我目前正在嘗試顯示我在putty中創建的表並編寫了一個代碼,以便我可以從數據庫中顯示該表,看起來我編寫了邏輯代碼但似乎有一部分失敗。不會顯示來自mysql數據庫的記錄表警告:mysqli_fetch_row()
它給我此警告
警告:mysqli_fetch_row()預計參數1被mysqli_result中,管線25,31和34
給定的boolean這是$行= mysqli_fetch_row($結果);和mysqli_free_result($ result);
require_once ("settings.php");
$conn = @mysqli_connect($host, $user, $pswd, $dbnm)
or die ("Failed to connect to server");
@mysqli_select_db($conn, $dbnm)
or die ("Database not available");
$query = "SELECT car_id, make, model, price, yom FROM cars";
$result = mysqli_query($conn, $query);
echo "<table width='100%' border='1'>";
echo "<tr><th>Car_id</th><th>Make</th><th>Model</th><th>Price</th></tr>";
$Row = mysqli_fetch_row($result);
do {
echo "<tr><td>{$Row[0]}</td>";
echo "<td>{$Row[1]}</td>";
echo "<td>{$Row[2]}</td>";
echo "<td>{$Row[3]}</td></tr>";
$Row = mysqli_fetch_row($result);
} while ($Row);
echo "</table>";
mysqli_free_result($result);
mysqli_close($conn);
?>
如果查詢失敗,因爲糟糕的連接設置也將返回false ......因此,布爾。嘗試「或死(MySQL_error())後查詢顯示錯誤。 – Gavin