Question

<form  action="original_photo_upload.php" method="POST" id="form" name="form"   class="form-horizontal" style="margin-top: 20px;">
<div class="form-group">
    <label class="control-label col-xs-4">File Upload</label>
    <div class="col-md-8">
        <div class="fileupload fileupload-new" data-provides="fileupload">
            <div class="input-append">
                <div class="uneditable-input">
                    <i class="fa fa-file fileupload-exists"></i>
                    <span class="fileupload-preview"></span>
                </div>
                <span class="btn btn-default btn-file">
                    <span class="fileupload-exists">Change</span>
                    <span class="fileupload-new">Select file</span>
                    <input type="file">
                </span>
                <a href="#" class="btn btn-default fileupload-exists" data-dismiss="fileupload">Remove</a>
            </div>
        </div>
    </div>
</div>
</form>

我希望在Db中插入图像值并将照片移动到一个文件夹中，我知道普通形式但是靴子我有困惑，如何在original_photo_upload.php中获取图像值

Answer 1

将表单更改为：

<form form action="original_photo_upload.php" method="post" enctype="multipart/form-data" class="form-horizontal" style="margin-top: 20px;">

同时将您的输入更改为：

<input name="file" type="file">

您的表单，需要提交才能发送数据。所以，添加：

<input type="submit" value="ok">

在他的档案中：original_photo_upload.php提出了一个条件。类似的东西：

<?php
if (isset($_FILES['file'])){
    var_dump($_FILES);
}

仅供测试。

<强>更新
要执行上传：

move_uploaded_file($_FILES['file']['tmp_name'], 'upload/' . $_FILES['file']['name']);

Answer 2

<form  action="original_photo_upload.php" enctype='multipart/form-data' method="POST" id="form" name="form"   class="form-horizontal" style="margin-top: 20px;">
<div class="form-group">
    <label class="control-label col-xs-4">File Upload</label>
    <div class="col-md-8">
        <div class="fileupload fileupload-new" data-provides="fileupload">
            <div class="input-append">
                <div class="uneditable-input">
                    <i class="fa fa-file fileupload-exists"></i>
                    <span class="fileupload-preview"></span>
                </div>
                <span class="btn btn-default btn-file">
                    <span class="fileupload-exists">Change</span>
                    <span class="fileupload-new">Select file</span>
                    <input type="file" name="image">
                </span>
                <input type="button" class="btn btn-default fileupload-exists" data-dismiss="fileupload" value="submit">
                <input type="button" class="btn btn-default fileupload-exists" data-dismiss="fileupload" value="remove">
                </div>
        </div>
    </div>
</div>
</form>

和php查询

<?php if(isset($_POST['submit']))
{
$ImageName = $_FILES['image']['name'];
$path = 'images/'; 
$location = $path . $_FILES['image']['name']; 
move_uploaded_file($_FILES['image']['tmp_name'], $location); 
} ?>

更详细的搜索图片上传功能。

如何在下一页的bootstrap中获取图像值

2 个答案: