使用php从SQL数据库传递select值 - Passing select value from SQL database with php

Passing select value from SQL database with php

本文关键字：select 数据库 php SQL 使用 | 更新日期: 2023-09-27

我正试图使用一个从SQL数据库中填充字段的select表单。数据库中有一个存储表，我使用这种方法将这些不同的字段放入选择工具中。

echo '<select name="StoreName" />'."'n";
while ($row = mysqli_fetch_assoc($distinctResult)){
echo '<option value="'.$row["StoreID"].'">';
echo $row["StoreName"];
echo '</option>'."'n";
};
echo '</select>'."'n";

正确的商店名称显示在选择工具上，但当我提交表单数据时，表单的值没有正确传递。在首页中：

<?php

require_once 'login.php';
$connection = mysqli_connect(
   $db_hostname, $db_username,
   $db_password, $db_database);
if(mysqli_connect_error()){
die("Database Connection Failed: " .
    mysqli_connect_error() .
    " (" . mysqli_connect_errno() . ")"
    );
};
$query = "SELECT * from PURCHASE";
//echo $query;
$result = mysqli_query($connection,$query);
if(!$result) {
die("Database Query Failed!");
};
$distinct = "SELECT DISTINCT StoreName FROM PURCHASE";

$distinctResult = mysqli_query($connection,$distinct);
if(!$result) {
die("Database Query Failed!");
};
echo '<title>Output 1</title>';
echo '</head>';
echo '<body>';
echo '<h1>Required Output 1</h1>';
echo '<h2>Transaction Search</h2>';
echo '<form action="output1out.php" method="get">';
echo 'Search Store:<br/>';
echo '<br/>';

echo '<select name="StoreName" />'."'n";
while ($row = mysqli_fetch_assoc($distinctResult)){
echo '<option value="'.$row["StoreID"].'">';
echo $row["StoreName"];
echo '</option>'."'n";
};
echo '</select>'."'n";
echo '<input name="Add Merchant" type="submit" value="Search">';
echo '</form>';
mysqli_free_result($result);
mysqli_close($connection);

?>

这是输出页面：

<?php
$transaction = $_REQUEST["StoreName"];
    require_once 'login.php';
    $connection = mysqli_connect(
           $db_hostname, $db_username,
           $db_password, $db_database);

       $sql = "SELECT * FROM PURCHASE WHERE StoreName LIKE '%".$transaction."%'";
        $result = $connection->query($sql);
    ?>
    Purchases Made From <?php echo $transaction ?>
<table border="2" style="width:100%">
    <tr>
                  <th width="15%">Item Name</th>
                  <th width="15%">Item Price</th> 
                  <th width="15%">Purchase Time</th>
                  <th width="15%">Purchase Date</th>
                  <th width="15%">Category</th>
                  <th width="15%">Rating</th>
                </tr>
</table>
     <?php
        if($result->num_rows > 0){
            // output data of each row
            while($rows = $result->fetch_assoc()){ ?>
               <table border="2" style="width:100%">
                <tr>
                    <td width="15%"><?php echo $rows['ItemName']; ?></td>
                    <td width="15%"><?php echo $rows['ItemPrice']; ?></td>
                    <td width="15%"><?php echo $rows['PurchaseTime']; ?></td>
                    <td width="15%"><?php echo $rows['PurchaseDate']; ?></td>
                    <td width="15%"><?php echo $rows['Category']; ?></td>
                    <td width="15%"><?php echo $rows['Rating']; ?></td>
                </tr>
                <?php
            }
        }
        ?>

我在输出页面中调用select名称StoreName，但它没有从下拉列表中传递值。我需要在第二页上运行一个带有该术语的查询，以便提取相关数据。如果有人能看到我做错了什么，请给出建议。非常感谢。

要回答的注释：

您正在执行SELECT DISTINCT StoreName FROM PURCHASE，但$row["StoreID"]不在您的选择集中。

将其添加到SELECT中