有人可以向我解釋爲什麼這不起作用。試圖轉換到準備好的語句按照每個人的意見,但在開始時被卡住的權利... ...的連接是好的,它不返回任何消息,但目前還沒有進入我的表(稱爲nametable)在mysqli編寫的語句不起作用
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "fidelio";
$dbname = "test";
$con = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_errno()) {
die("Database connection failed: " .
mysqli_connect_error() .
" (" . mysqli_connect_errno() . ")");
}
$query = "INSERT INTO nametable (fname, lname) values (?,?)";
$stmt = mysqli_prepare($con, $query);
$firstName = "simon";
$lastName = "morris";
mysqli_stmt_bind_param($stmt,"ss",$firstname, $lastname);
mysqli_stmt_execute($stmt);
printf("Error: %s.\n", $stmt->error);
$stmt->close();
?>
我添加的最後2行,並且回來是
Error: .
這工作得很好,但準備的語句不....誰知道爲什麼這個錯誤嗎?
$sql="INSERT INTO nametable (fname, lname)
VALUES ('$firstName', '$lastName')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
您是否嘗試過調試'mysqli_stmt_execute($語句)'來看看是否有相關的錯誤信息來幫助你? –
嘗試'mysqli_stmt_execute($ stmt)或死(mysqli_stmt_error($ stmt))'查看錯誤。 – Barmar