从mysql中选择Single Value

Question

I'm trying to select a single value from mysql - but however it does not work:

public function doStuff($posted_name){    
    $result = mysql_query("SELECT id FROM people WHERE people_name = $posted_name");
    $singleValue = mysql_fetch_object($result);
}

posted_name, people_name is TEXT, id is INT

I am posting a $posted_name , this I let compare with column people_name from table people, and selecting the column "id" ...

this id i want to store as value , here I tried to store it in $singleValue... but however it does not work - is there any mistake?

Solved: see comments

EDIT:

I changed some code to mysqli for testing- but now I have some error:

my code looks like:

private $db;


    // constructor
    function __construct() {
        include_once './db_connect.php';
        // connecting to database
        $db = new DB_Connect();
          $this->db = $db->connect();
    }

. . . . .

$result = this->db->query("INSERT INTO.....

The error appears for the above line ... Parse error: syntax error, unexpected '->' (T_OBJECT_OPERATOR)...

what I am doing wrong

my db_connect.php

class DB_CONNECT {

    private $db;

    // constructor
    function __construct() {
        // connecting to database
        $this->connect();
    }


    /**
     * Function to connect with database
     */
    function connect() {
        // import database db variables
        require_once __DIR__ . '/db_config.php';

        // Connecting to mysql database
       $this->db = new mysqli(DB_HOST, DB_USER, DB_PASSWORD);

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

        // returing connection cursor
        return $this->db;
    }



}

Answer 1

You need to add single quote in $posted_name like this '$posted_name'. and you should know how mysql_fetch_object() works and what it return.

mysql_fetch_object() fetches the result row as an object.It returns an object with properties that correspond to the fetched row and moves the internal data pointer ahead.

I have edited your code which looks like this ...

public function doStuff($posted_name){    
    $result = mysql_query("SELECT id FROM people WHERE people_name = '$posted_name'");
    if($result){
         while ($row = mysql_fetch_object($result)){
              $id =  $row->id;
              echo $id;
          }
     }
 }