mysql 中每一行的页面 [英] page for each row in mysql

问题描述

我是 php 的新手，为了为所有文章制作一个独特的页面，我已经努力了几个小时.无论我做什么都行不通.

我知道我应该将 mysql 转换为 mysqli，但首先我只是想让它工作.

当我转到 url/video.php?id=3 时，它只显示一个空白页面，它不会从文章表中打印出名称"

有什么建议吗?

video.php

我从 index.php 链接到以下内容

 <a href="/video.php?id=<? echo $row[id]; ?>

哪个有效.

解决方案

我看到的另一件事，但我假设如果你看到了:

• 您是否已成功连接到 MySQL 服务器?
• 您是否选择了要使用的数据库?
• 您知道 PHP 中的 mysql_result 函数吗?和
• 您知道 MySQL 中的 LIMIT 吗?

试试:

MySQL:

/*** 连接并定义您的连接变量*/$connection = mysql_connect("localhost","root","");/*** 选择您的数据库*/mysql_select_db("mydatabase",$connection);/*** 检查 $_GET 中是否存在id"*/if( array_key_exists( "id" , $_GET ) ){$id = mysql_real_escape_string( $_GET [ "id" ],$connection );/*** 记住使用 LIMIT*/$source = mysql_query("SELECT `name` FROM `article` WHERE `id` = '$id' LIMIT 1",$connection);/*** 如果您只需要打印名称*/if(mysql_num_rows($source)){//只打印名字回声 mysql_result( $source );}/*** 但如果你需要检索数组*/$row = mysql_fetch_assoc( $source );回声 $row["name"];}

MySQLi:

/*** 连接并定义您的连接变量*/$connection = mysqli_connect("localhost","root","");/*** 检查是否已连接[ http://us2.php.net/manual/en/mysqli.connect-errno.php ]*/如果(mysqli_connect_errno() == 0){/*** 选择您的数据库*/mysqli_select_db( $connection , "mydatabase" );/*** 检查 $_GET 中是否存在id"*/if( array_key_exists( "id" , $_GET ) ){$id = mysqli_real_escape_string($connection,$_GET ["id"]);/*** 记住使用 LIMIT*/$source = mysqli_query($connection,"SELECT `name` FROM `article` WHERE `id` = '$id' LIMIT 1");/*** 如果您只需要打印名称*/if(mysqli_num_rows($source)){//只打印名字$row = mysqli_fetch_array( $source );回声 $row[0];}/*** 但如果你需要检索数组*/$row = mysqli_fetch_assoc( $source );回声 $row["name"];}}

并且，阅读 MySQL 上的 LIMIT>

祝你好运！

i'm new to php and i've been struggling for several hours with making a unique page for all the articles. whatever i do it wont work.

i know i should convert the mysql to mysqli, but to begin with i just wanna make this work.

when i go to url/video.php?id=3 it just show a blank page it do not print "name" from the article table out

any suggestions?

video.php

<?php 

include "connect.php"; 


$id = mysql_real_escape_string($_GET['id']);
$query = "SELECT `name` FROM `article` WHERE `id` = '.$id.'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);

// Echo page content
echo $row['name'];
?>

im linking from index.php with following

 <a href="/video.php?id=<? echo $row[id]; ?> 

which works.

解决方案

Another thing that I see, but I assume that if you did:

• Have you managed to connect to the MySQL server?
• Did you select the database to use?
• you know the function mysql_result in PHP? And
• you know LIMIT in MySQL?

Try:

MySQL:

/**
* Connect and define your connection var
*/
$connection = mysql_connect("localhost","root","");
/**
*   Select your database
*/
mysql_select_db("mydatabase",$connection);

/**
* Check if exists "id" in $_GET
*/
if( array_key_exists( "id" , $_GET ) )
{
    $id = mysql_real_escape_string( $_GET [ "id" ],$connection );
    /**
    *   Remember use LIMIT
    */
    $source = mysql_query("SELECT `name` FROM `article` WHERE `id` = '$id' LIMIT 1",$connection);

    /**
    * if you need print only name
    */
    if( mysql_num_rows( $source ) )
    {
        // print only name
        echo mysql_result( $source );
    }
    /**
    * but if you need retrive array
    */
    $row = mysql_fetch_assoc( $source );
    echo $row["name"];
}

MySQLi:

/**
* Connect and define your connection var
*/
$connection = mysqli_connect("localhost","root","");
/**
* Check if is connected[ http://us2.php.net/manual/en/mysqli.connect-errno.php ]
*/
if( mysqli_connect_errno() == 0 )
{
    /**
    *   Select your database
    */
    mysqli_select_db( $connection , "mydatabase" );

    /**
    * Check if exists "id" in $_GET
    */
    if( array_key_exists( "id" , $_GET ) )
    {
        $id = mysqli_real_escape_string($connection,$_GET [ "id" ]);
        /**
        *   Remember use LIMIT
        */
        $source = mysqli_query($connection,"SELECT `name` FROM `article` WHERE `id` = '$id' LIMIT 1");

        /**
        * if you need print only name
        */
        if( mysqli_num_rows( $source ) )
        {
            // print only name
            $row = mysqli_fetch_array( $source );
            echo $row[0];
        }
        /**
        * but if you need retrive array
        */
        $row = mysqli_fetch_assoc( $source );
        echo $row["name"];
    }

}

And, read bout the LIMIT on MySQL

Good luck!

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