_error());
}

mysqli_real_escape_string需要有效的连接，而您似乎没有.

mysqli_real_escape_string needs a valid connection, which you don't seem to have.

这篇关于PHP代码中的错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！