Question

我有一个画布，我需要将其保存到目录并将URL存储在数据库中。

当我保存文件而不将URL存储在数据库中时，它可以正常工作，反之亦然。

然而，当我把两者放在一起并通过AJAX指定PHP文件时，由于某种原因它无法识别会话变量？

当我尝试在AJAX上调用“成功”时，没有任何显示。我没有回复。

这可能是一个简单的解决方案！我想我已经盯着这段代码太久了。

JavaScript的：

function doodleSave() {
  var canvas = document.getElementById("doodle-canvas");
  var canvasData = canvas.toDataURL("image/png");

  $.ajax({
    url:'doodleupload.php', 
    type:'POST', 
    data:{ data:canvasData },
    success: function(response){
        alert(response);
        //echo what the server sent back...
    }
  });
}

PHP：

<?php
  session_start();

  /* AUTOMATED VARIABLES */
  $url             = md5(uniqid(rand(), true));
  $unique_user_id  = $_SESSION['unique_user_id'];
  $unique_post_id  = md5(uniqid(rand(), true));
  $timestamp       = time();
  $nature          = "doodle";
  $imageUrl        = $upload_dir.$url.'.png';

  $upload_dir = "images/external/doodles/";
  $img = $_POST['data'];
  $img = substr($img,strpos($img,",")+1);
  $data = base64_decode($img);
  $file = $upload_dir . $url . ".png";
  $success = file_put_contents($file, $data);

  echo $success ? $file : 'Unable to save the file.';

  require_once 'php/connect.php';

  try
  {

    $stmt = $pdo->prepare("INSERT INTO posts (unique_user_id, unique_post_id, nature, image_url, timestamp) VALUE (:unique_user_id, :unique_post_id, :nature, :image_url, :timestamp)");
    $stmt->bindParam(":unique_user_id",$unique_user_id);
    $stmt->bindParam(":unique_post_id",$unique_post_id);
    $stmt->bindParam(":nature",$nature);
    $stmt->bindParam(":image_url",$imageUrl);
    $stmt->bindParam(":timestamp",$timestamp);

    if($stmt->execute())
    {
      echo "File in database";
    }
    else
    {
      echo "Not in database";
    }
  }
  catch(PDOException $e){
    echo $e->getMessage();
  }

?>

Answer 1

将$upload_dir移到顶部，就像在初始化之前调用它一样。

$upload_dir = "images/external/doodles/";
$url             = md5(uniqid(rand(), true));
$unique_user_id  = $_SESSION['unique_user_id'];
$unique_post_id  = md5(uniqid(rand(), true));
$timestamp       = time();
$nature          = "doodle";
$imageUrl        = $upload_dir.$url.'.png';

Canvas不会保存服务器端PHP

1 个答案: