StackOverflow2013

Note that there are some explanatory texts on larger screens.

plurals

PONotice: Undefined variable: mysqli in... on line 63
primarykey
Id
10482574
data
AcceptedAnswerId
0
AnswerCount
4
ClosedDate
CommentCount
9
CommunityOwnedDate
CreationDate
2012-05-07T13:01:18.147
FavoriteCount
0
LastActivityDate
2014-05-30T01:55:21.087
LastEditDate
2012-05-07T15:23:54.997
LastEditorUserId
1369722
OwnerUserId
1369722
ParentId
0
PostTypeId
1
Score
0
ViewCount
17549
LastEditorDisplayName
text
Body
<p>I received this error I really don't know what to do about it. I'm tyring to build a 'form' with which you can add information to a database. Because I want no double records in my database i'm checking some fields with AJAX.</p> <p>Here you see the code of my class where I receive my error from. (I use mysqli - language)</p> <pre><code><?php class Places { private $m_sName; private $m_sStreet; private $m_sHouseNumber; private $m_sCity; private $m_sCategory; public function __set($p_sProperty, $p_vValue) { switch($p_sProperty) { case "Name" : $this -> m_sName = $p_vValue; break; case "Street" : $this -> m_sStreet = $p_vValue; break; case "HouseNumber" : $this -> m_sHouseNumber= $p_vValue; break; case "City" : $this -> m_sCity = $p_vValue; break; case "Category" : $this -> m_sCategory = $p_vValue; break; } } public function __get($p_sProperty) { $vResult = null; switch($p_sProperty) { case "Name" : $vResult = $this -> m_sName; break; case "Street" : $vResult = $this -> m_sStreet; break; case "HouseNumber" : $vResult = $this -> m_sHouseNumber; break; case "City" : $vResult = $this -> m_sCity; break; case "Category" : $vResult = $this -> m_sCategory; break; } return $vResult; } public function addPlaces() { include_once("connection.php"); $sSql = "INSERT INTO tblPlaces (Name, Street, HouseNumber, City, Category) VALUES ('" . $mysqli -> real_escape_string($this -> m_sName) . "', '" . $mysqli -> real_escape_string($this -> m_sStreet) . "', '" . $mysqli -> real_escape_string($this -> m_sHouseNumber) . "', '" . $mysqli -> real_escape_string($this -> m_sCity) . "', '" . $mysqli -> real_escape_string($this -> m_sCategory) . "')"; if (!$mysqli -> query($sSql)) { throw new Exception("Er is iets mis gelopen bij het toevoegen van een plaats"); } } public function placeAvailable() { include("connection.php"); $sSql= "select Street from tblPlaces where Street = '".$this->m_sStreet."' AND HouseNumber = '".$this->m_sHouseNumber."'"; $vResult=$mysqli->query($sSql); if($vResult->num_rows>0) { return(false); } else { return(true); } $mysqli->close(); } } ?> </code></pre> <p>In my connection file I have this code:</p> <pre><code><?php $localhost = "localhost"; $user = //user hidden $password = //paswoord hidden $database = //database hidden $mysqli = new mysqli($localhost, $user, $password,$database); if ($mysqli->connect_error) { throw new Exception("Geen Databankconnectie"); } ?> </code></pre> <p>Does anyone have a solution? Or do you also want to see my ajax file and .php page? Thanks</p> <p>EDIT</p> <p>This is my add.php file (at least everything that matters)</p> <pre><code><?php $feedback = ""; include_once ('assets/classes/places.class.php'); if (isset($_POST['knop'])) { try { $place1 = new Places; $place1 -> Name = $_POST['Name']; $place1 -> Street = $_POST['Street']; $place1 -> HouseNumber = $_POST['HouseNumber']; $place1 -> City = $_POST['City']; $place1 -> Category = $_POST['Category']; if($place1->placeAvailable()) { $place1 -> addPlaces(); $feedback = $place1 -> Name . ", is met succes toegevoegd!"; } else { $feedback = "Sorry"; } } catch (Exception $e) { $feedback = $e -> getMessage(); } } ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8" />  <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" /> <title>Search | FoodSquare</title> <link rel="stylesheet" href="assets/css/reset.css" /> <link rel="stylesheet" href="assets/css/style.css" /> <script type="text/javascript" src="assets/javascript/geolocation.js"></script> <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script> <meta name="description" content="" /> <meta name="viewport" content="width=device-width; initial-scale=1.0" />  <link rel="shortcut icon" href="/favicon.ico" /> <link rel="apple-touch-icon" href="/apple-touch-icon.png" /> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("#klik").click(function(){ var naam = $("#naam").val(); var plaats = $("#straat").val(); var nummer = $("#huisnummer").val(); var block = "block"; $.ajax({ type: "POST", url: "assets/ajax/check_place.php", data: { eet:naam, place:plaats, number:nummer } }).done(function( msg ) { if(msg.status != "error") { if(msg.available == "yes") { $(".feedback").fadeOut(); } else { $(".feedback span").text(msg.message); $(".feedback").fadeIn(); $(".feedback").css("display", block); } } }); return(false); }) }); </script> </head> <body> </code></pre> <p>This is my ajax FILE</p> <pre><code><?php ini_set('display_errors', 1); include_once('../classes/places.class.php'); try { $oPlace = new Places(); $oPlace->Name = $_POST['eet']; $oPlace->Street = $_POST['place']; $oPlace->HouseNumber = $_POST['number']; if($oPlace->placeAvailable()) { $feedback['status'] = "success"; $feedback['available'] = "yes"; $feedback["message"] = "Go ahead, street is available"; } else { $feedback['status'] = "success"; $feedback['available'] = "no"; $feedback["message"] ="De zaak " . "'" . $_POST['eet'] . "'". " is reeds op dit adres gevestigd." ; } } catch(exception $e) { $feedback['status'] = "error"; $feedback["message"] =$e->getMessage(); } header('Content-type: application/json'); echo json_encode($feedback); ?> </code></pre> <p>Ok guys, I tried several of your solutions but none of them works. I will probably be my fault but I figured something out and i replaced the INCLUDE_ONCE by INCLUDE and now i can add something to my database BUT only without AJAX, when i use AJAX, the form checks if the values are already in the database but when I add them, nothing happens. I also receive no errors. I also receive the right information back from ajax. Can anyone help please? thank you</p>
Tags
<php><ajax><mysqli>
Title
Notice: Undefined variable: mysqli in... on line 63
singulars
PostAcceptedAnswerId
1. This table or related slice is empty.
PostParentId
1. This table or related slice is empty.
PostTypePostTypeId
1. PTQuestion
UserLastEditorUserId
1. USNiels
UserOwnerUserId
1. USNiels
plurals
PostLinksPostIdRelatedPostId
1. PL
  singulars
  LinkTypeLinkTypeId
  LTLinked
PostLinksRelatedPostIdPostId
1. This table or related slice is empty.
PostsAcceptedAnswerId
1. This table or related slice is empty.
PostsParentIdCreationDate
1. PO
  singulars
  PostTypePostTypeId
  PTAnswer
2. PO
  singulars
  PostTypePostTypeId
  PTAnswer
3. PO
  singulars
  PostTypePostTypeId
  PTAnswer
VotesPostIdCreationDate
1. This table or related slice is empty.
CommentsPostId

Querying!

Guidance

A row detail

Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

Related data can be related in a to-one or a to-many fashion. Captions of data related in a to-many fashion link to a list view showing a filtered view of the table.

Try moving around until you find a non-empty to-many entry and click on the label to get to one. You can move back to the root by clicking on the database name in the header.