How can I post form to mysqli without using an action page

Question

I don't wanna use an action page. I wanna post on the same page.

These are my codes. But there is an error called "Undefined index: baslik on line 5 and aciklama on line 6

<?php

include("baglan.php");

$site_basligi = mysqli_real_escape_string($conn, $_POST['baslik']);
$site_aciklamasi = mysqli_real_escape_string($conn, $_POST['aciklama']);

$ayarsql = "UPDATE ayar SET baslik='$site_basligi', aciklama='$site_aciklamasi' WHERE durum='1'";

if($conn->query($ayarsql)){
echo "Güncelleme başarılı";
}
?>

<form action="" method="post">

Site Başlığı: <input type="text" name="baslik" ><br><br>
Açıklama: <input type="text" name="aciklama"><br><br>

<input type="submit">

</form>

just leave it like that, but you should be checking if your the data is available from the `$_POST`, use `isset($_POST)` before doing your DB insert or so — Roljhon, Apr 01 '17 at 18:05

Hossam · Answer 1 · 2017-04-01T18:11:54.127

0

I can see the form IS submitting to the same page. You just should wrap the PHP code in if condition

<?php
include("baglan.php");
if(!empty($_POST['baslik'])) { // This is the new if condition
    $site_basligi    = mysqli_real_escape_string($conn, $_POST['baslik']  );
    $site_aciklamasi = mysqli_real_escape_string($conn, $_POST['aciklama']);
    $ayarsql = "UPDATE ayar SET baslik='$site_basligi', aciklama='$site_aciklamasi' WHERE durum='1'";
    if($conn->query($ayarsql)) {
        echo "Güncelleme başarılı";
    }
}
?>
<form action="" method="post">
    Site Başlığı: <input type="text" name="baslik" ><br><br>
    Açıklama:     <input type="text" name="aciklama"><br><br>
    <input type="submit">
</form>

edited Apr 01 '17 at 18:11

answered Apr 01 '17 at 18:05

Hossam

1,126
8
19

you will still be getting the undefined error from your code above. – Roljhon Apr 01 '17 at 18:08
@Roljhon : http://stackoverflow.com/questions/4559925/why-check-both-isset-and-empty – Hossam Apr 01 '17 at 18:09
Thank you so much, the problem resolved. – Samet Çiftci Apr 01 '17 at 18:14
@SametÇiftci You're welcome – Hossam Apr 01 '17 at 18:15
I could mark it correct. But It's a new account. :) – Samet Çiftci Apr 01 '17 at 18:20

score 0 · Answer 2 · answered Apr 01 '17 at 18:05

0

<?php

include("baglan.php");
// check for post request here 
if(isset($_POST['baslik']) && $_POST['baslik'] != "") {
$site_basligi = mysqli_real_escape_string($conn, $_POST['baslik']);
$site_aciklamasi = mysqli_real_escape_string($conn, $_POST['aciklama']);

$ayarsql = "UPDATE ayar SET baslik='$site_basligi', aciklama='$site_aciklamasi' WHERE durum='1'";

if($conn->query($ayarsql)){
echo "Güncelleme başarılı";
}
}
?>

<form action="" method="post">

Site Başlığı: <input type="text" name="baslik" ><br><br>
Açıklama: <input type="text" name="aciklama"><br><br>

<input type="submit">

</form>

answered Apr 01 '17 at 18:05

Sudhakar

324
3
9

hope this will work – Sudhakar Apr 01 '17 at 18:08
Thank you so much, resolved. :) – Samet Çiftci Apr 01 '17 at 18:13
@SametÇiftci, if this resolved your error, please mark it correct – Sudhakar Apr 01 '17 at 18:17
I wish, but I can't. It's a new account. :) – Samet Çiftci Apr 01 '17 at 18:20

score 0 · Answer 3 · answered Apr 01 '17 at 18:06

0

You are not checking if the POST data actually exists, so when you first load the page there is no $_POST

do something like

if (isset($_POST)){

//your php code here

}

answered Apr 01 '17 at 18:06

Luke143

16
2

How can I post form to mysqli without using an action page

3 Answers3