Php 当连接到MySQL时，我没有得到正确的结果

对于下面添加的代码，我不会打印任何结果

$con = @mysqli_connect("localhost","root","","temp");

if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$query="SELECT * FROM `login`";
echo $query;
$result=@mysqli_query($query) or die(mysql_error());
while($row=mysqli_fetch_array($result))
{
    echo $row["username"];
}

---

用这个。它可能会解决您的问题。

请尝试下面的代码，它会工作的

 //conection: 
$con = mysqli_connect("localhost","root","","temp") or die("Error " . mysqli_error($con)); 

//consultation: 

$query = "SELECT * FROM login" or die("Error in the consult.." . mysqli_error($con)); 

//execute the query. 

$result = $con->query($query); 

//display information: 

while($row = mysqli_fetch_array($result)) { 
  echo $row["username"] . "<br>"; 
}

//连接：
$con=mysqli_connect（“localhost”、“root”、“temp”）或die（“Error”。mysqli_Error（$con））；
//咨询：
$query=“SELECT*FROM login”或die（“咨询中的错误”）。mysqli_Error（$con））；
//执行查询。
$result=$con->query（$query）；
//显示信息：
而（$row=mysqli_fetch_数组（$result））{
echo$row[“username”]。“
”；
}

按原样使用此代码

$con=mysqli_connect("localhost","root","","temp");
        $result = mysqli_query($con,"SELECT * FROM login");
         while($row = mysqli_fetch_array($result))
        {
              echo $row["username"];
        }

---

不要混合使用mysql和mysqli函数。删除…，如果我删除…，我会收到警告信息。这些信息会帮助你，对于程序员来说，这没关系。但用户看到这些警告消息时会感到恼火。$con->query（）看起来像是在调用函数。它是如何工作的？你有没有试过上面的代码？它将执行$query变量query。

$con=mysqli_connect("localhost","root","","temp");
        $result = mysqli_query($con,"SELECT * FROM login");
         while($row = mysqli_fetch_array($result))
        {
              echo $row["username"];
        }

//connection 
$con = mysqli_connect("localhost","root","","my_db");

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// use this code  and plz check your db name
$host='localhost';
$user='root';
$pass='';
$db_name='temp';
$con=mysqli_connect($host,$user,$pass,$db_name);
if($con)
{
echo "db connect succecssfully";
}

$slt="select * from login";
$query=mysqli_query($slt,$con);
while($row=mysqli_fetch_array($query))
{
    echo $row["username"];
}