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。

代码：

<?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);
?>

你的东西顺序错误，需要在使用它之前构建sql查询。

Answer 2

您的操作顺序似乎有误。您应该首先创建$ sql变量，然后将其发送到数据库？类似的东西：

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

  $search_term = $_POST['searchquery'];

  $sql .= "WHERE Tech_F_Name = '{$search_term}'";

  $sql .= " OR Tech_L_Name = '{$search_term}'";

  $result = mysqli_query($con, $sql);

}

此外，对于初学者，请替换：

$search_term = $_POST['searchquery'];

与

$search_term = mysql_real_escape_string($_POST['searchquery']);

Answer 3

我觉得你对这个剧本很困惑:-) 如果你想在一些数据库coloumns中搜索，你试着写这样的东西：

include "connect.php";

$sql = "SELECT * FROM tech_info ";


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


$search_term = $_POST['searchquery'];

$sql .= "WHERE Tech_F_Name = '{$search_term}'";

$sql .= " OR Tech_L_Name = '{$search_term}'";

}

$result = mysqli_query($con, $sql);

while ($row = mysqli_fetch_array($result)) {}

PHP mysql $ _POST搜索问题

3 个答案: