Question

首先，我想说谢谢你让我再次提问。我知道我之前的问题是知识水平有点低。今天，我想问一下在ajax中将mysql转换为mysqli的原理是否与html相同。假设这是我的Connect.php

<?php
$host = "localhost";
$dbusername = "root";
$dbpassword = "765632";
$dbname = "student";
$link_id = mysqli_connect($host,$dbusername,$dbpassword,$dbname) or die("Error " . mysqli_error($link_id)); 
?>

我的ajax.php是

<?php
  //Connect to MySQL Server

include 'Connect.php';
mysql_connect($host, $dbusername, $dbpassword);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
 // Escape User Input to help prevent SQL Injection
$first_name = mysql_real_escape_string(trim($_GET['first_name']));
// Retrieve data from Query  
$query = "SELECT  student_id, LRN, first_name, last_name, grade, section  FROM student_information WHERE first_name LIKE '%{$first_name}%'";     
$result = mysql_query($query) or die(mysql_error());
 //Generate the output
$searchResults = '';
if(!mysql_num_rows($result))

我应该在不改变其逻辑方案的情况下将其转换为mysqli有什么变化。

Answer 1

你的意思是？

$link_id = mysqli_connect($host, $dbusername, $dbpassword);
//Select Database
mysqli_select_db($link_id, $dbname) or die(mysqli_error($link_id));
// Escape User Input to help prevent SQL Injection
$first_name = mysqli_real_escape_string($link_id, trim($_GET['first_name']));
// Retrieve data from Query
$query = "SELECT  student_id, LRN, first_name, last_name, grade, section  FROM student_information WHERE first_name LIKE '%{$first_name}%'";
$result = mysqli_query($link_id, $query) or die(mysqli_error($link_id));
//Generate the output
$searchResults = '';
if(!mysqli_num_rows($result))

在ajax环境中将mysql迁移到mysqli

1 个答案: