如何使用PHP检查MySQL中是否已经存在用户
问题描述：
我正在使用以下对我不起作用的代码.
I am using following code which is not working for me.
$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/>";
}
}
无法将mysqli_result类的对象转换为int C:\ xampp \ htdocs \ Exp \ welcome.php
Object of class mysqli_result could not be converted to int in C:\xampp\htdocs\Exp\welcome.php
答
此代码适合您...
$con=mysqli_connect("localhost","root","","my_db");
$check="SELECT * FROM persons WHERE Email = '$_POST[eMailTxt]'";
$rs = mysqli_query($con,$check);
$data = mysqli_fetch_array($rs, MYSQLI_NUM);
if($data[0] > 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/>";
}
}