Question

我知道这个问题已被多次回答，但似乎无法使其在我现有的代码中运行。请原谅我是个笨蛋。我的问题：我只需要在创建新记录时获取索引和自动递增的client_id值，因此我可以通过电子邮件将其与密码一起发送给新用户，以便他们可以登录到他们的帐户。这是我的代码，感谢您的帮助。我也明白我应该从MySQL转换为mysqli：

$conn = mysql_connect($servername, $username, $password, $dbname);

if (!$conn) { die('Could not connect: ' . mysql_error()); }

mysql_select_db($dbname,$conn);

$sql="INSERT INTO ricks_inserts (companyname, contact, email)

VALUES ('$_POST[companyname]','$_POST[contact],'$_POST[email]')";

$id = mysql_insert_id(); "My actual variable in my table is called client_id"

if (!mysql_query($sql,$conn))
{
die('Error: ' . mysql_error());
}echo "<br />";

$message= <<<EOD
Client Id:$_POST[client_id]
Email: $_POST[email]
Company: $_POST[companyname]
Contact: $_POST[contact]
EOD;

Answer 1

也许你可以尝试这个 MySQL版本

$conn = mysql_connect($servername, $username, $password, $dbname);

if (!$conn) { die('Could not connect: ' . mysql_error()); }

mysql_select_db($dbname,$conn);

$companyname = mysql_real_escape_string($_POST['companyname']);
$contact = mysql_real_escape_string($_POST['contact']);
$email = mysql_real_escape_string($_POST['email']);

$sql="INSERT INTO ricks_inserts (companyname, contact, email)

VALUES ('".$companyname."','".$contact."','".$email."' ) " or die(mysql_errno());

if (mysql_query($sql)) {
    $last_id = mysql_insert_id();
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
    echo "Error: " . $sql . "<br>" . mysql_error();
}

$message= <<<EOD
Client Id:$_POST[client_id]
Email: $_POST[email]
Company: $_POST[companyname]
Contact: $_POST[contact]
EOD;

MySQLi版

        <?php
$conn = mysqli_connect($servername, $username, $password, $dbname);

if (!$conn) { die('Could not connect: ' . mysqli_error($conn)); }

mysqli_select_db($conn,$dbname);

$companyname = mysqli_real_escape_string ($conn,$_POST['companyname']);
$contact = mysqli_real_escape_string ($conn,$_POST['contact']);
$email = mysqli_real_escape_string ($conn,$_POST['email']);

$sql="INSERT INTO ricks_inserts (companyname, contact, email)

VALUES ('".$companyname."','".$contact."','".$email."' ) " or die(mysqli_errno($conn));

if (mysqli_query($conn,$sql)) {
    $last_id = mysqli_insert_id($conn);
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

$message= <<<EOD
Client Id:$_POST[client_id]
Email: $_POST[email]
Company: $_POST[companyname]
Contact: $_POST[contact]
EOD;

?>

PHP获取上次插入记录的ID

1 个答案: