如何使用php创建搜索结果详细信息页面

Question

我是一名药剂师，致力于一个项目，用户可以从我的数据库中获取药品信息。 我已经进入搜索结果页面，但是无法进入详细信息页面。 这是我的search.php页面：

<?php
$db_hostname = 'localhost';
$db_username = 'root';
$db_password = '';
$db_database = 'drug';

// Database Connection String
$con = mysql_connect($db_hostname,$db_username,$db_password);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($db_database, $con);
?>

<!DOCTYPE html>
<html lang="en">
    <head>
        <meta charset="utf-8" />
        <title></title>
    </head>
    <body>
<div>

  <?php
if (!empty($_REQUEST['term'])) {

$term = mysql_real_escape_string($_REQUEST['term']);     

$sql = "SELECT * FROM drugs WHERE GenericName LIKE '%".$term."%' OR     BrandName LIKE '%".$term."%' OR Pharmacologicalclass LIKE '%".$term."%' OR     ManufacturedBy LIKE '%".$term."%'"; 
$r_query = mysql_query($sql); 

while ($row = mysql_fetch_array($r_query)){ 
echo '<a href="results.php?id='.$row['GenericName'].'">        <h4>'.$row['BrandName'].'</h4></a>'; 
echo '<br />Generic Name: ' .$row['GenericName'];  

echo '<br /> Dosage Form: '.$row['Dosage form'];  
echo '<br /> Pharmacological Class: '.$row['Pharmacologicalclass'];  
echo '<br /> Indications: '.$row['Indications']; 
echo '<br /> Manufactured By: '.$row['ManufacturedBy'];   
}  

}
?>
</div>
    </body>
</html>

在详细信息页面上，我想包括GenericName，BrandName，Dosageform，Strength，适应症，相互作用，副作用，怀孕类别，药理学类别，作用方式和制造者（所有药物表中的行）。

Answer 1

<html>
    <head><title>results - details</title></head>
    <body>
        <?php
            /* 
                "mysql_*" - the old mysql functions are deprecated and their use is discouraged
                You would be wise to take time now to implement mysqli in conjunction with
                prepared statements to avoid sql injection attacks.

                That said, below semi-pseudo code might help you get started

            */

            if( $_SERVER['REQUEST_METHOD']=='POST' && !empty( $_GET['id'] ) ){

                $sql='select * from `drugs` where `id`='.$_GET['id'];
                $res=mysql_query( $sql, $con );

                if( $res ){
                    while( $rs=mysql_fetch_object( $res ) ){
                        echo $rs->GenericName, $rs->BrandName, $rs->Dosageform, $rs->Strength; /* etc */
                    }
                } else {
                    echo 'no results';
                }
                mysql_close( $con );
            }
        ?>
    </body>
</html>

Answer 2

我能够做到。 如果有人遇到同样的问题，这是我的代码。 Search.php

// Database Connection String
$con = mysql_connect($db_hostname,$db_username,$db_password);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($db_database, $con);
?>
 <form action="search.php" method="post">


<input type="text" class="form-control" name="term" placeholder="Search" /> 

<input type="submit" value="Submit" />  
</form> 
</div></div>
<div>

  <div class="container">
     <section class="col-xs-12 col-sm-6 col-md-12">

    <article class="search-result row">

  <?php
if (!empty($_REQUEST['term'])) {

$term = mysql_real_escape_string($_REQUEST['term']);     

$sql = "SELECT * FROM drugs WHERE GenericName LIKE '%".$term."%' OR     BrandName LIKE '%".$term."%' OR Pharmacologicalclass LIKE '%".$term."%' OR     ManufacturedBy LIKE '%".$term."%' "; 
?>

<div style="">
      Results for: <font color="blue" style="font:bold 20px 'Aleo';"><?php     echo     $term;?></font>
      </div>
      <?php $r_query = mysql_query($sql); 

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

echo '<a href="results.php?id='.$row['id'].'"><h4>'.$row['BrandName']. ":".     $row['GenericName']. " ".$row['Dosageform']. " ".$row['Strength'].'</h4></a>';     //links to results.php by id since id is unique    
echo ' '.$row['Pharmacologicalclass'];  
echo '<br /> uses: '.$row['Indications'];    
}  

}
?>

这是results.php

<?php
$connect = mysqli_connect("localhost","root","mypassword","drug");
?>
<?php
if(isset($_GET['id'])){ // checks if id was posted

$id = (int)$_GET['id'];

$query_fetch = mysqli_query($connect,"SELECT * FROM drugs WHERE ID = $id");    // id that was posted by search.php

 while($show = mysqli_fetch_array($query_fetch)){ 
  echo  " ".$show['BrandName']. ": &nbsp;" . $show['GenericName']. "      &nbsp;".$show['Strength']." <br></a>";

echo '<br /> <h3> <h3>Indications:</h3>'.$show['Indications'];  // these     rows are displayed to the user
echo '<br /> <h3> Pharmacological Class:</h3> '.$show['Precautions'];  
echo '<br /> <h3> Drug interactions:</h3> '.$show['Interactions']; 
echo '<br /> <h3> Undesirable effects:</h3> '.$show['SideEffects'];
echo '<br /> <h3> Use in pregnacy:</h3> '.$show['PREGNANCYcategory'];  
echo '<br /> <h3> Pharmacological Class:</h3>     '.$show['Pharmacologicalclass'];  
echo '<br /> <h3> Mode of action:</h3> '.$show['ModeOfAction']; 
echo '<br /> <h3> Manufactured By:</h3> '.$show['ManufacturedBy'];   
}  

}
?>

如何使用php创建搜索结果详细信息页面

问题描述

2 个解决方案

解决方案1
0 2016-02-01 13:12:50

解决方案2
0 2016-11-15 16:41:26

如何使用php创建搜索结果详细信息页面

问题描述

2 个解决方案

解决方案1 0 2016-02-01 13:12:50

解决方案2 0 2016-11-15 16:41:26

解决方案1
0 2016-02-01 13:12:50

解决方案2
0 2016-11-15 16:41:26