CODEWITHSUNDEEP
php
Moloi Mokete
Moloi Mokete

Reputation: 31

Warning: mysqli_connect(): (HY000/2002): php_network_getaddresses: getaddrinfo failed: No such host is known

i have the code im trying to upload data read from two sensors to in arduino to the WAMP server but when running the localhost\project,i get that Warning.THE php file are located in the projects directory.The Connect.php is used when accessing the database.The Connect.php code is the one used mainly for connection here it is,the add.php proccess the POST from arduino.index.php displays the sensor values in a table:

Below is Connect.php

****<?php
    function Connection(){
        $server="server";
        $user="user";
        $pass="pass";
        $db="database";

        $connection = mysqli_connect($server, $user, $pass);
        //$db = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'user', 'pass');
        if (!$connection) {
            die('MySQL ERROR: ' . mysql_error());
        }

        mysql_select_db($db) or die( 'MySQL ERROR: '. mysql_error() );
        return $connection;
    }
?>****

Below is add.php:

**<?php
    include("connect.php");

    $link=Connection();
    $temp1=$_POST["temp1"];
    $hum1=$_POST["hum1"];
    $query = "INSERT INTO `tempLog` (`temperature`, `humidity`) 
        VALUES ('".$temp1."','".$hum1."')"; 

    mysql_query($query,$link);
    mysql_close($link);
    header("Location: index.php");
?>**

lastly index.php:

**<?php
    include("connect.php");     

    $link=Connection();
    $result=mysql_query("SELECT * FROM `tempLog` ORDER BY `timeStamp` DESC",$link);
?>
<html>
   <head>
      <title>Sensor Data</title>
   </head>
<body>
   <h1>Temperature / moisture sensor readings</h1>
   <table border="1" cellspacing="1" cellpadding="1">
        <tr>
            <td>&nbsp;Timestamp&nbsp;</td>
            <td>&nbsp;Temperature 1&nbsp;</td>
            <td>&nbsp;Moisture 1&nbsp;</td>
        </tr>
      <?php 
          if($result!==FALSE){
             while($row = mysql_fetch_array($result)) {
                printf("<tr><td> &nbsp;%s </td><td> &nbsp;%s&nbsp; </td><td> &nbsp;%s&nbsp; </td></tr>", 
                   $row["timeStamp"], $row["temperature"], $row["humidity"]);
             }
             mysql_free_result($result);
             mysql_close();
          }
      ?>
   </table>
</body>
</html>**

Upvotes: 3

Views: 6440

Answers (2)

Jonast92
Jonast92

Reputation: 4967

You're mising mysqli_ and mysql_ (It's not the same thing..)

Connection returns a mysqli_ connection:

$connect = mysqli_connect($server, $user, $pass);

But you're trying to use it to select db and query via mysql_:

mysql_select_db($db) or die( 'MySQL ERROR: '. mysql_error() );

Also, using the mysqli_ connection as a link to a mysql_ query makes no sense:

$link=Connection();
$result=mysql_query("SELECT * FROM `tempLog` ORDER BY `timeStamp` DESC",$link);

Please just take a look at the mysqli docs and scrape the mysql_code. -- Or even better, use PDO!

Upvotes: 1

RiggsFolly
RiggsFolly

Reputation: 94642

Try either using the mysqli_select_db with the correct params like this

$connection = mysqli_connect($server, $user, $pass);
if (!$connection) {
    die('MySQL ERROR: ' . mysql_error());
}

mysql_select_db($connection,$db) or die( 'MySQL ERROR: '. mysql_error() );
return $connection;

Or use the mysqli_connect() to also select the database by adding it as the 4th parameter, then you can forget the mysql_select_db() completely.

$connection = mysqli_connect($server, $user, $pass, $db);
if (!$connection) {
    die('MySQL ERROR: ' . mysql_error());
}

Upvotes: 0

Related Questions