php - 表單包含圖片上傳

Question

我嘗試製作包含圖片上傳的表單。讓說，我有表participant我想插入到：

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]) 

我想要做的是插入數據並上傳圖片。它的屬性是Pic_Participant。

我使用ajax Ajax Image Upload and Resize with jQuery and PHP搜索關於上傳。然後我認爲流程，填寫表單，然後在同一頁面上傳圖像，然後上傳圖像後，圖像發送到數據庫的數據，但表單尚未提交。如何從表格圖像中獲取屬性以添加到表格參與者中？

請幫幫我。我對此很新。

---

EDIT

我嘗試這種代碼，但得到一個錯誤：未定義的變量

<?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

如果我得到它的正確，你想要的是插入圖像插入（可能是MYSQL）數據庫。

您可以通過使用base64_encode()圖像二進制數據並將結果明文插入數據庫來實現此目的。

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

然而，該方法具有由於BASE64_ENCODE缺點（）佔用了更多的約33％，那麼原始存儲器，一些時間來處理輸入二進制，並且對於這種要求在MySQL專用數據類型 - BLOB。

Answer 2

在閱讀您的問題時，您想要使用您找到的教程/代碼上傳圖像，然後在圖像已經上傳後單獨提交表單。

要知道你在哪裏可以將文件上傳後，找到你的形象，你有兩個選擇：

1. 具有上載完成，包括當圖像上傳回的文件路徑，在您的表單變量（如例如一個隱藏的輸入）;
2. 將圖像路徑存儲在會話變量中，以便當您提交數據字段時，可以訪問該變量以獲取圖像的信息。

編輯：您需要檢查表單插件的文檔獲取更多細節，但你可以從你上傳的PHP腳本返回的東西。例如，您可以echo的文件名，然後你會在你的success功能可供選擇：

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 
    }