PHP - Can't use the fetched data

Question

Hello monsters of programming, Good day! What i want to do is, when the user click the file button and upload the image it will update, else if the user didn't change it just echo that image. But when the user didn't change the image, im having an error **Warning**: file_get_contents(): Filename cannot be empty in C:\\xampp\\htdocs\\studentportal\\edit2.php on line 27 Can someone help me? The if(isset($_FILES['image'])) is working properly but the else statement is not. How can i just echo that image if the user didn't change it? Im new to php and starting to learn it please give me ideas.

this is cause of the error the $newsimages = $row['news_image']; in the else statement.

else{
  $title = $_POST['titles'];
  $date = $_POST['dates'];
  $content = $_POST['contents'];
  $newsimages = $row['news_image'];
  $sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimages' WHERE news_id = '$newsid'";
  mysqli_query($con, $sql);
  echo "oh it worked again ";
}

this is all of the php code

<?php

include_once('connection.php');

 $newsid = $_GET['news_id'];

    if(isset($_POST['esubmit'])){
        /* create a prepared statement */
        if ($stmt = mysqli_prepare($con, "SELECT * FROM news WHERE news_id = ? LIMIT 1")) {
            /* bind parameters */
            mysqli_stmt_bind_param($stmt, "s", $newsid);

            /* execute query */
            mysqli_stmt_execute($stmt);

            /* get the result set */
            $result = mysqli_stmt_get_result($stmt);

            /* fetch row from the result set */
            $row = mysqli_fetch_array($result);
        }
    }

    if(isset($_POST['update'])){

        if(isset($_FILES['image'])){
          $file=$_FILES['image']['tmp_name'];
          $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
          $image_name= addslashes($_FILES['image']['name']);
          move_uploaded_file($_FILES["image"]["tmp_name"],"img/" . $_FILES["image"]["name"]);
          $newsimage="img/" . $_FILES["image"]["name"];

          $title = $_POST['titles'];
          $date = $_POST['dates'];
          $content = $_POST['contents'];
          $sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimage' WHERE news_id = '$newsid'";
          mysqli_query($con, $sql);
          echo "oh it worked ";
        }
        else{
          $title = $_POST['titles'];
          $date = $_POST['dates'];
          $content = $_POST['contents'];
          $newsimages = $row['news_image'];
          $sql ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content', news_image ='$newsimages' WHERE news_id = '$newsid'";
          mysqli_query($con, $sql);
          echo "oh it worked again ";
        }

    }

?>
<!DOCTYPE HTML>
<html>
<head>
</head>
<body>

<?php

    if(isset($_POST['esubmit'])){
        ?>

        <form method="post" action ="edit2.php?news_id=<?php echo $row['news_id']; ?>" enctype="multipart/form-data">
            Title<input type ="text" name ="titles" value="<?php echo $row['news_title']; ?>"/><br>
            Date<input type ="text" name="dates" value="<?php echo $row['news_date']; ?>" /><br>
            Content<textarea name="contents"><?php echo $row['news_content']; ?></textarea>
            <input class="form-control" id="image" name="image" type="file" accept="image/*" onchange='AlertFilesize();'/>
            <img id="blah" src="<?php echo $row['news_image']; ?>" alt="your image" style="width:200px; height:140px;"/>

            <input type="submit" name="update" value="Update" />
        </form>

        <?php
    }

?>

<script src="js/jquery-1.12.4.min.js"></script>
<script src="js/bootstrap.min.js"></script>
<script type="text/javascript">
    function readURL(input) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();

            reader.onload = function (e) {
                $('#blah').attr('src', e.target.result);
            }

            reader.readAsDataURL(input.files[0]);
        }
    }

    $("#image").change(function(){
        readURL(this);
    });
    </script>
</body>
</html>

Answer 1

What you can do is remove <img></img> out of the first form and put it in separate <form> with separate submit button and add more php code to just update image only.You will have two forms and two updates like

 $sql1 ="UPDATE news SET news_title ='$title', news_date ='$date', news_content = '$content' WHERE news_id = '$newsid'";
$sql2="update new SET new_image='$newsimages' WHERE new_id='$newsid'";

I Hope You undestand.I have done same thing for one of my websites.I think this is the best solution.Try it.For further query you can comment.

PHP - Can't use the fetched data

Question

1 answers

solution1
1 ACCPTED 2016-07-01 05:12:31

PHP - Can't use the fetched data

Question

1 answers

solution1 1 ACCPTED 2016-07-01 05:12:31

solution1
1 ACCPTED 2016-07-01 05:12:31