Question

我试图允许用户使用3个下拉框从我的数据库中选择限制。我已经设置好了，我已连接到我的数据库。但是，一旦用户点击提交按钮，我就无法将数据显示在表格中。这是我的代码：

<?php
require_once 'connection.php';
?>


<form action="stats.php" method ="post">
<input type="hidden" name="submitted" value="true" />

<fieldset>
<legend>
Specify Date, Month, and County
</legend>
<p>
<label for="year">
Please select a year
</label>

<select name= 'year'>
<?php
$query = "select distinct year from unemployed";

$result = $conn->query($query);
while($row = $result->fetch_object()) {
  echo "<option value='".$row->year."'>".$row->year."</option>";
 }
?>
</select>
</p>

<p>
<label for="month">
Please select a month
<label>

<select name= 'month'>
<?php
$query = "select distinct month from unemployed";

$result = $conn->query($query);
while($row = $result->fetch_object()) {
  echo "<option value='".$row->month."'>".$row->month."</option>";
 }
?>
</select>
</p>

<p>
<label for="location">
Please specify a location
</label>

<select name='select'>
<?php
$query = "select * from unemployed";

$result = $conn->query($query);

while ($finfo = $result->fetch_field()) {
  echo "<option value='".$finfo->name."'>".$finfo->name."</option>";
 }

?>
</select>
</p>


<input type ="submit" />

</fieldset>
</form>

<?php

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

include('connection.php');

$gYear = $_POST["year"];
$gMonth = $_POST["month"];
$gSelect = $_POST["select"];

$query = "select $gSelect from unemployed where year='$gYear' and month='$gMonth'";

$result = $conn->query($query) or die('error getting data');


echo"<table>";
echo "<tr><th>Year</th><th>Time</th><th>$gSelect</th></tr>";

while ($row = $result->fetch_object()){

echo "<tr><td>";
echo $row['Year'];
echo "</td><td>";
echo $row['Month'];
echo "</td><td>";
echo $row['$gSelect'];
echo "</td></tr>";

}




echo "</table";

} // end of main if statement

?>

我根本无法将数据显示在表格中。我尝试了多种方法，但我仍然遇到错误。为了确保我连接到我的数据库，我使用了var_dump（$ row）来确保，这样就可以了。所以这不是问题。有谁知道我的代码有什么问题？

Answer 1

从结果集中检索数据时，您将其作为对象获取：

while ($row = $result->fetch_object()){

但是当你来显示它时，你将它称为数组：

echo "<tr><td>";
echo $row['Year'];   // array syntax.

您应该使用对象语法：

echo "<tr><td>";
echo $row->Year;   // object syntax.

如果你检查错误日志，你应该看到很多这样的消息。

无法显示数据库中的数据

1 个答案: