Question

我有以下代码，它应该返回某个区域内的所有团队。我有一个足球队数据库，其中包含球队和国家队的表格。 teams表具有状态表的外键引用，states表具有区域（北，南，东，西）的属性。

我的主页上有以下html / php代码：

<div>
  <form method="post" action="regions_filter.php">
    <fieldset>
        <legend>Filter Teams By Region</legend>
        <select name="Region">
            <?php 
            if(!($stmt = $mysqli->prepare("SELECT DISTINCT region FROM states"))){
                echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
                }

            if(!$stmt->execute()){
                echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }
            if(!$stmt->bind_result($region)){
                echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }
            while($stmt->fetch()){
                echo '<option value=" ' . $region . ' "> ' . $region . '</option>\n';
            }
            $stmt->close();
            ?>
        </select>
        <input type="submit" value="Run Filter"/>
    </fieldset>
  </form>
</div>

下面是regions_filter.php文件代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<body>
<div>
<table>
    <tr>
        <td>Teams By Region</td>
    </tr>
    <tr>
        <td>School Name</td>
        <td>State Name</td>
        <td>State Capital</td>
        <td>State Population</td>
        <td>Region</td>
    </tr>
<?php
if(!($stmt = $mysqli->prepare("SELECT teams.school_name, states.name,       states.capital, states.population, states.region FROM teams
                            INNER JOIN states ON states.id = teams.state_id
                            WHERE states.region = ?"))){
echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
}

if(!($stmt->bind_param("s",$_POST['Region']))){
echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
}

if(!$stmt->execute()){
echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli- >connect_error;
}
if(!$stmt->bind_result($school, $state, $capital, $population, $region)){
echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli- >connect_error;
}
while($stmt->fetch()){
 echo "<tr>\n<td>" . $school . "\n</td>\n<td>" . $state . "\n</td>\n<td>" .   $capital . "\n</td>\n</td>"
    . $population . "\n</td>\n<td>" . $region . "\n</td>\n</tr>";
}
$stmt->close();
?>
</table>
</div>

</body>
</html>

当我在主页面上运行过滤器时，我被带到regions_filter.php页面，但没有结果。唯一显示的是regions_filter.php页面顶部的预编码html表。我相信错误发生在下面的代码段中。我已尝试使用选项值进行不同的变化，但似乎无法破解它：

while($stmt->fetch()){
                echo '<option value=" ' . $region . ' "> ' . $region . '</option>\n';
            }

任何指向正确方向的人都会非常感激。

Answer 1

我相信你的where语句中的states.id不是region id。如果要将$ region作为参数从标记过滤到select语句state.id =？（state.id = $ region）它不会给出任何结果。因为你是通过状态过滤的。 注意：这只是我对上述代码的第一次理解

我认为你没有任何错误......只是你可能会过滤错误的属性^ _ ^

Answer 2

填充区域时出现错误选择区域名称前后有一个额外的空白区域。应该是这样的：

while($stmt->fetch()){
   echo '<option value="' . $region . '">' . $region . '</option>\n';
}

PHP / MYSQL - 选择未发送的选项值？

2 个答案: