This article在此使用mysqli_prepare()
,但mysqli_stmt_*
用于其他所有内容。
但是,在PHP手册的this article中,它使用mysqli_stmt_prepare()
这些有什么区别?我可以使用吗?如果mysqli_prepare()
有效,那是否意味着mysqli_bind_param()
有效?
好的,mysqli_bind_param()
mysql_stmt_bind_param()
表示它是mysqli_prepare()
的别名。我认为可以安全地假设它与mysqli_*
和string QuoteMessageBody (MimeMessage message)
{
using (var quoted = new StringWriter ()) {
quoted.WriteLine ("On {0}, {1} wrote:", message.Date.ToString ("f"), message.From.ToString ());
using (var reader = new StringReader (message.TextBody)) {
string line;
while ((line = reader.ReadLine ()) != null) {
quoted.Write ("> ");
quoted.WriteLine (line);
}
}
return quoted.ToString ();
}
}
函数的其余部分相同吗?
答案 0 :(得分:6)
mysqli_prepare()
基本上只是
$stmt = mysqli_stmt_init();
$stmt->prepare($sql);