如何使用PHP将照片插入MySQL数据库？

Question

Good day... I have to insert a photo into MySQL database,,,, But this message appears 美好的一天...我必须将一张照片插入MySQL数据库，,,,但是此消息出现

Notice: Undefined index: fileToUpload in C:\\wamp\\www\\great\\index.php on line 41 注意：未定义的索引：第41行的C：\\ wamp \\ www \\ great \\ index.php中的fileToUpload

Line 41: 第41行：

  GetSQLValueString($_POST['fileToUpload'], "int"));

 <?php $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION); // Check if image file is a actual image or fake image if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false) { echo "File is an image - " . $check["mime"] . "."; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } // Check if file already exists if (file_exists($target_file)) { echo "Sorry, file already exists."; $uploadOk = 0; } // Check file size if ($_FILES["fileToUpload"]["size"] > 500000) { echo "Sorry, your file is too large."; $uploadOk = 0; } // Allow certain file formats if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) { echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed."; $uploadOk = 0; } // Check if $uploadOk is set to 0 by an error if ($uploadOk == 0) { echo "Sorry, your file was not uploaded."; // if everything is ok, try to upload file } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) { echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded."; } else { echo "Sorry, there was an error uploading your file."; } } ?>

 <?php require_once('Connections/greatCONN.php'); ?> <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") { if (PHP_VERSION < 6) { $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue; } $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue); switch ($theType) { case "text": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "long": case "int": $theValue = ($theValue != "") ? intval($theValue) : "NULL"; break; case "double": $theValue = ($theValue != "") ? doubleval($theValue) : "NULL"; break; case "date": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "defined": $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue; break; } return $theValue; } } $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form")) { $insertSQL = sprintf("INSERT INTO great (id) VALUES (%s)", GetSQLValueString($_POST['fileToUpload'], "int")); mysql_select_db($database_greatCONN, $greatCONN); $Result1 = mysql_query($insertSQL, $greatCONN) or die(mysql_error()); } ?> <!DOCTYPE html> <html> <body> <form name="form" action="<?php echo $editFormAction; ?>" method="POST" enctype="multipart/form-data"> Select image to upload: <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> <input type="hidden" name="MM_insert" value="form"> </form> </body> </html>

enter image description here 在此处输入图片说明

I'll be very thankful for helping me :) 我将非常感谢您的帮助:)

Answer 1

选择数据类型varchar而不是整数来上传照片。

Answer 2

如果是任何文件，应该有$ _FILES ['fileToUpload']

Answer 3

使用$_FILES['fileToUpload']代替$_POST['fileToUpload']

如何使用PHP将照片插入MySQL数据库？

问题描述

3 个解决方案

解决方案1
0

解决方案2
0 2016-01-05 06:08:03

解决方案3
0 2016-01-05 06:19:27

如何使用PHP将照片插入MySQL数据库？

问题描述

3 个解决方案

解决方案1 0

解决方案2 0 2016-01-05 06:08:03

解决方案3 0 2016-01-05 06:19:27

解决方案1
0

解决方案2
0 2016-01-05 06:08:03

解决方案3
0 2016-01-05 06:19:27