[英]Trying to show results from MySQL query in a HTML table
我試圖從數據庫表中獲取一些簡單的信息,並將其發布在HTML頁面上。 看起來非常簡單明了,沒有錯誤,這意味着我連接到數據庫和表。 我已經有所有數據了(手動輸入)。
我假設這是我調用列名進行查詢的方式。
<?php
require_once("settings.php");
//Open Connection
$conn = @mysqli_connect("$host","$user","$pswd")
or die ('Failed To Connect to Server');
@mysqli_select_db("$conn", "$dbnm")
or die ('Database Not Available');
//Set up SQL string and excecute
$car_id = mysqli_escape_string($_GET['car_id']);
$make = mysqli_escape_string($_GET['make']);
$model = mysqli_escape_string($_GET['model']);
$price = mysqli_escape_string($_GET['price']);
$query = "SELECT car_id, make, model, price FROM cars";
$results = mysqli_query($conn, $query);
echo "<table width ='100%' border='1'>
<tr>
<th>car_id</th>
<th>make</th>
<th>model</th>
<th>price</th>
</tr>";
// $row = mysqli_fetch_row($query);
while ($row = mysqli_fetch_array($results)) {
echo "<tr>";
echo "<td>" . $row['car_id'] . "</td>";
echo "<td>" . $row['make'] . "</td>";
echo "<td>" . $row['model'] . "</td>";
echo "<td>" . $row['price'] . "</td>";
echo "</tr>";
// $row = mysqli_fetch_row($query);
}
echo "</table>";
//Close Connection
mysqli_free_result($results);
mysqli_close($conn);
?>
settings.php保存所有連接信息,並全部簽出。 我什至需要($ _GET ['car_id'])等嗎? 我可以用字段名稱來稱呼他們嗎?
答案將是如此明顯……
我沒有錯誤,這意味着我已連接到數據庫和表
這並不意味着。
mysqli_select_db("$conn", "$dbnm") // That `$conn` should not be inside those quotes.
應該
mysqli_select_db($conn, $dbnm); // that $conn has to be a MySQLi link identifier, not an interpolated one.
還要刪除所有@,並對這些函數返回的內容進行一些錯誤檢查。
請如下更改以下行。
$conn = @mysqli_connect("$host","$user","$pswd")
至
$conn = mysqli_connect($host,$user,$pswd)
和
@mysqli_select_db("$conn", "$dbnm")
至
mysqli_select_db($conn, $dbnm)
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