PHP

Question

我在执行代码时遇到问题...我不知道为什么它不起作用...我想显示数据库标题和许多其他按 ID...文件名是 view.php...如果我想显示 id 1 然后我输入 url view.php?id=1 或 id 2 view.php?id=2...here是我的代码吗，请帮帮我...

<?php
$getid = $_GET["id"];

$dbuser = 'user';
$userpass = 'pass';
$database = 'db';

error_reporting(0);
$con = mysql_connect("localhost",$dbuser,$userpass);
if(!$con){die('Could not connect:'.mysql_error());}
mysql_select_db($database, $con);
mysql_query("SET NAMES utf8");
date_default_timezone_set('Europe/Athens');


$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
if(mysql_num_rows($query)){
    $result = mysql_fetch_assoc($query);
    $id = $result['id'];
    $title = $result['title'];
    $description = $result['description'];
    echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
}else{
    echo 'Error';
    exit;
}
?>

Answer 1

改变这个：

if(mysql_num_rows($query)){

对此：

if(mysql_num_rows($query) >= 1){

请记住，您的数据库中必须包含您提供的 ID，
否则它会再次转到其他地方。

(mysql_num_rows returns FALSE 或带有行数的整数，并且由于 if 语句不检查整数而是检查 true，它总是沿 ELSE 路径向下移动)

Source: PHP.net

Answer 2

<?php
$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
$count = mysql_num_rows($query);
if($count != 0){
    while($result = mysql_fetch_array($query)){
        $id = $result['id'];
        $title = $result['title'];
        $description = $result['description'];
        echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
    }
}else{
    echo 'Error';
}
?>

PHP - MYSQL - 按来自一个 php 文件的 id 显示页面

PHP - MYSQL - Display Pages by id from One php file