如何在引导程序中获取下一页的图像值[重复]

Posted 2023-02-24

【中文标题】如何在引导程序中获取下一页的图像值[重复]

【英文标题】：How to get the image value in bootstrap for next page [duplicate]

【发布时间】：2016-02-28 05:26:26

【问题描述】：

<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>

我想在 Db 中插入图像值并将照片移动到一个文件夹中，我知道正常形式但引导我很困惑，如何在 original_photo_upload.php 中获取图像值

【参考方案1】：

<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); 
 ?>

还有更详细的搜索图片上传功能。

【参考方案2】：

将您的表单更改为：

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

还将您的输入更改为：

<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">

在他的文件中： original_photo_upload.php 做了一个条件。比如：

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

仅用于测试。

更新 要执行上传：

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

【讨论】：

var_dump($_FILES);//array(1) ["file"]=> array(5) ["name"]=> string(12) "DSC05019.JPG" [ "type"]=> string(10) "image/jpeg" ["tmp_name"]=> string(24) "C:\xampp\tmp\phpFBD5.tmp" ["error"]=> int(0) [ "size"]=> int(4165988) 如何只获取文件名

我明白了，谢谢...

要获取文件名，请使用：echo $_FILES['file']['name']; 但是，如果要执行上传，则需要 tmp_name。我更新了我的答案。看！