得到错误“ mysqli_query()期望参数1为mysqli,资源在“ [英] getting error " mysqli_query() expects parameter 1 to be mysqli, resource given in "
问题描述
嗨朋友
我的php代码出错了
mysqli_query()期望参数1成为mysqli,我的PHP代码中给出的资源
Quote:警告: mysqli_query()期望参数1为mysqli,第16行的C:\ xampp \\\\\\\\\\\\\\\\\\\\\\\ b警告:mysqli_num_rows()要求参数1为mysqli_result,在第17行的C:\ xampp \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
我的代码在这里
Quote:if(isset($ _ POST [usernamecheck])){
$ db_conx = mysqli_connect('localhost','root');
$ username = preg_replace('#[^ a-z0-9] #i','',$ _ POST ['usernamecheck']);
$ sql = SELECT id FROM users WHERE username ='$ username'LIMIT 1;
$ query = mysqli_query($ db_conx,$ sql);
$ unamecheck = mysqli_num_rows($ query);
if(strlen($用户名)< 3 || strlen($ username)> 16)
{
echo' 3-16个字符请强>';
exit();
}
if(is_numeric($ username [0]))
{
echo'用户名必须以字母开头';
exit();
}
if($ unamecheck< 1){
echo''。$ username。 '没问题';
exit();
}
else
{
echo''。$ username。'is taken ';
exit();
}
}
?>
解决方案
_POST [usernamecheck])){
db_conx = mysqli_connect('localhost','root');
username = preg_replace('#[^ a-z0-9] #i','',
Hi friends
I getting error in my php code
mysqli_query() expects parameter 1 to be mysqli, resource given in my php code
Quote:Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\Sites\freak\SignUp.php on line 16
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\Sites\freak\SignUp.php on line 17
my code is here
Quote:if(isset($_POST["usernamecheck"])){
$db_conx = mysqli_connect('localhost','root');
$username = preg_replace('#[^a-z0-9]#i','',$_POST['usernamecheck']);
$sql = "SELECT id FROM users WHERE username = '$username'LIMIT 1";
$query = mysqli_query($db_conx,$sql);
$unamecheck = mysqli_num_rows($query);
if(strlen($username)< 3 || strlen($username) > 16)
{
echo'3-16 characters please';
exit();
}
if(is_numeric($username[0]))
{
echo' Usernames must begin with a letter';
exit();
}
if($unamecheck < 1) {
echo ''.$username. 'is OK';
exit();
}
else
{
echo''.$username.'is taken ';
exit();
}
}
?>
解决方案_POST["usernamecheck"])){
db_conx = mysqli_connect('localhost','root');
username = preg_replace('#[^a-z0-9]#i','',
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