php - 尝试从 MySQL 数据库中调用 longblob 信息以进行文件输入控制

Question

各位程序员大家好，

我在网页设计方面相当新手，所以请以实物方式回复。我无法在任何论坛中找到针对我的独特问题的解决方案，因此我在这里提出我的问题。

我已经设置了 MySQL 数据库，可以很好地上传数据和图像。但是，我有一个管理员帐户，我希望能够通过网络表单编辑数据库中的信息。我尝试了很多方法来让表单从数据库中读取信息并自动填充表单，但是我无法让文件输入控件“记住”或调用已上传的图像。所有其他数据，例如 ID、名称等工作文件和自动填充，但不是图像。

这是我用来呈现和填写表单的一些代码。我强烈感觉错误出在 $row else 语句中。我不确定如何自动填充文件输入元素。我省略了部分内容，因为整个代码相当长(用省略号表示)。如果您需要更多信息，请与我们联系。任何想法或例子都会很棒。谢谢。

PS - 我知道 mysql_query 已被贬值...我稍后会修复它。

//runs when the form needs to display, retaining the values <- abbreviated to make shorter
function renderForm($id, $name, $image, $description, $salePrice, $listPrice, $shipping, $company, $category, $subCategory, $quantity, $type, $newArrival, $vintage, $error){
//if there are any errors...
if ($error != ''){
    //display errors
    echo "<div id=\"newMemberError\">$error</div>\n";
} 

//display the form
echo "<form method=\"post\" enctype=\"multipart/form-data\">\n";
echo "<input type=\"hidden\" name=\"max_file_size\" value=\"6000000\">\n";
echo "<table id=\"newMemberTable\">\n";
echo "<tr>\n";
echo "  <th><label>ID: <span class=\"aster\">*</span></label></th>\n";
echo "  <td><input type=\"text\" name=\"id\" value=\"$id\"/></td>\n";
echo "</tr>\n";
echo "<tr>\n";
echo "  <th><label>Name: <span class=\"aster\">*</span></label></th>\n";
echo "  <td><input type=\"text\" name=\"name\" value=\"$name\"></td>\n";
echo "</tr>\n";
...
echo "<tr>\n";
echo "  <th><label>Image: <span class=\"aster\">*</span></label></th>\n";
echo "  <td><input type=\"file\" name=\"image\" value=\"data:image/$type;base64,$image\"></td>\n";
echo "</tr>\n";
echo "<tr><td><img class=\"paginatedImg\" height=\"40\" width=\"50\" src=\"data:image/$type;base64,$image\"></td></tr>\n";
echo "<tr>\n";
echo "  <th><label>Type of Image: <span class=\"aster\">*</span></label></th>\n";
echo "  <td><input type=\"text\" name=\"type\" value=\"$type\"></td>\n";
echo "</tr>\n";
echo "<tr>\n";
echo "  <th>&nbsp;</th>\n";
echo "  <td><input id=\"newMemberBtn\" type=\"submit\" name=\"submit\" value=\"Submit\"><span id=\"requiredMsg\">* required</span></td>\n";
echo "</table>\n";
echo "</form>\n";
}

//check if the form has been submitted. If it has, process the form and save it to the database
if(isset($_POST['submit'])){
...
}
//if the form hasn't been submitted, get the data from the database and display the form
else{
//get the id value from the URL (if it exists)
if (isset($_GET['id'])){
    //query database
    $id = $_GET['id'];
    $result = mysql_query("SELECT * FROM tbl WHERE id='$id'")
              or die(mysql_error()); 
    //set variable to hold amount of rows
    $row = mysql_fetch_array($result);

    //if the ID matches up with the database...
    if($row){
        //get the data from database
        $id = $row['id'];
        $name = $row['name'];
        $image = $row['image'];
        $description = $row['description'];
        $salePrice = $row['salePrice'];
        $listPrice = $row['listPrice'];
        $shipping = $row['shipping'];
        $company = $row['company'];
        $category = $row['category'];
        $subCategory = $row['subCategory'];
        $quantity = $row['quantity'];
        $type = $row['type'];
        $newArrival = $row['newArrival'];
        $vintage = $row['vintage'];

        //display form
        renderForm($id, $name, $image, $description, $salePrice, $listPrice, $shipping, $company, $category, $subCategory, $quantity, $type, $newArrival, $vintage, '');
    }
    //if no match...
    else{
        //display error message to user
        echo "No results!";
    }
}
//if the 'id' in the URL isn't valid, or if there is no 'id' value...
else{
    //display error message to user
    echo 'Error!';
}

}

Answer 1

您无法使用 input 标签显示图像，因为 value 属性不适用于 input[type=file]。请参阅http://www.w3.org/TR/html-markup/input.file.html 。但是，您可以在图像标签中显示图像:

echo "<img src=\"data:image/$type;base64,$image\" />";