使用HTML表單輸入和php更新MySQL

Question

我在頁面打開時以HTML表單輸入形式輸入MYSQL傳入數據。 更改信息后按更新按鈕時，更新操作成功，但是數據未更新。

<?php


$servername = "";
$username = "";
$password = "";
$dbname = "";

$baglan = mysqli_connect($servername,$username,$password,$dbname) or die(myslqi_error());


$place = $_POST['n_place'];
$description = $_POST['n_description'];
$latitude = $_POST['lat'];
$longitude = $_POST['long'];
$kayitliyer = $_POST['search_new_places'];
$yetkiliad = $_POST['n_yetkiliad'];
$magazaad = $_POST['n_magazaad'];
$telefon = $_POST['n_telefon'];
$yetkilitelefon = $_POST['y_telefon'];
$derece = $_POST['derece'];
$country = $_POST['country'];


$sql =  "UPDATE tbl_places SET place='$place',description='$description',lat='$latitude', lng='$longitude', kayitliyer='$kayitliyer',
yekiliad='$yetkiliad', magazaad='$magazaad', telefon='$telefon', yetkilitelefon='$yetkilitelefon', derece='$derece' WHERE place_id='$gID' ";



if (mysqli_query($baglan,$sql)) {
  echo '<br/><div class="container"><div class="alert alert-success" role="alert">
              Kayıt Güncelleme <b>Başarılı :-)</b> <a href=javascript:history.back(-1)>Geri Dön</a>
             </div></div> ';
}else {
  echo '<br/><div class="container"><div class="alert alert-danger" role="alert">
              Güncelleme İşlemi <b>Başarız!</b> <a href=javascript:history.back(-1)>Geri Dön</a>
             </div></div>'. mysqli_error($baglan);
}



?>

HTML頁面表單SELECT查詢。感謝您的答復。 我從HTML表單頁面獲得了“ $ gID”的值，在該頁面中，我使用SELECT查詢輸入了數據庫數據。

  <?php
 //UPDATE ID
   $gID = $_GET["place_id"];

   include ("baglanti.php");

 $sorgu = mysqli_query($baglan,"select * from tbl_places where place_id='$gID'");


  while($goster = mysqli_fetch_array($sorgu)){
  $grupadi = $goster["place"];
  $id = $goster["place_id"];
  $desc = $goster["description"];
  $adres = $goster["kayitliyer"];
  $yetkiliad = $goster["yekiliad"];
  $magazaad = $goster["magazaad"];
  $telefon = $goster["telefon"];
  $yetkilitelefon = $goster["yetkilitelefon"];
  $derece = $goster["derece"];
  $country = $goster["country"];
  $lat =$goster["lat"];
  $long = $goster["lng"];
  }
 ?>
<!DOCTYPE html>
<html lang="tr">
<head>

Answer 1

您應該為此使用PDO ，以確保安全和代碼干凈。 除此之外，由於沒有更多關於$gID來源的信息，我會嘗試刪除與其比較的值周圍的單引號，因為它可能是整數而不是字符串：

$sql =  "... WHERE place_id=$gID";

使用HTML表單輸入和php更新MySQL

問題描述

1 個解決方案

解決方案1
-3 2018-02-22 09:50:27

使用HTML表單輸入和ph​​p更新MySQL

問題描述

1 個解決方案

解決方案1 -3 2018-02-22 09:50:27

使用HTML表單輸入和php更新MySQL

解決方案1
-3 2018-02-22 09:50:27