Question

到目前为止，我有以下php，它允许用户搜索数据库的特定列并显示所有结果。

<?php
require("header.php");
if(isset($_REQUEST['searching'])){ //check if form has been submitted
echo"<h2>Results</H2><p>";
connect('final');//connect to DB
//set the values from search form
$field = $_POST['field'];
$query = $_POST['query']; 

    $query = htmlspecialchars($query); // stop HTML characters
    $query = mysql_real_escape_string($query); //stop SQL injection

     $data = mysql_query("SELECT *
     FROM customer 
     INNER JOIN address ON customer.ID = address.customer_ID
     LEFT OUTER JOIN sites ON address.ID = sites.address_ID 
     WHERE customer.ID IN (SELECT customer.ID 
     FROM customer
     INNER JOIN address ON customer.ID = address.customer_ID 
     LEFT OUTER JOIN sites ON address.ID = sites.address_ID
     WHERE upper(customer.$field) LIKE'%$query%')") ;//query the DB with search field in colleumn selected//

     //$data = mysql_query("SELECT * FROM customer INNER JOIN address ON customer.ID = address.Customer_ID LEFT OUTER JOIN sites ON address.ID = sites.address_ID WHERE upper(customer.$field) LIKE'%$query%'") ;

    if($data === FALSE) {
    $error = 'Query error:'.mysql_error();
    echo $error;
    }
    else
    {
    while($results = mysql_fetch_array($data)){// puts data from database into array, loops until no more
            echo "<br>"; 
            echo $results['First_Name']; 
            echo " "; 
            echo $results['Surname']; 
            echo " "; 
            echo $results['Company_Name']; 
            echo " "; 
            echo $results['Telephone']; 
            echo " "; 
            echo $results['Alt_Telephone']; 
            echo " "; 
            echo $results['line_1']; 
            echo " "; 
            echo $results['line2']; 
            echo " "; 
            echo $results['town']; 
            echo " "; 
            echo $results['postcode'];
            echo " "; 
            echo $results['site_name'];


            //posts results from db query
        }
    }


        $anymatches=mysql_num_rows($data); //checks if the querys returned any results
            if ($anymatches == 0) 
                    { 
                        echo "Sorry, but we can not find an entry to match your query<br><br>"; 
                    } 

    }

然而，当我使用echo输出时，它只显示页面顶部的所有结果，并将所有html内容推送到它下面。

我如何格式化它以便它出现在我的html内容中？例如，在搜索按钮下方。

Answer 1

解决方案是不回显它将在顶部显示的内容，因为PHP首先处理代码然后加载html因此它将回显页面顶部的数据。

解决方案是将结果分配到while循环中的数组中，然后使用该数组在html中显示它。

Answer 2

也许，关闭标签'p'会有帮助吗？尝试添加

echo "</p>";

最后。

Answer 3

你在那里缺少一些html标签，例如：

        echo "<div>"; 
        echo $results['First_Name']; 
        echo " "; 
        echo $results['Surname']; 
        echo ", "; 
        echo $results['Company_Name']; 
        echo "</div><address>"; 
        echo $results['Telephone']; 
        echo "<br>"; 
        echo $results['Alt_Telephone']; 
        echo "<br>"; 
        echo $results['line_1']; 
        echo "<br>"; 
        echo $results['line2']; 
        echo "<br>"; 
        echo $results['town']; 
        echo " "; 
        echo $results['postcode'];
        echo "</address><div>"; 
        echo $results['site_name'];
        echo "</div>";

如果你允许稍微加时答案，你的代码就会告诉我你正在学习PHP。

如果是这样，请记住，在线PHP教程经常在几年前编写，并且处于过时状态，完整的示例代码显示了您可以做的最糟糕的事情。

假设我的评估是正确的，请考虑学习PHP以及设计良好的框架。在这方面，Silex相对容易接受：

您可能希望了解的其他框架：

还有很多很多。

格式化数据库查询的结果

3 个答案: