if(move_uploaded_file($_FILES[$thumbnail_fieldname]['tmp_name'], $thumbnail_filename) &&
//move desktop files
move_uploaded_file($_FILES[$desktop_fieldname_1280x800]['tmp_name'], $desktop_filename_1280x800) &&
move_uploaded_file($_FILES[$desktop_fieldname_1366x768]['tmp_name'], $desktop_filename_1366x768) &&
move_uploaded_file($_FILES[$desktop_fieldname_1920x1080]['tmp_name'], $desktop_filename_1920x1080)){
echo "<p>We can move all files.</p>";}在我上面的代码中,当其中一个条件失败时不会打印文本,这是因为&这意味着所有需要为真,我在这里要实现的是,当选择任何一个文件上传时,它应该经过条件并打印文本,如果没有选择其中的一个或两个文件,它可以跳过并仍然打印文本,因为有一个文件是存在的,如果没有选择文件,那么这个条件应该是假的。因为如果一个是真的,那么什么组合都是对的呢?
例如,大拇指文件被移动=真
桌面文件1280x800移动=真
桌面文件1366x768 未移动= false //从这里应该测试下一个条件
桌面文件1920x1080 moved = true
//并应打印输出。
发布于 2013-11-09 14:18:02
那这个呢?
// trys to upload the thumbnail
if(move_uploaded_file($_FILES[$thumbnail_fieldname]['tmp_name'], $thumbnail_filename)
// and tries to upload all the other images and store the return values.
// if one or more of them succeeded print the message
&& in_array(TRUE, array(
move_uploaded_file($_FILES[$desktop_fieldname_1280x800]['tmp_name'], $desktop_filename_1280x800),
move_uploaded_file($_FILES[$desktop_fieldname_1366x768]['tmp_name'], $desktop_filename_1366x768),
move_uploaded_file($_FILES[$desktop_fieldname_1920x1080]['tmp_name'],$desktop_filename_1920x1080)
))) {
echo 'upload done';
} else {
echo 'no upload happened';
} https://stackoverflow.com/questions/19876871
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