PHP和mysql，选择查询返回null。

Question

这是PHP代码：

<?php
// header('Content-type: application/json');
header('Content-Type: text/html; charset=UTF-8');

$d=$_GET["userid"];

 $servername = "*******";
 $username = "***";
 $password = "****";
 $dbname = "*****";
 $arr = array();

  $My_Connection = mysql_connect ( $servername, $username , $password ) ;
     if ( ! $My_Connection )
       {
           die( ' Could not connect : ' . mysql_error( ) ) ;
       }
    //pick data base
    mysql_select_db ( $dbname, $My_Connection );

    mysql_set_charset('utf8',$My_Connection);

         $sql_tempcreate="CREATE TABLE tmp(id INT(11) UNSIGNED AUTO_INCREMENT PRIMARY KEY,DID VARCHAR(20) NOT NULL)";
         if(mysql_query($sql_tempcreate,$My_Connection))
         {
            $sql_convert="ALTER TABLE tmp CONVERT TO CHARACTER SET utf8";
            mysql_query($sql_convert);

           $sql_inserttotemp="INSERT INTO tmp (DID) SELECT DID FROM user WHERE 1 Order By HIGHSCORE DESC";
                if(mysql_query($sql_inserttotemp,$My_Connection))

                {
                 //***************************************** from here problem start
                    $sql_rank="SELECT * FROM `tmp` WHERE DID =".$d."";//    |
                    $r=mysql_query($sql_rank,$My_Connection);             //    |
                                                                          //    |
                    $rank= $r;                                            //    V
                }//******************************************************** to here

         }
         else
         {
            $rank= array("ERROR","ERROR");
         }
            $output = json_encode(array('top' => $arr,'rank' =>$rank));
            echo ($output);
    }
    else
    {
        echo "there is an error :(";
    }
    mysql_query("DROP TABLE tmp",$My_Connection);
mysql_close($My_Connection); 

?>

表tmp创建成功，数据插入tmp表成功，但“选择查询”返回me null!实际上$r是空，我尝试了(类似&=)，但结果相同

这个查询有什么问题？

编辑：

我甚至将查询更改为：

$sql_rank="SELECT * FROM tmp WHERE DID=352136069213581" 

而不再工作：

油箱

编辑:正确答案：

坦克到gaurav kumar这是正确的代码:D

$sql_rank="SELECT * FROM `tmp` WHERE DID LIKE ".$d."";
$res=mysql_query($sql_rank,$My_Connection);
$r=mysql_fetch_assoc($res);
$rank= $r;

Answer 1

坦克到gaurav kumar这是正确的代码:D

$sql_rank="SELECT * FROM `tmp` WHERE DID LIKE ".$d."";
$res=mysql_query($sql_rank,$My_Connection);
$r=mysql_fetch_assoc($res);
$rank= $r;

Answer 2

$d可能是一个string.So，您需要将它们括在引号(单引号或双引号)中，也可以从$d中修剪空格。

$d=trim($_GET["userid"]);
$sql_rank="SELECT * FROM `tmp` WHERE DID ='".$d."'";

也尝试使用mysqli或PDO而不是mysql，因为 functions are deprecated

Answer 3

$d=trim($_GET["userid"]);
$sql_rank="SELECT * FROM `tmp` WHERE ID ='".$d."'";

您的代码是这样的：$d=$_GET["userid"];

$sql_rank="SELECT * FROM `tmp` WHERE DID ='".$d."'";

没有使用ID