Question

我目前正在尝试创建一个搜索菜单，并以表格形式显示结果。我设法只将搜索结果作为字符串，但无法使用它在表格中显示完整的行。这是我的代码： - 创建表单：

<form action = "profile.php" method = "post">
            <input type="text" name="search" placeholder = "Search">
            <input type="submit" value = "SEARCH"><br>
        </form>

这是获取搜索结果的php

if(isset($_POST['search']))
        {
            $searchq = $_POST['search']; 
            $searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

            $query = mysql_query("SELECT * FROM identity WHERE fullname LIKE '%$searchq%'") or die("Could not Find!");
            $count = mysql_num_rows($query);

            if($count == 0)
            {
                $output = 'There is no result';
            }
            else
            {
                while($row = mysql_fetch_array($query))
                {
                    $fullname =  $row['fullname'];
                    $output .= '<div> '.$fullname.' </div>';
                    $records = mysql_query("SELECT * FROM identitas_pengadu WHERE fullname='$output'");
                }

            }           
        }

这是我用来在数据库中显示结果的代码：

while($identity = mysql_fetch_assoc($records))
                {           
                    echo "<tr>";
                    echo "<td>".$identity['fullname']."</td>";
                    echo "<td>".$identity['nip']."</td>";
                    echo "<td>".$identity['email']."</td>";
                    echo "<td>".$identity['position']."</td>";
                    echo "<td>".$identity['status']."</td>";
                    echo "<td><a href=\"profiledetails.php?nip=".$identity['nip']."\">EDIT</a></td>";
                    echo "</tr>";
                }

如果$ records的值不使用WHERE函数，它可以正常工作。请帮忙。谢谢

PHP使用搜索结果显示数据库

0 个答案: