PHP mysql $ _POST搜索问题

Question

我无法使我的搜索功能正常工作.我没有看到任何PHP错误,但页面刷新并重新显示数据库列.我希望它只显示搜索查询的结果.

例如,如果我搜索"Mike",我想在"Tech_Num","Tech_F_Name","Tech_L_Name"和"Mobile_Num"中显示所有Mikes.

码:

<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');

//Include the connection
include "connect.php";

$sql = "SELECT * FROM tech_info ";

if (isset($_POST['searchquery'])) {

    $result = mysqli_query($con, "SELECT * FROM tech_info");
    $search_term = $_POST['searchquery'];
    $sql .= "WHERE Tech_F_Name = '{$search_term}'";
    $sql .= " OR Tech_L_Name = '{$search_term}'";
}
?>

<form action="test2.php" method="POST">
    Search: <input type="text" name="searchquery" />
    <input type="submit" name="searchname" value="Search Me">
</form>

<table width="70%" cellpadding="5" cellspace="5">

    <tr>
        <td><strong>Tech_Num</strong></td>
        <td><strong>First Name</strong></td>
        <td><strong>Last Name</strong></td>
        <td><strong>Mobile Number</strong></td>
    </tr>

<?php
while ($row = mysqli_fetch_array($result)) {
    ?>

        <tr>
            <td><?php echo $row['Tech_Num']; ?>
            <td><?php echo $row['Tech_F_Name']; ?>
            <td><?php echo $row['Tech_L_Name']; ?>
            <td><?php echo $row['Mobile_Num']; ?>


<?php } ?>

</table>  

Answer 1

<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');

//Include the connection
include "connect.php";

$sql = "SELECT * FROM tech_info ";

if (isset($_POST['searchquery'])) {
    $search_term = mysql_real_escape_string($_POST['searchquery']);
    $sql .= "WHERE Tech_F_Name = '{$search_term}'";
    $sql .= " OR Tech_L_Name = '{$search_term}'";



}
$result = mysqli_query($con, $sql);
?>

你的东西是错误的顺序,需要在使用之前构建sql查询.