我需要帮助检查是否存在行.我收到"电子邮件不再存在publisher@example.com".
有没有更好的方法来检查mysqli是否存在行？
email no longer exists publisher@example.com
我找了类似的问题没有成功.
我有这段代码:
form1.php
$query = "INSERT INTO table1 ";
$query .= "(fname, lname, mail)";
$query .= " VALUES ";
$query .= "('".$_POST[fname]."', '".$_POST[lname]."', '".$_POST[mail]."')";
$result = mysql_query($query) or die ("Query Failed: " . mysql_error());
我希望脚本将检查插入的值是否存在于相应的列中,如果存在则抛出错误.有任何想法吗？
我正在使用以下代码,这对我不起作用.
$con=mysqli_connect("localhost","root","","my_db");
$check="SELECT COUNT(*) FROM persons WHERE Email = '$_POST[eMailTxt]'";
if (mysqli_query($con,$check)>=1)
{
echo "User Already in Exists<br/>";
}
else
{
$newUser="INSERT INTO persons(Email,FirstName,LastName,PassWord) values('$_POST[eMailTxt]','$_POST[NameTxt]','$_POST[LnameTxt]','$_POST[passWordTxt]')";
if (mysqli_query($con,$newUser))
{
echo "You are now registered<br/>";
}
else
{
echo "Error adding user in database<br/>";
}
}
mysqli_result类的对象无法在C:\ xampp\htdocs\Exp\welcome.php中转换为int