php wrote a login, server error 500.

The code is as follows:
Table structure

id         int phone      charpassword   char

conn.php

  
   错误原因：'.mysql_error());    //设置字符集，如utf8和gbk等    mysql_query("set names 'utf8'");    //选定数据库    mysql_select_db($db_name,$conn) or die('数据库选定失败！
错误原因：'.mysql_error());    //执行SQL语句(查询)    //$result = mysql_query($sql) or die('数据库查询失败！
错误原因：'.mysql_error());?>

login.php

  ' Unlawful request ');        return false; }//Create SQL statement $sql = "SELECT * from lms_users WHERE phone = '". $userName. "'            and ' password ' = ' ". $userPassword." '; ";        Call the conn.php file for database operation require (' conn.php ');                Global $conn;                $result =mysql_query ($sql, $conn) or Die (' database query failed! ' ERROR reason: '. mysql_error ());        $flag = 0; while (!!        $item =mysql_fetch_assoc ($result)) {$flag + +;        };        Only one result can actually be returned.                           if ($flag >= 1) {$result = Array (echo json_encode ($row);            );            echo Json_encode ($result);        Exit ();            }else {$result = array (' id ' = = 0);        echo Json_encode ($result); } return true;? >

Login.html

    
  
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The code is as follows:
Table structure

id         int phone      charpassword   char

conn.php

  
   错误原因：'.mysql_error());    //设置字符集，如utf8和gbk等    mysql_query("set names 'utf8'");    //选定数据库    mysql_select_db($db_name,$conn) or die('数据库选定失败！
错误原因：'.mysql_error());    //执行SQL语句(查询)    //$result = mysql_query($sql) or die('数据库查询失败！
错误原因：'.mysql_error());?>

login.php

  ' Unlawful request ');        return false; }//Create SQL statement $sql = "SELECT * from lms_users WHERE phone = '". $userName. "'            and ' password ' = ' ". $userPassword." '; ";        Call the conn.php file for database operation require (' conn.php ');                Global $conn;                $result =mysql_query ($sql, $conn) or Die (' database query failed! ' ERROR reason: '. mysql_error ());        $flag = 0; while (!!        $item =mysql_fetch_assoc ($result)) {$flag + +;        };        Only one result can actually be returned.                           if ($flag >= 1) {$result = Array (echo json_encode ($row);            );            echo Json_encode ($result);        Exit ();            }else {$result = array (' id ' = = 0);        echo Json_encode ($result); } return true;? >

Login.html

    
  
       PHP 测试    
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The development of the words to open the wrong display of PHP (in the php.ini search configuration display_errors=On ) in order to know what the specific error.

In addition, it is recommended not to use related functions mysql , with mysqli related functions or objects instead, the mysql correlation function is not recommended, and in the latest PHP7 formally removed.

It's a grammatical error,

$result = array(    echo json_encode($row););                       

Here is the wrong one. Can actually open the server on the detailed error output, so you can see the 500 error specific cause of the error

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