另一个

时间：2015-11-22 09:29:46

标签： php mysql

如何链接两个mysql表。 一个（已经）显示客户记录 还有一个专门针对每个记录的评论系统的表格。？

即 帐户＃1 //显示特定的客户信息.. 我的需要是...... 我需要一个系统来发布关于该记录的评论（帐户＃1）..等等以及所有未来的记录。

我有两张桌子 客户表和 评论表......

结论。

如何链接它们以便我可以单独评论每个客户表记录？

非常感谢

这是我的代码...... 

<?php
include('../includes/mysql_connect.php');

$query = "CREATE TABLE `customers` (

`id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,

`cust_name` VARCHAR(50) NOT NULL,

PRIMARY KEY (`id`)

)engine=innodb";
if(@mysql_query($query,$dbc)){
echo '<p>Table for customers has been successfully created!</p><br/>';
} else {
echo '' . mysql_error($dbc)  .'<br/>';
}
mysql_close($dbc);
?>

<?php 
include('../includes/mysql_connect.php');
$query = "CREATE TABLE `comments` (

`id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,

`customer` INT(10) UNSIGNED NOT NULL,

`comment` VARCHAR(255)NOT NULL,

PRIMARY KEY  (`id`),

INDEX `customer` (`customer`)

)engine=innodb";

if(@mysql_query($query,$dbc)){
echo '<p>Table for comments has been successfully created!</p><br/>';
} else {
echo '' . mysql_error($dbc)  .'<br/>';
}
mysql_close($dbc);
?>
/*  ABOVE CODE CREATES THE TABLES */

/* CODE BELOW INSERTS*/

    <?php


if($_SERVER['REQUEST_METHOD'] == 'POST'){
// NEEDS DATABAE CONNECTION
include('includes/mysql_connect.php');
//CREATE VARIBLE IF PROBLEM DOES NOT OCCUR 
$problem =  FALSE;
// VALIDATION
if(!empty($_POST['cust_name'])){
// REAL ESCAPE STRING AGAINST SQL INJECTION
$cust_name = mysql_real_escape_string(trim(strip_tags($_POST['cust_name'])), $dbc);
} else { // IF PROBLEM
echo '<p style="color:red;">Please enter name!</p>';
$problem = TRUE;
}
// IF NO PROBLEM
if(!$problem){
//RUN QUERY
$query = "INSERT INTO `customers` (`cust_name`) values ('$cust_name')";
// EXECUTE QUERY
if(@mysql_query($query, $dbc)){ // NO PROBLEM
echo '<p style="color:blue;">This name has been added!<p>';
} else {// IF PROBLEM
echo '<p style="color:red;">This name was not successfully added!</p>';
}
} // END OF VARIABLE IN NO PROBLEM

mysql_close($dbc); // CLOSING CONNECTION

} // END OF MAIN IF

?>

/* CODE BELOW RECEIVES */

<?php
include('includes/mysql_connect.php');

//$query = 'SELECT * FROM customers';

$query = 'SELECT * FROM customers';
if($y = mysql_query($query,$dbc)){
while($row = mysql_fetch_array($y)){
//echo " {$row['id']} <br/>";


echo "<div id='container'>
<div id='first_wrapper'>
<a style='text-decoration:none'; href=\"fetch_join.php?id={$row['id']}\">
Stolve # " . $row['user_id'] ."
</a></div>";

}
}

mysql_close($dbc);
?>

/* CODE BELOW SELECTS */  

<?PHP 

include('includes/mysql_connect.php');
 /* Select records for specific user based upon their ID */
$query = "SELECT c.`id` as 'customer_id', ct.`id` as 'comment_id', c.`cust_name` as 'customer', ct.`comment` from `customers` c
    inner join `comments` ct on ct.`customer`=c.`id`
     where c.`id`=1";

if($y = mysql_query($query,$dbc)){
while($row = mysql_fetch_array($y)){

echo "{$row['customer']} {$row['comment']}<br/>";

//echo " {$row['id']} <br/>";

}
}

mysql_close($dbc);  

?>

但无论我输入多少记录，我都会继续获取所有记录 而不只是特定ID的那个。

在我输入的每条记录下方，但用户ID仍显示为一个。

    1 This is a test for joining tables
    1 this is a second test
    1 third test
    1 another comment here
    1 fourth test
    1 last test 4

我应该只能看到每次点击一条记录。

2 个答案:

答案 0 :(得分：1)

enter image description here

这是我对你的MySQL表的解释。从我所看到的，您正在尝试建立一个系统，您可以轻松地点击客户并撰写/查看有关它们的书面评论。此查询将分别从“数据库= 1”的客户的数据库中获取评论以及客户用户名。

SELECT comments.comment, users.username FROM comments, users WHERE comments.customerid = users.customerid AND users.customerid = 1

由于提供的信息量很大，这是一个非常通用的答案。如果您有任何问题或疑虑，请发表评论！

答案 1 :(得分：0)

回答您的问题"do both tables have to be created at the same time so their Id's start both from 1?"〜否。

下面的sql创建了两个表 - comments表有一个键控字段（customer），它链接到id表中的customers字段。

create table `customers` (
    `id` int(10) unsigned not null auto_increment,
    `cust_name` varchar(50) not null,
    primary key (`id`)
)engine=innodb;

create table `comments` (
    `id` int(10) unsigned not null auto_increment,
    `customer` int(10) unsigned not null,
    `comment` varchar(255) not null,
    primary key (`id`),
    index `customer` (`customer`)
)engine=innodb;

