PHP搜索结果错误
我已经写了下面的代码作为测试,但是结果没有出现在我创建的表格下,表格显示在顶部,所有结果显示在左侧,即使我没有在搜索中添加任何内容,所有结果都会显示。我看不出哪里出了错。我并不担心实际的HTML外观,因为它只是一个测试。谢谢你的帮助。错误是 index.phpPHP搜索结果错误,php,mysqli,Php,Mysqli,我已经写了下面的代码作为测试,但是结果没有出现在我创建的表格下,表格显示在顶部,所有结果显示在左侧,即使我没有在搜索中添加任何内容,所有结果都会显示。我看不出哪里出了错。我并不担心实际的HTML外观,因为它只是一个测试。谢谢你的帮助。错误是 index.php <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <link rel= "stylesheet" href="style
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<link rel= "stylesheet" href="style5.css"/>
</head>
<boby>
<div class="third_bar">
<div class="second_image">
</div>
<div class="form"><form action= "search1.php" method="post">
<input type="text" name="search" id="search_bar" placeholder="" value="Search for your whisky here" max length="30" autocomplete="off" onMouseDown="active();" onBlur="inactive();"/><input type="submit" id="search_button" value="Go!"/>
</form>
</div>
</body>
</div>
</html>
<?php
if (isset($_POST['search'])){
include ('connect.php');
$search = $_POST['search'];
$query = "SELECT * FROM whisky_results";
$result = mysqli_query($conn, $query) or die ('error getting data');
echo "<table>";
echo "<tr> <th>Whisky Name</th> <th>Whisky Desription</th> <th>Highest Price Recorded</th> <th>Lowest Price Recorded</th> </tr>";
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr><td>";
echo $row['name'];
echo "</td><td>";
echo "<tr><td>";
echo $row['description'];
echo "</td><td>";
echo "<tr><td>";
echo $row['highest_price'];
echo "</td><td>";
echo "<tr><td>";
echo $row['lowest_price'];
echo "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
?>
search1.php
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<link rel= "stylesheet" href="style5.css"/>
</head>
<boby>
<div class="third_bar">
<div class="second_image">
</div>
<div class="form"><form action= "search1.php" method="post">
<input type="text" name="search" id="search_bar" placeholder="" value="Search for your whisky here" max length="30" autocomplete="off" onMouseDown="active();" onBlur="inactive();"/><input type="submit" id="search_button" value="Go!"/>
</form>
</div>
</body>
</div>
</html>
<?php
if (isset($_POST['search'])){
include ('connect.php');
$search = $_POST['search'];
$query = "SELECT * FROM whisky_results";
$result = mysqli_query($conn, $query) or die ('error getting data');
echo "<table>";
echo "<tr> <th>Whisky Name</th> <th>Whisky Desription</th> <th>Highest Price Recorded</th> <th>Lowest Price Recorded</th> </tr>";
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr><td>";
echo $row['name'];
echo "</td><td>";
echo "<tr><td>";
echo $row['description'];
echo "</td><td>";
echo "<tr><td>";
echo $row['highest_price'];
echo "</td><td>";
echo "<tr><td>";
echo $row['lowest_price'];
echo "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
?>
首先,您的表单没有名为submittedOop的输入,谢谢。已如上所述进行更改,现在出现致命错误。听起来您可能没有安装mysqli。当你这样做的时候你会得到什么:echo扩展加载(“mysqli”)结果是1aha!数组有两个r。你只有一个。另外,在尝试使用$result之前,您应该检查$result是否为FALSE。首先,您的表单没有名为submittedOop的输入,谢谢。已如上所述进行更改,现在出现致命错误。听起来您可能没有安装mysqli。当你这样做的时候你会得到什么:echo扩展加载(“mysqli”)结果是1aha!数组有两个r。你只有一个。另外,在尝试使用$result之前,应该检查$result是否为FALSE。