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我想用PHP和Ajax創建一個小照片上傳。所以我不是一個PHP的傢伙，我有一些問題，總是Something went wrong with your upload!狀態。但是，當我將此設置爲表單操作時，我有成功的動作。 Anyboody可以幫忙嗎？Ajax和PHP的照片文件上傳
HTML：
<form onsubmit="return submitForm()" method="POST" enctype="multipart/form-data">
<input type="file" name="pic" id="pic" multiple/>
<input type="submit" value="Upload" for="pic"/>
</form>
PHP：
<?php
$demo_mode = false;
$upload_dir = 'uploads/';
$allowed_ext = array('jpg','jpeg','png','gif');
if(strtolower($_SERVER['REQUEST_METHOD']) != 'post'){
exit_status('Error! Wrong HTTP method!');
}
if(array_key_exists('pic',$_FILES) && $_FILES['pic']['error'] == 0){
$pic = $_FILES['pic'];
if(!in_array(get_extension($pic['name']),$allowed_ext)){
exit_status('Only '.implode(',',$allowed_ext).' files are allowed!');
}
if($demo_mode){
$line = implode(' ', array(date('r'), $_SERVER['REMOTE_ADDR'], $pic['size'], $pic['name']));
file_put_contents('log.txt', $line.PHP_EOL, FILE_APPEND);
exit_status('Uploads are ignored in demo mode.');
}
if(move_uploaded_file($pic['tmp_name'], $upload_dir.$pic['name'])){
exit_status('File was uploaded successfuly!');
}
}
exit_status('Something went wrong with your upload!');
function exit_status($str){
echo json_encode(array('status'=>$str));
exit;
}
function get_extension($file_name){
$ext = explode('.', $file_name);
$ext = array_pop($ext);
return strtolower($ext);
}
?>
和JS：
function submitForm() {
console.log("submit event");
$.ajax({
type: "POST",
url: "post_file.php",
enctype: 'multipart/form-data'
}).done(function(data) {
console.log("PHP Output:");
console.log(data);
});
return false;
}
很多THX！
而不是'exit_status（）'使用'die（）'或'exit（）'並且你的錯誤可能在html形式而不是'onsubmit =「return submitForm（）」'use 'onsubmit =「submitForm（）」'它可能不是固定的東西:) – roun512
可能的重複[如何可以上傳文件與jQuery異步？]（http://stackoverflow.com/questions/166221/how-can-我上傳 - 文件 - 異步與 - jQuery的） – showdev