表单包括图片上传

Question

我尝试制作一个包含图像上传的表单。假设我有一个表participant，我想把它插入到:

INSERT INTO `participant`(`Matric`, `Name`, `IC`, `Address`, `Tel`, `Phone`, 
`Email`, `Phone_Ref`, `Institute`, `Course`, `Pic_Participant`, `Exp_Work`) 
VALUES ([value-1],[value-2],[value-3],[value-4],[value-5],[value-6],[value-7],
[value-8],[value-9],[value-10],[value-11],[value-12])

我要做的是插入数据并上传图像。它的属性是Pic_Participant。

我搜索关于使用ajax上传ajax图片上传和调整大小与jQuery和PHP。然后我想流程，填写表单然后在同一页面上传图像，然后上传图像后将图像数据发送到db，但表单还没有提交。我如何从表图像中获得属性以添加表参与者?

请帮帮我。我是新手。

---

编辑

我尝试这个代码，但得到一个错误:Undefined变量

 <?php
session_start();
include 'dbconnect.php';
function is_valid_type($file)
{
$valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif");
if (in_array($file['type'], $valid_types))
return 1;
return 0;
}

function showContents($array)
{
echo "<pre>";
print_r($array);
echo "</pre>";
}

$TARGET_PATH = "upload/";
//ERROR START HERE
$Matric = $_POST['Matric'];
$Name = $_POST['Name'];
$IC = $_POST['IC'];
$Address = $_POST['Address'];
$Tel = $_POST['Tel'];
$Phone = $_POST['Phone'];
$Email = $_POST['Email'];
$Phone_Ref = $_POST['Phone_Ref'];
$Institute = $_POST['Institute'];
$Course = $_POST['Course'];
/* $fname = $_POST['fname'];
$lname = $_POST['lname']; */
$image = $_FILES['image'];
$Exp_Work =$_POST['Exp_Work'];
//ERROR END HERE

$Matric = mysql_real_escape_string($Matric);
$Name = mysql_real_escape_string($Name);
$IC = mysql_real_escape_string($IC);
$Address = mysql_real_escape_string($Address);
$Tel = mysql_real_escape_string($Tel);
$Phone = mysql_real_escape_string($Phone);
$Email = mysql_real_escape_string($Email);
$Phone_Ref = mysql_real_escape_string($Phone_Ref);
/* $Total_sales = addslashes($_POST['Total_sales']);
$Date = addslashes($_POST['Date']); */
/* $Cer_name = mysql_real_escape_string($Cer_name); */
$Institute = mysql_real_escape_string($Institute);
$Course = mysql_real_escape_string($Course);
/* $Cat_name = addslashes($_POST['Cat_name']);
$Product_name = addslashes($_POST['Product_name']); */
/* $fname = mysql_real_escape_string($fname);
$lname = mysql_real_escape_string($lname); */
$image['name'] = mysql_real_escape_string($image['name']);
$Exp_Work = mysql_real_escape_string($Exp_Work);

$TARGET_PATH .= $image['name'];

if ( $Matric == "" ||$Name == "" ||$IC == "" ||$Address == "" ||$Tel == "" ||$Phone == "" ||$Email == "" ||$Phone_Ref == "" || $Institute == "" || $Course == ""|| $image['name'] == ""|| $Exp_Work == "" )
{
$_SESSION['error'] = "All fields are required";
echo "All fields are required";
exit;
}

if (!is_valid_type($image))
{
$_SESSION['error'] = "You must upload a jpeg, gif, or bmp";
echo"You must upload a jpeg, gif, or bmp";
exit;
}

if (file_exists($TARGET_PATH))
{
$_SESSION['error'] = "A file with that name already exists";
echo"A file with same name exists already";
exit;
}

if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))
{
$sql = "insert into participant (Matric, Name, IC, Address, Tel, Phone, Email, Phone_Ref, Institute, Course, image, Exp_Work) values ('$Matric','$Name','$IC','$Address','$Tel','$Phone','$Email','$Phone_Ref','$Institute', '$Course','" . $image['name'] . "','$Exp_Work')";
$result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());
echo"Imgage uploaded successfully";
exit;
}
else
{
$_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory";
header("Location: fail.php");
exit;
}
?>

Answer 1

当我读到你的问题时，你想要上传已经使用教程/代码找到的图像，然后在图像已经上传时单独提交表单。

要知道在文件上传后在哪里可以找到您的图像，您有两个选项:

1. 让图片上传在上传完成后返回文件路径，并在表单中包含该变量(例如，作为隐藏输入);
2. 将图像路径存储在会话变量中，以便在提交数据字段时，可以访问该变量以获取图像的信息。

编辑:你需要检查表单插件的文档了解更多细节，但你可以从你的上传php脚本返回一些东西。例如，您可以使用echo作为文件名，然后您可以在success函数中使用它:

    function afterSuccess(return_value)  {
        console.log(return_value);  // here you have what was echoed out by php
        $('#UploadForm').resetForm();  // reset form
        $('#SubmitButton').removeAttr('disabled'); //enable submit button
    }

Answer 2

如果我得到了正确的，你想要的是插入图像到(可能是MYSQL)数据库。

您可以通过使用图像二进制数据的base64_encode()并在数据库中插入结果明文来实现这一点。

$image = 'path/to/image/image.png';
$imagefordbs  =  base64_encode($image);
/*now your image is ready to be stored in database*/

然而，这种方法有缺点，因为base64_encode()比原始内存多占用约33%，一些时间来处理输入二进制，并且在mysql - BLOB中有专门的数据类型来满足这种需求。