/* Create a dummy / test user in the customers table */
insert into `customers` (`cust_name`) values ('Bob');

/* Insert a dummy / test record in the comments table */
insert into `comments` (`customer`,`comment`) values (1,'Bob is the best customer in the World, ever! However, he rarely pays on time and has bad body odour');

/* Select ALL records from both tables using the customer id as a key */
select c.`id` as 'customer_id', ct.`id` as 'comment_id', c.`cust_name` as 'customer', ct.`comment` from `customers` c
    inner join `comments` ct on ct.`customer`=c.`id`;

/* 
   Note the JOIN type is `inner` - this shows only records where there is
   a value in both. If the JOIN type was `left outer` there would be records 
   for all regardless of whether or not a comment exists for that customer
*/

 /* Select records for specific user based upon their ID */
select c.`id` as 'customer_id', ct.`id` as 'comment_id', c.`cust_name` as 'customer', ct.`comment` from `customers` c
    inner join `comments` ct on ct.`customer`=c.`id`
     where c.`id`=1;

@AndyChavarria：sql旨在说明如何将评论链接到客户，而不打算运行最终版本！也就是说，我稍微整理了一下php并进行了一些小修改......如果有关于sql注入的评论或关于使用已弃用的mysql_*函数系列，请不要感到惊讶。如果这还处于生产/构建的早期阶段，那么你真的应该考虑实施mysqli或PDO〜mysqli需要更少的更改但是有一些微妙的变化语法差异。从长远来看，PDO可能更强大，而且肯定更灵活 - 但是，如果正确实现，这两种方法都会阻止SQL注入，而你的代码即使在转义时仍然容易受到攻击。

例如，mysql和amp;之间的许多细微差别之一mysqli的：

mysqli_query( $conn, $sql )而不是mysql( $sql, $conn ) 

<?php
    include('../includes/mysql_connect.php');

    /* To build the tables: this should only need to be run once! */
    $query = "CREATE TABLE `customers` (
                `id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
                `cust_name` VARCHAR(50) NOT NULL,
                PRIMARY KEY (`id`)
                )engine=innodb";
    $res=mysql_query( $query, $dbc );
    echo $res ? '<p>Table for customers has been successfully created!</p><br/>' : mysql_error( $dbc )  .'<br/>';

    $query = "CREATE TABLE `comments` (
                `id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
                `customer` INT(10) UNSIGNED NOT NULL,
                `comment` VARCHAR(255)NOT NULL,
                PRIMARY KEY  (`id`),
                INDEX `customer` (`customer`)
                )engine=innodb";
    $res=mysql_query( $query, $dbc );
    echo $res ? '<p>Table for comments has been successfully created!</p><br/>' : mysql_error($dbc)  .'<br/>';
    mysql_close( $dbc );
?>



<?php
    /* Insert comment */
    if( $_SERVER['REQUEST_METHOD'] == 'POST' ){

        include('includes/mysql_connect.php');

        $problem =  FALSE;

        if( isset( $_POST['cust_name'] ) && !empty( $_POST['cust_name'] ) ){
            $cust_name = mysql_real_escape_string( trim( strip_tags( $_POST['cust_name'] ) ), $dbc );
        } else {
            echo '<p style="color:red;">Please enter name!</p>';
            $problem = TRUE;
        }

        if( !$problem ){
            $query = "INSERT INTO `customers` (`cust_name`) values ('$cust_name')";
            $res=mysql_query( $query, $dbc );
            echo $res ? '<p style="color:blue;">This name has been added!<p>' : '<p style="color:red;">This name was not successfully added!</p>';
        }
        mysql_close($dbc);
    }
?>



<?php
    /* Fetch customers and create hyperlink */
    include('includes/mysql_connect.php');

    $query = 'SELECT * FROM `customers`';
    $res=mysql_query( $query, $dbc );
    if( $res ){

        /* Assumed that the container does not get repeated */
        echo "<div id='container'>";

        while( $row = mysql_fetch_array( $res ) ){

            /* Here you have the same id used for each iteration of the loop.... this is not valid html */
            echo "
                <div id='first_wrapper'>
                    <a style='text-decoration:none' href='fetch_join.php?id={$row['id']}'>Stolve # {$row['user_id']}</a>
                </div>";            
        }
        echo "</div>";
    }

    mysql_close($dbc);
?>



<?php 
    /* Show specific customer ( based upon id ) and associated comments */
    include('includes/mysql_connect.php');

    $id=isset( $_GET['id'] ) ? intval( mysql_real_escape_string( trim( strip_tags( $_GET['id'] ) ), $dbc ) ) : 1;

    $query = "SELECT c.`id` as 'customer_id', ct.`id` as 'comment_id', c.`cust_name` as 'customer', ct.`comment` from `customers` c
    inner join `comments` ct on ct.`customer`=c.`id`
    where c.`id`='{$id}';";

    $res=mysql_query( $query, $dbc );
    if( $res ){
        while( $row = mysql_fetch_array( $res ) ){
            echo "{$row['customer']} {$row['comment']}<br/>";
        }
    } else {
        echo 'No comments for selected user';
    }
    mysql_close( $dbc );  
?>
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