Question

我试图在使用mysql_insert_id();的查询后获取id，但是我仍然在0，尽管

我把它放在查询本身后面
我确保id（称为P_Id）具有AUTO_INCREMENT。

以下代码：

$con=mysqli_connect(-connectiondetails-);
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO clients (Name, Email, Address, Phone, Date, Service, ExtraOne, ExtraTwo, ExtraThree, ExtraFour, ExtraFive) VALUES ('$_POST[name]','$_POST[email]','$_POST[address]','$_POST[phone]','$_POST[date]','$_POST[service]','$_POST[extra1]','$_POST[extra2]','$_POST[extra3]','$_POST[extra4]','$_POST[extra5]')";
$idit = mysql_insert_id();
echo $idit;



if (mysqli_query($con,$sql))
{

}
else
{
    echo "Error message goes here: " . mysqli_error($con);
}

思考“哦好吧也许我必须先实际查询它”，我做了以下但得出了相同的结果：

$con=mysqli_connect(-connectiondetails-);
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO clients (Name, Email, Address, Phone, Date, Service, ExtraOne, ExtraTwo, ExtraThree, ExtraFour, ExtraFive) VALUES ('$_POST[name]','$_POST[email]','$_POST[address]','$_POST[phone]','$_POST[date]','$_POST[service]','$_POST[extra1]','$_POST[extra2]','$_POST[extra3]','$_POST[extra4]','$_POST[extra5]')";



if (mysqli_query($con,$sql))
{
    $idit = mysql_insert_id();
    echo $idit;

}
else
{
    echo "Error message goes here: " . mysqli_error($con);
}

知道我哪里出错了吗？我已经完成了其他线程，但似乎没有任何东西可以帮助我。谢谢你提前。

编辑：我更改了mysql_insert_id（）;到mysqli_insert_id（）;而这次它甚至没有返回0，只是空白。

编辑2 ：谢谢mbouzahir - 您的解决方案有效：）

Answer 1

在第二个例子中使用mysqli_insert_id($con)代替mysql_insert_id()进行了尝试？

Answer 2

您的查询容易受到SQL注入攻击。您应该使用带参数绑定的预准备语句。试试这个（是的，我正在使用OOP API，因为它是一个该死的网站清洁工）

$con = new mysqli(-connectiondetails-);
if ($con->connect_errno) {
    throw new Exception($con->connect_error, $con->connect_errno);
}

$stmt = $con->prepare(
    'INSERT INTO clients (Name, Email, Address, Phone, Date, Service, ExtraOne, ExtraTwo, ExtraThree, ExtraFour, ExtraFive)
    VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)');

if ($stmt === false) {
    throw new Exception($con->error);
}

// you should probably check here that all required POST params are present

$stmt->bindParam('sssssssssss',
    $_POST['name'],
    $_POST['email'],
    $_POST['address'],
    $_POST['phone'],
    $_POST['date'],
    $_POST['service'],
    $_POST['extra1'],
    $_POST['extra2'],
    $_POST['extra3'],
    $_POST['extra4'],
    $_POST['extra5']);

if (!$stmt->execute()) {
    throw new Exception($stmt->error);
}

echo $con->insert_id;

Answer 3

试试这个

$con=mysqli_connect(-connectiondetails-);
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO clients (Name, Email, Address, Phone, Date, Service, ExtraOne, ExtraTwo, ExtraThree, ExtraFour, ExtraFive) VALUES ('$_POST[name]','$_POST[email]','$_POST[address]','$_POST[phone]','$_POST[date]','$_POST[service]','$_POST[extra1]','$_POST[extra2]','$_POST[extra3]','$_POST[extra4]','$_POST[extra5]')";



if (mysqli_query($sql,$con))
{
    $idit = mysqli_insert_id();
    echo $idit;

}
else
{
    echo "Error message goes here: " . mysql_error($con);
}

mysql_insert_id（）;继续返回'0'

3 个答案: