Php Mysql Select，其中要回送的语法来自多个表

我就是不知道下面的语法有什么错

$con = mysqli_connect("localhost","*****","*****");
$rank = 78;
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_select_db($con,"******") or die ("no database"); 

$result = mysqli_query($con,"
SELECT first_name FROM table_1 WHERE reputation=$rank;
 SELECT first_name FROM table_2 WHERE reputation=$rank;
SELECT first_name FROM table_3 WHERE reputation=$rank;
SELECT first_name FROM table_4 WHERE reputation=$rank
");


while($row = mysqli_fetch_row($result)) {

echo "$row[0] ";

echo "$row[1] ";

echo "$row[2] ";

echo "$row[3] ";

}

我没有收到任何错误，也没有收到任何结果。如何解决此错误？谢谢。

试着做一个

首先更换

$con = mysqli_connect("localhost","*****","*****");
$rank = 78;
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_select_db($con,"******") or die ("no database"); 

与

$con = mysqli_connect(hostname,user,password,database);
$rank = 78;
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

现在要么使用UNION，要么使用多个查询，要么使用mysqli\u multi\u查询

$sql = "SELECT first_name FROM table_1 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_2 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_3 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_4 WHERE reputation='$rank'";


if (mysqli_multi_query($con,$sql))
{
  do
    {

    if ($result=mysqli_store_result($con))
      {
      while ($row=mysqli_fetch_row($result))
        {
        printf("%s\n",$row[0]);
        }
      mysqli_free_result($con);
      }
    }
  while (mysqli_next_result($con));
}

---

您需要使用

联合

您确定分号吗？非常有效。非常感谢。

$sql = "SELECT first_name FROM table_1 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_2 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_3 WHERE reputation='$rank';";
$sql .= "SELECT first_name FROM table_4 WHERE reputation='$rank'";


if (mysqli_multi_query($con,$sql))
{
  do
    {

    if ($result=mysqli_store_result($con))
      {
      while ($row=mysqli_fetch_row($result))
        {
        printf("%s\n",$row[0]);
        }
      mysqli_free_result($con);
      }
    }
  while (mysqli_next_result($con));
}