Question

由于某种原因，我的800x800px图像被修剪了。但是，当我使用80x80px时，它似乎没有被修剪。我该如何解决？它还说我在SQL中有一个斑点-64kib。如何增加限额？仅显示图像的20％。解决了，我用blob代替了mediumblob。它不是重复的，因为与What is the maximum length of data I can put in a BLOB column in MySQL?的问题不同，但是它们都有相同的解决方案。

if(isset($_POST["insert"]))
{
     $file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
     $query = "UPDATE users SET image='$file' WHERE email= '$param_email'";
     if(mysqli_query($connect, $query))
     {
echo '<script>alert("Image Inserted into Database")</script>';
     }
}
?>
Html:
               <form method="post" enctype="multipart/form-data">
                    <input type="file" src=
                    <?php
                            $query = "SELECT image FROM users where email='$param_email'";
                            $result = mysqli_query($connect, $query);
                            while($row = mysqli_fetch_array($result))
                            {echo'"data:image/jpeg;base64,'.base64_encode($row['image'] ).'"';}?> name="image" id="image" />
                    <input type="submit" name="insert" id="insert" value="Insert" class="btn btn-info" />
               </form>

               <table>
               <?php
               $query = "SELECT image FROM users where email='$param_email'";
               $result = mysqli_query($connect, $query);
               while($row = mysqli_fetch_array($result))
               {
                    echo '
                         <tr>
                              <td>
                                   <img src="data:image/jpeg;base64,'.base64_encode($row['image'] ).'" height="200" width="200" class="img-thumnail" />
                              </td>
                         </tr>
                    ';
               }
               ?>
               </table>
JS:
<script>
$(document).ready(function(){
     $('#insert').click(function(){
          var image_name = $('#image').val();
          if(image_name == '')
          {
               alert("Please Select Image");
               return false;
          }
          else
          {
               var extension = $('#image').val().split('.').pop().toLowerCase();
               if(jQuery.inArray(extension, ['gif','png','jpg','jpeg']) == -1)
               {
                    alert('Invalid Image File');
                    $('#image').val('');
                    return false;
               }
          }
     });
});
</script>

谢谢大家！请也分享此内容，以便可以更快地解决它：）

高质量的图像输出在php和mysql中被修剪

0 个答案: