data not inserted in table used php code and mysqli

Question

i did all right but not getting rquierd results. using php and mysqli commands i want to insert data. actually im making a CMS system as practice work. not sure how to do it please HELP!

some screen shots here

post page

this is error i got

code here

<!DOCTYPE html>
<html lang="en">
  <head>
    <meta charset="utf-8">
 <meta http-equiv="X-UA-Compatible" content="IE=edge">
 <meta name="viewport" content="width=device-width, initial-scale=1">
    <title>Personal Home Page</title>
    <!-- Bootstrap -->
 <link href="css/bootstrap.css" rel="stylesheet">
    <link href="css/custom.css" rel="stylesheet">
    <link rel="stylesheet" href="css/footer-distributed-with-address-and-phones.css">

 <!-- HTML5 shim and Respond.js for IE8 support of HTML5 elements and media queries -->
 <!-- WARNING: Respond.js doesn't work if you view the page via file:// -->
 <!--[if lt IE 9]>
    <script src="https://oss.maxcdn.com/html5shiv/3.7.2/html5shiv.min.js"></script>
    <script src="https://oss.maxcdn.com/respond/1.4.2/respond.min.js"></script>
  <![endif]-->
  </head>
  <body>

  <div class="jumbotron-2">
    <h1 style="text-align:left">Write your New Post</h1>
  <form class="form-horizontal" method="post" action="new_post.php" enctype="multipart/form-data">
    <div class="form-group">
        <label for="inputEmail3" class="col-sm-2 control-label">Post Title*:</label>
        <div class="col-sm-10">
            <input type="name" class="form-control" name="title" placeholder="Title of Post">
        </div>
    </div>
    <div class="form-group">
        <label for="inputPassword3" class="col-sm-2 control-label">Post Author*</label>
        <div class="col-sm-10">
            <input type="name" class="form-control" name="author" placeholder="Published By">
        </div>
    </div>

    <div class="form-group">
        <label for="inputPassword3" class="col-sm-2 control-label">File(img/vid:)*</label>
        <div class="col-sm-10">
            <input type="file" name="image" class="form-control" placeholder="image or video file">
        </div>
    </div>
    <div class="form-group">
        <label for="inputPassword3" class="col-sm-2 control-label">Post Content:*</label>
        <div class="col-sm-10">
           <textarea rows="15" cols="100"  name="content" class="form-control" placeholder="content goes here"></textarea>
        </div>
    </div>
    
    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-10">
            <button type="submit" name="submit" class="btn btn-success btn-lg" style="align-items:center">Publish</button>
            <button type="reset" class="btn btn-danger btn-lg" style="align-items:center">Cancle</button>
        </div>
    </div>
</form>
  
</div>


<div style="background-color:#000000">
  <div  style="color:#FFFFFF" class="btn btn-danger btn-lg"><a href="backend.php">Backend</a></div>
  </div>
  
 <!-- jQuery (necessary for Bootstrap's JavaScript plugins) --> 
 <script src="js/jquery-1.11.2.min.js"></script>

 <!-- Include all compiled plugins (below), or include individual files as needed --> 
 <script src="js/bootstrap.js"></script>
  </body>
</html>

<?php
include("includes/connect.php");

if (isset($_POST['submit']))
{
  $title=$_POST['title'];
  
  $author=$_POST['author'];
  $content=$_POST['content'];
  $image_name = $_FILES['image']['name'];
  $image_type = $_FILES['image']['type'];
     $image_size = $_FILES['image']['size'];
     $image_tmp = $_FILES['image']['tmp_name'];
   
   
   if ($title=='' or $author=='' or $content==''){
    echo "<script>alert('Do not let any field empty')</script>";
     exit();
   }
  if($image_type=="image/jpeg" or $image_type=="image/png" or $image_type=="image/gif"){
   
   if($image_size<=5000000){
    move_uploaded_file($image_tmp, "img/$image_name");
   }
  else{
   echo "<script>alert('image is greater')</script>";
  }
  }
  else{
   echo "<script>alert('image type is invalid')</script>";
  }
  
  $sql="INSERT INTO posts(post_title, post_author, post_img, post_content) VALUES($title,$author,$image_name,$content)";
  $link = mysqli_connect("localhost", "root", " ", "firstwebsitedb");
  if (mysqli_query($link,$sql)){
   echo "<script>alert('Post is Published')</script>";
  }
  else{
   echo "<script>alert('Post is not Published')</script>";
  }
}
?>

$sql="INSERT INTO posts(post_title, post_author, post_img, post_content) VALUES($title,$author,$image_name,$content)"; $link = mysqli_connect("localhost", "root", " ", "firstwebsitedb"); — Badr Rehman, Apr 18 '17 at 07:11
check this code in coding of page i have put it already i think. @shafiqul Islam — Badr Rehman, Apr 18 '17 at 07:13

mageDev0688 · Answer 1 · 2017-04-18T07:14:38.917

0

Add the $link in your Query, below is insert statement:

$link = new mysqli_connect("localhost", "root", " ", "firstwebsitedb");

$sql="INSERT INTO posts(post_title, post_author, post_img, post_content) VALUES($title,$author,$image_name,$content)";

if ($link->query($sql)){
    echo "<script>alert('Post is Published')</script>";
}
else{
    echo "<script>alert('Post is not Published')</script>";
}

Hope this will help!!!

edited Apr 18 '17 at 07:14

answered Apr 18 '17 at 07:11

mageDev0688

566
4
27

He has already included a connect.php file. – LalaByte Apr 18 '17 at 07:12
Yes but if you this question and it has added connection string after the `$sql`. so that'w why I have added this line, – mageDev0688 Apr 18 '17 at 07:14
It doesn't matter whether he connects above or below a variable as long as he is not calling query function. – LalaByte Apr 18 '17 at 07:18

score 0 · Answer 2 · answered Apr 18 '17 at 07:27

0

screen shot Now its showing this error---sec-shot

answered Apr 18 '17 at 07:27

Badr Rehman

1
2

score -1 · Answer 3 · answered Apr 18 '17 at 07:15

$link = mysqli_connect("localhost", "root", " ", "firstwebsitedb");
if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}

if (mysqli_query($link,$sql)){
    echo "<script>alert('Post is Published')</script>";
}
else{
    echo "<script>alert('Post is not Published')</script>";
}

I think got problem in your database connection. You use " " as your database password. There is space there.

that a point.... i appreciate dude . thanx – Badr Rehman Apr 18 '17 at 07:20 — Badr Rehman, Apr 18 '17 at 07:20

data not inserted in table used php code and mysqli

3 Answers3