连接到数据库时收到错误

时间:2014-10-04 04:21:12

标签: php mysql

警告:mysqli_query()需要至少2个参数,1在第27行的C:\ Apache24 \ htdocs \ TheWorld \ index.php中给出

警告:mysqli_num_rows()要求参数1为mysqli_result,在第29行的C:\ Apache24 \ htdocs \ TheWorld \ index.php中给出null

警告:mysqli_query()需要至少2个参数,1在第31行的C:\ Apache24 \ htdocs \ TheWorld \ index.php中给出

警告:mysqli_num_rows()要求参数1为mysqli_result,在第33行的C:\ Apache24 \ htdocs \ TheWorld \ index.php中给出null

警告:mysqli_query()需要至少2个参数,1在第55行的C:\ Apache24 \ htdocs \ TheWorld \ index.php中给出

....  我的代码是:

第27行:

   $u_check = mysqli_query("SELECT username FROM users WHERE username='$un'");

第29行:

   $check = mysqli_num_rows($u_check);

第31行:

   $e_check = mysqli_query("SELECT email FROM users WHERE email='$em'");

第33行:

   $email_check = mysqli_num_rows($e_check);

并在第55行:

   $query = mysqli_query("INSERT INTO users VALUES ('','$un','$fn','$ln','$em','$pswd','$d','0','Write something about yourself.','','','no')");

有人可以帮忙吗?

3 个答案:

答案 0 :(得分:2)

这条线你必须改变

$u_check = mysqli_query("SELECT username FROM users WHERE username='$un'");

应该是这个

$u_check = mysqli_query($connect,"SELECT username FROM users WHERE username='$un'");

同样适合您插入查询

应该是这个

$query = mysqli_query($connect,"INSERT INTO users VALUES ('','$un','$fn','$ln','$em','$pswd','$d','0','Write something about yourself.','','','no')");

$connect是您用来连接数据库的变量。我不必说你可能 这个变量有不同的名称。

查看链接以获取更多信息http://php.net/manual/en/mysqli.query.php

答案 1 :(得分:0)

您应该将数据库连接传递给“mysqli_query”函数。

例如:

$link = mysqli_connect("localhost", "my_user", "my_password", "world");
$result = mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City");

http://php.net/manual/en/mysqli.query.php

答案 2 :(得分:0)

连接:

$username= "user";
$password= "password";
$host= "ip local host";
$database= "databasename";

$sql1 = mysqli_connect("$host", "$username", "$password", "$database");

if(!$sql1){
  echo "error to connect server.";
  die();
}

if (mysqli_connect_errno()) {
  printf("Connect failed: %s\n", mysqli_connect_error());
  exit();
}

然后如果你想执行你的查询,它应该是这样的。

$mysql = mysqli_query($sql1,"select * from table");