Question

the interface of uploading recipe这是我的代码，用于上传配方（new.php），一旦单击“提交”按钮，它将被上传至mysql。我试图通过将它们更改为mysqli来检查是否有任何mysql代码已过时。因为这个项目有点旧。因此，在其他页面上，我将许多mysql更改为mysqli。

<form method="post" action="new.php" enctype="multipart/form-data" class="form3">


***I'm suspecting that the mistake is from here. As all the other parts of the code seem to be okay to me.***


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

     $title = $_POST['title'];
     $date = date('Y.m.d');
     $author = $_POST['author'];
     $desc = $_POST['desc'];
     $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 $desc ==''){
        echo"<script>alert('Some Field/fields is/are Empty')</script>";
        exit();
    }
    if($image_type=="image/jpeg" or $image_type=="image/png" or   $image_type=="image/gif"){

        if($image_size<=50000){
            move_uploaded_file($image_tmp,"uploads/$image_name");
        }
        else{
            echo"<script>alert('Image is Larger, Only 50kb size is   allowed')</script>";
        }
    }
    else{
        echo"<script>alert('Image Type is Invalid')</script>";
    }
    $query = "insert into    new_recipe(post_title,post_date,post_author,post_image,post_desc
    ) values('$title','$date','$author','$image_name','
    $desc')";

    if (mysqli_query($con,$query)){
        echo "<center><h1>Recipe Has Been Submitted!</h1></center>";
    }
}
 **Once all the details have been typed. When I click on submit , it doesn't give me any error/s. There were some previous recipes uploaded before I face these errors, and I can see them perfectly under the view.php page.**
?>
<?php } ?>

PHP POST方法问题，上传配方

0 个答案: