我正在使用以下不适合我的代码。
$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/>";
}
}
类mysqli_result的对象无法转换为int in 的 C:\ XAMPP \ htdocs中\精通\的welcome.php
答案 0 :(得分:6)
此代码适用于您...
$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/>";
}
}
答案 1 :(得分:3)
mysqli_query
函数返回结果集句柄。然后,您需要从中读取行:
$rs = mysqli_query($con,$check);
$data = mysqli_fetch_array($rs, MYSQLI_NUM);
if($data[0] > 1) {
//user exists;
}
另请注意,SELECT count(1) FROM ...
将比SELECT count(*) FROM ...
更快您在小表中看不到多少差异,但是对于数十万行的大表,差异可能很大
答案 2 :(得分:2)
mysqli_query
会返回mysqli_result
个对象。不要将此与整数进行比较,请尝试使用num_rows
:
$res = mysqli_query($con,$check);
if($res->num_rows){
//User exists
}
修改:上面假设查询使用SELECT *
,这不适用于SELECT COUNT(*)
。 Checkout Aleks G的答案。
答案 3 :(得分:0)
您可以尝试以下代码:
$query_code = "SELECT COUNT(itemCode) FROM masterData WHERE itemCode='{$itemCode}'";
$result_login = mysqli_query($conn,$query_code);
$anything_found = mysqli_num_rows($result_login);
if($anything_found > 0) {
$formOk = false;
echo "ITEM CODE ALREADY EXISTS! Please try again.";
}