PHP - 上传多张图片 [英] PHP - Upload multiple images

问题描述

我需要通过表单上传多张图片。我认为我会做没有问题的，但我有一个。

当我尝试做foreach并逐个获取图像时，它并不像我希望的那样行事

HTML

 < form method =post action =enctype =multipart / form-dataid =frmImgUpload> 
< input name =fileImage []type =filemultiple =true/> 
< br /> 
< input name =btnSubmittype =submitvalue =Upload/> 
< / form> 
  

PHP

 <？php 
 if（$ _POST）
 {
 echo< pre>; 
 foreach（$ _FILES ['fileImage'] as $ file）
 {
 print_r（$ file）; 
 die（）; //我想要它打印第一个图像内容，然后死去测试... 
 // imgUpload（$ file） - 我已经有了可以上传一张图片的工作函数
} 
 } 
  

我期望从中打印出第一张图像，而不是打印所有图像的名称。

示例

 数组
（
 [0] => 002.jpg 
 [1] => 003.jpg 
 [2] => 004.jpg 
 [3] => 005.jpg 
 
 $ / code> 

我要输出的内容

 数组
（
 [name] =&>; 002.jpg 
 [type] => image / jpeg 
 [tmp_name ] => php68A5.tmp 
 [error] => 0 
 [size] => 359227 
）
   
 
 
 Okey我找到了解决方案这是我如何做的，可能不是最好的方式，但它的工作原理。
  foreach（ $ _FILES ['fileImage'] ['name'] as $ f）
 {
 $ file ['name'] = $ _FILES ['fileImage'] ['name'] [$ i]; 
 $ file ['type'] = $ _FILES ['fileImage'] ['type'] [$ i]; 
 $ file ['tmp_name'] = $ _FILES ['fileImage'] ['tmp_name'] [$ i]; 
 $ file ['error'] = $ _FILES ['fileImage'] ['error'] [$ i]; 
 $ file ['size'] = $ _FILES ['fileImage'] ['size'] [$ i]; 
 imgUpload（$ file）; 
 $ i ++; 
} 
  
 
 
解决方案
重建 $ _ FILES 数组以访问它们的子项目作为一个数组。
  $ index = 0; 
 $ field ='fileImage'; 
 $ keys = array_keys（$ _ FILES [$ field]）; 
 $ file = array（）; 
 foreach（$ key为$ key）
 {
 $ file [$ key] = $ _FILES [$ field] [$ key] [$ index]; 
} 
 print_r（$ file）; 
  
将 $ index 更改为您需要的值选择一个特定的文件。
 
I need to upload multiple images via form. I thought that I will do it with no problem, but I have one.

When I try to do foreach and get image by image it is not acting like I hoped it will.

HTML
<form method="post" action="" enctype="multipart/form-data" id="frmImgUpload">
    <input name="fileImage[]" type="file" multiple="true" />
    <br />
    <input name="btnSubmit" type="submit" value="Upload" />
</form>
PHP
<?php
if ($_POST)
{
    echo "<pre>";
    foreach ($_FILES['fileImage'] as $file)
    {
        print_r($file);
        die(); // I want it to print first image content and then die to test this out...
        //imgUpload($file) - I already have working function that uploads one image
    }
}
What I expected from it to print out first image, instead it prints names of all the images.

Example
Array
(
    [0] => 002.jpg
    [1] => 003.jpg
    [2] => 004.jpg
    [3] => 005.jpg
)
What I want it to output
Array
(
    [name] => 002.jpg
    [type] => image/jpeg
    [tmp_name] => php68A5.tmp
    [error] => 0
    [size] => 359227
)
So how can I select image by image in the loop so I can upload them all?

Okey I found solution and this is how I did it, probably not the best way but it works.
foreach ($_FILES['fileImage']['name'] as $f)
{
    $file['name'] = $_FILES['fileImage']['name'][$i];
    $file['type'] = $_FILES['fileImage']['type'][$i];
    $file['tmp_name'] = $_FILES['fileImage']['tmp_name'][$i];
    $file['error'] = $_FILES['fileImage']['error'][$i];
    $file['size'] = $_FILES['fileImage']['size'][$i];
    imgUpload($file);
    $i++;
}

解决方案
You are basically asking of how to rebuild the $_FILES array to access subitems of them as one array.
$index = 0;
$field = 'fileImage';
$keys = array_keys($_FILES[$field]);
$file = array();
foreach($keys as $key)
{
    $file[$key] = $_FILES[$field][$key][$index];
}
print_r($file);
change $index to the value you need to pick a specific file.

                        
这篇关于PHP  - 上传多张图片的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！