Question

我遇到这个问题，当我搜索一个单词时，它不会出现并且会弹出此错误

警告：mysqli_fetch_array（）期望参数1为mysqli_result，在213行的C：\ xampp \ htdocs \ public_html \ filterdata.php中给出的布尔值）。

仅在过滤之前显示表格

<?php
include("auth_admin.php");


if(isset($_POST['search']))
{
$valueToSearch = $_POST['valueToSearch'];
// search in all table columns
// using concat mysql function
$query = "SELECT * FROM `lcho_dengue_activities` CONCAT(`id`, `month`, `year`, `dengue_ind1`) where `month`= '".$valueToSearch."'";
$search_result = filterTable($query);


}
else {
$query = "SELECT * FROM `lcho_dengue_activities`";
$search_result = filterTable($query);

 }

// function to connect and execute the query
function filterTable($query)
 {
$connect = mysqli_connect("localhost", "root", "", "lcho_login");
$filter_Result = mysqli_query($connect, $query);
return $filter_Result;
 }


 ?>

<form action="filterdata.php" method="post">
        <input type="text" name="valueToSearch" placeholder="Value To Search"><br><br>
        <input type="submit" name="search" value="Filter"><br><br>

        <table>
            <tr>
                <th>Id</th>
                <th>Month</th>
                <th>Year</th>
                <th>dengue_ind1</th>
            </tr>

  <!-- populate table from mysql database -->
         <?php while($row = mysqli_fetch_array($search_result)):?>
            <tr>
                <td><?php echo $row['id'];?></td>
                <td><?php echo $row['month'];?></td>
                <td><?php echo $row['year'];?></td>
                <td><?php echo $row['dengue_ind1'];?></td>

            </tr>
            <?php endwhile;?>
        </table>
    </form>

//while($row = mysqli_fetch_array($search_result))是我的第213行

我尝试将$search_result更改为$query，并且发生相同的错误。

Answer 1

该查询失败并返回false

将此内容放在mysqli_query()之后，以查看发生了什么情况。

if (!$filter_Result) {
   printf("Error: %s\n", mysqli_error($connect));
   exit();
}

有关更多信息：

http://www.php.net/manual/en/mysqli.error.php

无法从数据库过滤

1 个答案: