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我有一个应用程序，人们可以在其中发布错误。他们可以添加屏幕截图，唯一的事情是，所有内容都写入我的数据库中，但图像不会保存，我只看到图像的编号。在地图“上传”中没有保存图像。

这是错误：

Notice: Undefined index: upload in /Applications/MAMP/htdocs/2IMD_KellyVerhaegen_Project2013/loggedIn.php on line 29
Notice: Undefined index: upload in /Applications/MAMP/htdocs/2IMD_KellyVerhaegen_Project2013/loggedIn.php on line 30
Notice: Undefined index: upload in /Applications/MAMP/htdocs/2IMD_KellyVerhaegen_Project2013/loggedIn.php on line 32

php代码：

if (isset($_POST['btnBug'])) 
{
// wanneer er op de knop geklikt is proberen we user te saven in de databank
if (!empty($_POST['btnBug'])) 
{
    /* controleren of titel en beschrijving velden zijn ingevuld */
    if(!empty($_POST['subject']) && !empty($_POST['post']))
    {
        /* zien of de bug kan gesaved worden */
    try
    {
            //nieuwe bug aanmaken en gegevens wegschrijven
            $bug = new Bug();
            $screenshot = time() . $_FILES['upload']['name'];
            move_uploaded_file($_FILES['upload']['tmp_name'],
            "uploads/" . $screenshot);
            $bug->Bug_Name  = $_FILES['upload']['name'];
            $bug->Screenshot = $screenshot;
            $bug->Post = htmlspecialchars($_POST['post']);
            $bug->Subject = htmlspecialchars($_POST['subject']);
            //kijken of de bug solved/unsolved is
            $keuze = $_POST['myradio'];
                if ($keuze == "Unsolved") {
                    $bug -> Status = "Unsolved";
                } else {
                    $bug -> Status = "Solved";
                }
            $bug -> User_id = htmlspecialchars($_POST['getUser']);
            $bug -> Project_id = htmlspecialchars($_POST['getProject']);
            // bug wordt gesaved
            $bug->saveBug();    
            }
        catch(Exception $e)
        {
            $feedback = $e->getMessage();
            echo "Vul alle velden in!"; 
        }
    }
}
}

我的表格：

<div id="bugform">
        <div class="control-group">
        <h4>Vul hier u bug in.</h4>
        <br />
            <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" class="form-horizontal">
                <label class="control-label" for="inputOnderwerp">Onderwerp</label> 
            <div class="controls">
                <input type="text" name="subject" />
            </div>
        <br />
            <label class="control-label" for="inputBeschrijving">Beschrijving</label>
            <div class="controls">
                <textarea name="post" name="post"></textarea>
            </div>
        <br />
            <label class="control-label" for="inputUnsolved">Unsolved</label>
            <div class="controls">
                <input type="radio" name="myradio" value="Unsolved" id="Unsolved" checked="true">
            </div>
        <br />
            <label class="control-label" for="inputSolved">Solved</label>
            <div class="controls">
                <input type="radio" name="myradio" value="Solved" id="Solved">
            </div>
            <?php
            if(isset($_POST['btnBug']))
            {
                echo "<img src='uploads/".$screenshot."'/>";

            }
            ?>
            <input type="file" name="upload" id="upload"  />
        <br />
        <br />
        <br />
    <!-- projects ophalen -->
        <label class="control-label" for="inputProjectKiezen">Project kiezen</label>
        <div class="controls">
        <?php
            if(mysqli_num_rows($allProjects) > 0)
            {
                echo "<select name= getProject>";   
                while ($row = mysqli_fetch_assoc($allProjects)) 
                {
                    echo "<option value=" . $row['project_id'] . ">" . $row['project_name'] . "</option>";
                }
                echo "</select>";
            }
            ?>
        </div>
        <br />
        <!-- users ophalen -->
        <label class="control-label" for="inputUserToekennen">Aan welke user toekennen?</label>
        <div class="controls">
        <?php
            if(mysqli_num_rows($showUser) > 0)
            {
                echo "<select name= getUser>";  
                while ($row = mysqli_fetch_assoc($showUser)) 
                {
                    echo "<option value=" . $row['user_id'] . ">" . $row['username'] . "</option>";
                }
                echo "</select>";
            }
        ?>
        </div>
        <br />
        <div class="controls">
            <input class="btn btn-info dropdown-toggle" type="submit" name="btnBug" id="btnBug" value="Verzenden" />    
        </div>
        </form>
    </div>

我一直在寻找一整天，有人可以帮助我吗？谢谢。

4

1 回答 1

1

您需要将enctype="multipart/form-data"属性添加到<form>元素。

原因是默认enctype（数据的编码方式）是application/x-www-form-urlencoded，它不允许对整个文件进行编码。

于 2013-08-20T20:57:22.440 回答