php - 使用PHP将文件路径名保存到mySQL数据库

Question

抱歉，如果这个问题真的很啰嗦，但我想尽可能详细地回答我的问题。

基本上我有一个多部分表单，我想向其中添加文件上传功能。我创建了表单，一个用于将图像保存到目录的函数，还有一个用于处理将表单值发布到数据库的脚本。我的问题是，我现在在提交表单时将图像存储在我想要的位置，但没有任何表单值被发送到数据库。我认为我的问题是我试图将新路径名存储为一个变量，然后我在 POST 脚本中调用它，但我认为这是错误的？

抛出的错误是:

注意:未定义索引:C:\xampp\htdocs\web_design_cms\create_wireframe.php 第13行的图片

表单代码如下:

<form action="create_wireframe.php" method="post" enctype="multipart/form-data">


        <!-- page_title -->



        <p>
            <label>Title:</label><br/>
            <input type="text" class="text small" name="wireframe_title" id="wireframe_title" value=""  />
            <span class="note">*required</span>
        </p>    

        <!-- page_meta_title -->


        <p>
            <label>Browser Title:</label><br/>
            <input type="text" class="text small" name="browser_title" id="browser_title" value=""  >
            <span class="note">*required</span>
        </p>

        <!-- url_key -->            



        <p>
            <label>Permanent Link:</label><br/>
            <input type="text" class="text small" name="url_key" id="url_key" value=""/>
        </p>

        <!-- page_image -->

        <div style="float:left" >
            <!-- wireframe_type -->

            <p>
                <label>Type:</label><br/>
                <select name="wireframe_type" id="wireframe_type" class="styled" style="width:240px">
                    <option value="design"  > Design Draft</option>
                    <option value="wireframe" selected > Wireframe</option>                          
                </select>
                <span class="note">*required</span>
            </p>                                            

        </div>

        <div style="clear:both"></div>

        <div class="message info"><p>
            Allowed file types for upload: jpg,jpeg,gif,png.<br/>
            Max file size: 10Mb<br/>
            Picture size: 4096x4096 px
        </p></div>


        <p>
            <label>Upload Image:</label><br/>                   
            <input type="file" name="page_main_image" id="page_main_image" value=""  />
        </p>


        <!-- page_bg_color -->
        <p>
            <label>Color:</label><br/>
            <input type="text" class="text small" maxlength="6" size="6" style="width:60px" id="colorpickerField"  name="page_bg_color" value="ffffff" />
            <span id="colorSelector" style="background-color:#ffffff;padding:7px 10px;">&nbsp;</span>
            <span class="note">*required</span>
        </p>


        <p>
            <input type="submit" class="submit small" value="Save" name="submit" />
        </p>
     </form>

这是处理表单数据的 php 脚本“create_wireframe.php”:

<?php require_once("includes/db_connection.php"); ?>
<?php require_once("includes/functions.php"); ?>

<?php
if (isset($_POST['submit'])) {
    //Process the form
    $image = upload_file();

    $wireframe_title = mysql_prep($_POST["wireframe_title"]);
    $browser_title = $_POST["browser_title"];
    $url_key = $_POST["url_key"];
    $wireframe_type = $_POST["wireframe_type"];
    $image = $_POST["page_main_image"];
$page_bg_color = $_POST ["page_bg_color"];

    $query  = "INSERT INTO wireframes (";
    $query .= " wireframe_title, browser_title, url_key, wireframe_type, page_main_image, page_bg_color";
    $query .= " ) VALUES (";
    $query .= " '{$wireframe_title}', '{$browser_title}', '{$url_key}', '{$wireframe_type}', '{$image}', '{$page_bg_color}' ";
    $query .= ")";

    echo $query;

    try { $result = mysqli_query($connection, $query);
    } catch (Exception $e) {
        return 'Caught exception: '+  $e->getMessage()+ "\n";
    }
    //Test if there was a query error
    if ($result) {
        //Success
        // would normally use a redirect ie redirect_to("somepage.php");
        //$message = "Subject created.";
        redirect_to("wireframes.php");
    }else {
        //failure
        //$message = "Subject creation failed.";
        //redirect_to("add_project.php");
        echo $query;
    }
} else {
    // This is probably a GET request
    redirect_to("add_edit_wireframe.php");
}
?>

<?php
// Close database connection
if(isset($connection)){ mysqli_close($connection); }
?>

最后是我创建的用于处理目录中图像存储的函数:

function upload_file(){
    $allowedExts = array("gif", "jpeg", "jpg", "png");
    $temp = explode(".", $_FILES["page_main_image"]["name"]);
    $extension = end($temp);
    if ((($_FILES["page_main_image"]["type"] == "image/gif")
        || ($_FILES["page_main_image"]["type"] == "image/jpeg")
        || ($_FILES["page_main_image"]["type"] == "image/jpg")
        || ($_FILES["page_main_image"]["type"] == "image/pjpeg")
        || ($_FILES["page_main_image"]["type"] == "image/x-png")
        || ($_FILES["page_main_image"]["type"] == "image/png"))
        && ($_FILES["page_main_image"]["size"] < 200000)
        && in_array($extension, $allowedExts))
    {
        if ($_FILES["page_main_image"]["error"] > 0) {
            echo "Return Code: " . $_FILES["page_main_image"]["error"] . "<br>";;
        }
        else {
            echo "Upload: " . $_FILES["page_main_image"]["name"] . "<br>";
            echo "Type: " . $_FILES["page_main_image"]["type"] . "<br>";
            echo "Size: " . ($_FILES["page_main_image"]["size"] / 1024) . " kb<br>";

            if (file_exists("uploads/" . $_FILES["page_main_image"]["name"]))
            {
                echo $_FILES["page_main_image"]["name"] . " already exists. ";
            }
            else
            {
                move_uploaded_file($_FILES["page_main_image"]["tmp_name"],
                    "uploads/" . $_FILES["page_main_image"]["name"]);
                echo "Stored in: " . "uploads/" . $_FILES["page_main_image"]["name"];
                $image="{$_FILES['page_main_image']['name']}";
            }
        }
    }
    else {
        echo "Invalid file";
    }

    return $image;
}

我不完全确定，但我认为问题与我试图在其中存储路径名的变量 $image 有关？在函数结束时，我返回变量，然后在后脚本中尝试获取该值并将其发布到数据库中的“page_main_image”字段中，但显然我做错了什么？

再次抱歉发了这么长的帖子，但如果您能给我任何帮助，我将不胜感激!谢谢

Answer 1

在处理上传的文件(例如图片)时，您需要使用 $_FILES 而不是 $_POST 来访问数据。

看看这个 documentation.

编辑

您需要更改主要逻辑。

$image = upload_file(); //Good

$wireframe_title = mysql_prep($_POST["wireframe_title"]);
$browser_title = $_POST["browser_title"];
$url_key = $_POST["url_key"];
$wireframe_type = $_POST["wireframe_type"];

//Delete this, you're throwing out the value from upload_file()
$image = $_POST["page_main_image"];