2012-09-19 57 views
-3

我有一段代碼(imageupload.php),其中php假設上傳文件到服務器並將數據插入數據庫。如何創建一個與此代碼關聯的html表格

<?php 
    /* check connection */ 
    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     die(); 
    } 

    $result = 0; 

    //UPLOAD IMAGE FILE 

    move_uploaded_file($_FILES["fileImage"]["tmp_name"], "ImageFiles/" . $_FILES["fileImage"]["name"]); 

    $result = 1; 

    //INSERT INTO IMAGE DATABASE TABLE 

    $imagesql = "INSERT INTO Image (ImageFile) VALUES (?)"; 

    if (!$insert = $mysqli->prepare($imagesql)) { 
     // Handle errors with prepare operation here 
    } 

    //Dont pass data directly to bind_param store it in a variable 
    $insert->bind_param("s", $img); 

    //Assign the variable 
    $img = 'ImageFiles/' . $_FILES['fileImage']['name']; 

    $insert->execute(); 

    $insertimagequestion->execute(); 

    //IF ANY ERROR WHILE INSERTING DATA INTO EITHER OF THE TABLES 
    if ($insert->errno) { 
     // Handle query error here 
    } 

    $insert->close(); 

    } 
    ?> 

但我想創建html表單來關聯這個,但我從來沒有在html中創建文件輸入。有誰知道如何創建可以與此相關的html表單嗎?

我想要一個html代碼,以便我可以測試上面的代碼,然後能夠測試任何錯誤,如果有的話。

+0

你可能會有所幫助:http://php.net/manual/en/features.file-upload.post-method.php – andrewsi

回答

0

可以使下面的HTML表單:

<form action="http://www.example.com/upload.php" 
enctype="multipart/form-data" method="post"> 
<p> 
Please specify a file, or a set of files:<br> 
<input type="file" name="fileImage" size="40"> 
</p> 
<div> 
<input type="submit" value="Send"> 
</div> 
</form> 
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