0

好吧，我有一个难题。我希望能够将最多六个图像和 2 个文本字段（标题和描述）上传到数据库中......实际上让我纠正自己，我想将图像的名称存储在数据库中，这样我就可以得到稍后的照片。我知道如何将多张图片上传到一个文件夹，我知道如何将行插入数据库，但我不知道如何将两者结合起来。如果其中一个文件有问题，我将如何将其与图像工作结合起来，取消整个过程？

我的数据库设置只是 id | 标题 | 描述 | 图片1 | 图片2 | 图片3 | 图片4 | 图片5 | img6

到目前为止我写的代码是：

if (isset($_POST['formsubmitted'])) { //if form was submitted
$error = array();//Declare An Array to store any error message  

if (empty($_POST['title'])) {//if no name has been supplied 
    $error[] = 'Please enter a title for your post.';//add to array "error"
    $show_errors = 'show';
} else {
    $title = $_POST['title'];//else assign it a variable
}



if (empty($_POST['desc'])) {
    $error[] = 'Please enter a short desc of your post.';//add to array "error"
    $show_errors = 'show';
} else {
    $desc = $_POST['desc'];//else assign it a variable
}



    if (empty($error)){ //if no error, insert into db

$new_post = "INSERT INTO `posts` ( `title`, `desc`) VALUES ( '$title', '$desc')";
    $result = mysql_query($new_post ) or die(mysql_error('error inserting post'));

}


}

那么html是：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php if (isset($show_errors)) { 
    //show user the errors if form cant be submitted
    echo '<div> <ol>';
        foreach ($error as $key => $values) { echo '    <li>'.$values.'</li>'; }
        echo '</ol></div><br />'; }?>

<br />
<form method="post" id="newpost" action="" enctype="multipart/form-data">
<div><input name="title" type="text" value="" class="title_input"></div>
<div><textarea id="area4" cols="40" rows="5" name="desc" class="desc_texbox"></textarea></div>
<div><input type="file" name="images1"></div>
<div><input type="file" name="images2"></div>
<div><input type="file" name="images3"></div>
<div><input type="file" name="images4"></div>
<div><input type="file" name="images5"></div>
<div><input type="file" name="images6"></div>
<input type="hidden" name="formsubmitted" value="TRUE" />
<input type="submit" id="upload" value="Upload">
</form>
</body>
</html>
4

2 回答 2

1

这绝不是完成，但它应该让你朝着正确的方向前进。请注意，我是根据更改文件字段来重命名的，<input type="file" name="images[1]" />以便将它们作为数组处理。

<?php

error_reporting(E_ALL);
ini_set('display_errors', true);

$db = new PDO('mysql:dbname=test;host=127.0.0.1', 'user', 'pass');


if (isset($_POST['formsubmitted'])) { //if form was submitted

    $error_array = array();//Declare An Array to store any error message

    // check all images for upload errors
    // you should probably add extra file validation here - check image type etc
    $upload_error = false;
    foreach($_FILES['images']['error'] as $error) {
        if ($error != 0 && $error != 4) {
            $upload_error = true;
        }
    }

    if ($upload_error) {
        $error_array[] = 'One of the image uploads failed!';
    }


    if (empty($_POST['title'])) {//if no name has been supplied
        $error_array[] = 'Please enter a title for your post.';//add to array "error"
    } else {
        $title = $_POST['title'];//else assign it a variable
    }

    if (empty($_POST['desc'])) {
        $error_array[] = 'Please enter a short desc of your post.';//add to array "error"
    } else {
        $desc = $_POST['desc'];//else assign it a variable
    }

    if (empty($error_array)){ //if no error, insert into db

        $new_post = "INSERT INTO `posts` ( `title`, `desc`) VALUES (?, ?)";
        $stmt = $db->prepare($new_post);
        $stmt->execute(array($title, $desc));

        $new_post_id = $db->lastInsertId();

        // now start processing the images
        $image_sql = "INSERT INTO `post_images` (`post_id`, `img_name`) VALUES (?, ?)";
        $stmt = $db->prepare($image_sql);

        for ($i = 1; $i <= 6; $i++) {

            // you need to add some code to vlaidate, move and rename the files

            // add the files to the db
            $file_name = $_FILES['images']['name'][$i];
            $stmt->execute(array($new_post_id, $file_name));

        }

    } else {
        print_r($error_array);
    }

}
?>
于 2012-04-14T16:49:03.723 回答
0

我觉得比较好，先把图片上传到服务器，如果 move_uploaded_file() 函数返回错误，在 $error 数组中添加一个值。这样，只有上传成功，数据才会插入数据库。

于 2012-04-14T16:10:49.687 回答