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I want to insert data to my DB, but the query didn't work. I was using this code in another page, but it work. And for this page it seems didn't work.

This my upload.php

<?php
include_once("connect.php");
//Fetching Values from URL
if (isset($_POST['submit'])) {
    $nama_program=$_POST['nama_program'];
    $deskripsi=$_POST['deskripsi'];
    $code=$_POST['link_youtube'];
    $link_youtube="<iframe width='420' height='315' src='https://www.youtube.com/embed/".$code."' frameborder='0' allowfullscreen></iframe>";
    $link_twitter=$_POST['link_twitter'];
    $rating=$_POST['rating'];
    $image=$_POST['uploafimage'];
    $target_dir = "images/";
    $target_file = $target_dir . basename($_FILES["uploadimage"]["name"]);

    $result=$mysqli->query("INSERT INTO program(nama, deskripsi, link_youtube, link_twitter, image, rating, slider, twtiter) VALUES('$nama_program','$deskripsi','$link_youtube','$link_twitter','$target_file','$rating', 0, 0)");
    if($result === TRUE){
        echo "Congrats Your data hase been saved";
        } else{
            echo"There is eror here".$result;
        }
} else{
    echo "Error";
    }
?>

This is my form look like

<form action="upload.php" method="post" enctype="multipart/form-data">
    <input type="text" class="form-control" id="nama_program" name="nama_program">
    <textarea class="form-control textarea" name="deskripsi" id="deskripsi"></textarea>
    <textarea class="form-control textarea" name="link_youtube" id="link_youtube"></textarea>
    <textarea class="form-control textarea" name="link_twitter" id="link_twitter"></textarea>
    <select name="rating" id="rating" class="form-control">
             <option>Rating</option>
             <option>1</option>
             <option>2</option>
             <option>3</option>
    <input type="file" name="uploadimage" id="uploadimage">
    <button type="submit" class="btn btn-info">
           submit
    </button>
</form>

What's wrong with mysqli ? I use this enctype="multipart/form-data" because i want to upload image, but first the query didn't work.

This is my connect.php

 <?php
 $host = "localhost";
 $user = "*****";
 $pass = "****";
 $dbnm = "*******";
 $mysqli = new mysqli($host, $user, $pass, $dbnm);
 if($mysqli->connect_errno > 0){
    die('Unable to connect to database [' . $mysqli->connect_error . ']');
}
?>
Wingke
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    Do you get any errors, is error reporting on? You should use prepared statements, http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?rq=1. – chris85 Sep 11 '15 at 03:25
  • What errors are you getting? – Blip Sep 11 '15 at 03:29
  • when i submit, the result in page echo "error" that's mean there is something wrong that make my query can't run? – Wingke Sep 11 '15 at 04:49
  • You get `error` because you haven't given the name attribute to your submit button so `if (isset($_POST['submit'])) {` is never met. This puts you into `} else{ echo "Error"; }` on every run. Nothing to do with `mysqli`, yet. You will be open to SQL injections once that is corrected.. – chris85 Sep 11 '15 at 13:20
  • Hey@chris85, i have been set, name to submit button. Then the result now is Echo "there is eror here". is that mean my query wrong at some point ? – Wingke Sep 14 '15 at 02:25

1 Answers1

0
mysqli_query($yourcon,"INSERT INTO program(nama, deskripsi, link_youtube, link_twitter, image, rating, slider, twtiter) VALUES ('$nama_program','$deskripsi','$link_youtube','$link_twitter','$target_file','$rating', 0, 0)");

Give more Explanation about question and errors you getting for batter solution.

Gaurang Joshi
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