从数据库中获取和显示数据（JavaScript XML HTTP请求）

Question

I'm really struggling with this task for my course and hope someone doesn't mind helping out or just offering guidance here. Basically I'm trying to create a simple Javascript XML Http Request to display basic information (the country_name & country_capital fields) from the database just in the html page. Below I just describe the apparent stages from the guide, and what I have done.

Firstly the 'database.html' page contains javascript XHR code which I think is mostly correct, but may have an error. To be honest I'm not 100% sure what else it does other than somehow refer to the getcountries.php file.

Secondly the getcountries.php file is where I'm really struggling as I've never coded in PHP. I think it's supposed to fetch the data from the local server (I'm running XAMPP) and echo the results on the web page.

The database on phpMyAdmin is simple with just a table of countries including a primary key ID number, the country name, capital and currency, with the details below: Database name = countries_db Table name = countries_table Table fields: country_ID (primary key) country_name country_capital country_currency An example entry: 2, USA, Washington DC, US Dollar

To summarise, my question is this: how can I edit what I've done to correctly fetch the data from the database and display it on the page?

Really appreciate any help or advice here, thanks a lot.

 <!-- Code on Page 1 (database.html) --> <p id="txtHint"></p> <p id="hint"></p> <script> function showUser(str) { if (str=="") { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) { // detects whether the browser has XMLHttpRequest functionality // code for modern browsers xmlhttp=new XMLHttpRequest(); // creates an XMLHttpRequest object } else { // code for old browsers xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { // onreadystatechange defines the function to be called when the readyState property changes if (this.readyState==4 && this.status==200) { document.getElementById("hint").innerHTML=this.responseText; } } xmlhttp.open("GET","getcountries.php?q="+str,true); xmlhttp.send(); } </script> <!-- Code on Page 2 (getcountries.php) --> <?php $q = intval($_GET['q']); $con = mysqli_connect('localhost','root',''); if (!$con) { die('Could not connect: ' .mysqli_error($con)); } mysqli_select-db($con,"countries_db"); $sql="SELECT country_name AND country_capital FROM records"; $result = mysqli_query($con,$sql); echo "Results:" error_reporting(E_ERROR | E_PARSE); \\ while($row = mysqli_fetch_array($result)) { echo $row['country_name'] . "<br>"; echo $row['country_capital'] . "<br>"; } mysqli_close($con); ?> 

Answer 1

Assuming that this is the structure of your data base:

Database name = countries_db 
Table name = countries_table 
Table fields: 
    country_ID (primary key) 
    country_name 
    country_capital 
    country_currency 

The problem is that you have some syntax error in your code change this lines:

mysqli_select-db($con,"countries_db");
$sql="SELECT country_name AND country_capital FROM records";

with:

mysqli_select_db($con,"countries_db");
$sql="SELECT country_name, country_capital FROM countries_table";

Alternative : using PDO:

Try this instead of your getcountries.php implementation

<?php
$driver = 'mysql';
$database = "dbname=countries_db";
$dsn = "$driver:host=localhost;unix_socket=/home/cg/mysql/mysql.sock;$database";

$username = 'root';
$password = 'root';

try {
   $conn = new PDO($dsn, $username, $password);
   echo "<h2>Database countries_db Connected<h2>";
}catch(PDOException $e){
   echo "<h1>" . $e->getMessage() . "</h1>";
}
$sql = 'SELECT country_name, country_capital FROM countries_table';
$stmt = $conn->prepare($sql);
$stmt->execute();

echo "Results:";
echo "<table style='width:100%'>";
while($row = $stmt->fetch(PDO::FETCH_ASSOC)){
  echo "<tr>";
  foreach($row as $value)
  {
    echo sprintf("<td>%s</td>", $value);
  }
  echo "</tr>";
}
echo "</table>";
?>

Answer 2

在您的getcountries.php中使用mysqli_select_db代替mysqli_select-db ：

mysqli_select_db($con,"countries_db");