警告：mysql_num_rows（）期望参数1是资源，在C：\ xampp \\\\\\\\\\\\\\\\\\\\\ [英] Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\

问题描述

我试图在将数据插入数据库之前检查我的数据，但是这个错误总是出现



我尝试过：

 <   pre  >  if（isset（$ _ POST ['submit']））{
 
 
 $ first_name = test_input（$ _ POST [firstname]）; 
 $ last_name = test_input（$ _ POST [lastname]）; 
 $ id = test_input（$ _ POST [member_id]）; 
 $ username = test_input（$ _ POST [username]）; 
 $ password = test_input（$ _ POST [password]）; 
 $ email = test_input（$ _ POST [email]）; 
 $ membertype = test_input（$ _ POST [membertype]）; 
 $ phone = test_input（$ _ POST [phonenumber]）; 
 $ department = test_input（$ _ POST [department]）; 
 
 
 $ query =SELECT * FROM lms WHERE ID ='。$ id。'OR email ='。$ email。'OR user_name ='。$ username 。'; 
 $ resultdata = mysqli_query（$ link，$ query）; 
 $ data = mysqli_fetch_array（$ resultdata，MYSQLI_NUM）; 
 
 if（$ data [0]> 1）{
 echo用户已存在<   br /  > ; 
} 
 
 else 
 {
 $ sql =INSERT INTO`member`（`ID`，`first_name`，`last_name`，`member_type`，` password`，`email`，`phoneNo`，`department`，`user_name`）
 VALUES（'$ id'，'$ first_name'，'$ last_name'，'$ membertype'，'$ password'， '$电子邮件'， '$电话'， '$部门'， '$用户名'）; 
 if（mysqli_query（$ link，$ sql））
 {
 echo您现在已注册<   br /  > ; 
} 
 else 
 {
 echoERROR：无法执行$ sql。。 mysqli_error（$链接）; } 
} 
 
 
} 
函数test_input（$ data）{
 $ data = trim（$ data）; 
 $ data = stripslashes（$ data）; 
 $ data = htmlspecialchars（$ data）; 
返回$ data; 
} 
 
？> 

i也试过这个

  $ results  = mysql_num_rows（$ resultdata）; 

和

  $ user  = mysqli_fetch_assoc（$ resultdata）; 
 

但它给了我同样的错误，我知道这个错误出现是因为查询返回错误的结果，但我不知道我还应该使用什么

解决方案

_POST ['submit']））{

first_name = test_input（

< BLOCKQUOTE> _POST [ 名字]）;

I'm trying to check my data before inserting it into the database, but this error always appear

What I have tried:

<pre>if(isset($_POST['submit'])){
    

$first_name = test_input($_POST["firstname"]);
$last_name = test_input($_POST["lastname"]);
$id = test_input($_POST["member_id"]);
$username = test_input($_POST["username"]);
$password = test_input($_POST["password"]);
$email = test_input($_POST["email"]);
$membertype = test_input($_POST["membertype"]);
$phone = test_input($_POST["phonenumber"]);
$department = test_input($_POST["department"]);

 
  $query = "SELECT * FROM lms WHERE  ID='".$id."' OR email='".$email."' OR  user_name='".$username."' ";
$resultdata = mysqli_query($link , $query);
$data = mysqli_fetch_array($resultdata, MYSQLI_NUM);
    
if($data[0] > 1) {
    echo "User Already in Exists<br/>";
}

else
{
    $sql = "INSERT INTO `member` (`ID`, `first_name`, `last_name`, `member_type`, `password`, `email`, `phoneNo`, `department`, `user_name`)
VALUES ('$id', '$first_name', '$last_name','$membertype','$password','$email','$phone','$department','$username')";
    if (mysqli_query($link,$sql))
    {
        echo "You are now registered<br/>";
    }
    else
    {
 echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);    }
}
    
    
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}

?>

i have tried also this

$results = mysql_num_rows($resultdata);

and

$user = mysqli_fetch_assoc($resultdata);

but it gives me the same error, I know this error appears because of the query return false result, but I don't what else should I use

解决方案

_POST['submit'])){

first_name = test_input(

_POST["firstname"]);

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