Javascript 创建所需阵列
我正在通过ajax上传文件,我需要安排我的Javascript 创建所需阵列,javascript,php,jquery,Javascript,Php,Jquery,我正在通过ajax上传文件,我需要安排我的$\u文件数组来定制我的需求 目前,这段代码: <input id='athletes_gid' type='file' name='athletes_gid[]' multiple value="1" /> <script type="text/javascript"> // Variable to store your files var files; // Add events $('#athletes_gid').on
$\u文件
数组来定制我的需求
目前,这段代码:
<input id='athletes_gid' type='file' name='athletes_gid[]' multiple value="1" />
<script type="text/javascript">
// Variable to store your files
var files;
// Add events
$('#athletes_gid').on('change', prepareUpload);
// Grab the files and set them to our variable
function prepareUpload(event)
{
files = event.target.files;
}
var data = new FormData();
$.each(files, function(key, value)
{
data.append(key, value);
});
如何主要在中编写上述代码。每个循环使数组如下所示:
array (size=5)
'name' =>
array (size=2)
0 => string '1398423008894.jpg' (length=17)
1 => string '1239049963136.jpg' (length=17)
'type' =>
array (size=2)
0 => string 'image/jpeg' (length=10)
1 => string 'image/jpeg' (length=10)
'tmp_name' =>
array (size=2)
0 => string 'C:\Program Files\wamp\tmp\php172F.tmp' (length=37)
1 => string 'C:\Program Files\wamp\tmp\php1740.tmp' (length=37)
'error' =>
array (size=2)
0 => int 0
1 => int 0
'size' =>
array (size=2)
0 => int 116496
1 => int 42415
简易钥匙交换功能
从一个foreach()
循环开始,php$\u文件的默认行为是你的第二个数组。你能用javascript在控制台中粘贴什么样的“数据”吗。这将节省我用JS写这篇文章的时间。谢谢你的逻辑,现在对我来说已经太晚了,我不能像现在这样敏锐地思考
array (size=5)
'name' =>
array (size=2)
0 => string '1398423008894.jpg' (length=17)
1 => string '1239049963136.jpg' (length=17)
'type' =>
array (size=2)
0 => string 'image/jpeg' (length=10)
1 => string 'image/jpeg' (length=10)
'tmp_name' =>
array (size=2)
0 => string 'C:\Program Files\wamp\tmp\php172F.tmp' (length=37)
1 => string 'C:\Program Files\wamp\tmp\php1740.tmp' (length=37)
'error' =>
array (size=2)
0 => int 0
1 => int 0
'size' =>
array (size=2)
0 => int 116496
1 => int 42415
function array_keys_flip($oldArray) {
foreach ($oldArray as $k1 => $v1) {
foreach ($v1 as $k2 => $v2) {
$newArray[$k2][$k1] = $v2
}
}
return $newArray;
}