我在一个表单中有2个按钮。当我单击第一个或第二个按钮时,都会写一个警告示例,但Ajax请求不会运行。我需要一个表单,因为我想上传图片。我不知道是什么问题。
page.php文件
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>jQuery Ajax two submit in one form</title>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
</head>
<body>
<form id="animal-upload" method="post" enctype="multipart/form-data">
<span>Name:</span>
<input type="text" name="animalname" id="animalname">
<span>Image:</span>
<input type="file" name="imagefile" id="imagefile">
<button type="submit" name="publish" id="publish">Publish</button>
<button type="submit" name="save" id="save">Save</button>
</form>
<script>
$(document).ready(function() {
$('#animal-upload').on('submit', function() {
return false;
});
$('#publish').click(function() {
alert("Test");
});
$('#save').click(function(e) {
e.preventDefault();
$.ajax({
url: "animal-upload.php",
type: "POST",
data: new FormData(this),
contentType: false,
processData: false,
success: function(data) {
alert(data);
}
});
});
});
</script>
</body>
</html>
动物upload.php的
<?php
$connect = mysqli_connect("localhost", "root", "", "test");
mysqli_set_charset($connect,"utf8");
$status = '';
$animalname = $connect->real_escape_string($_POST["animalname"]);
if ($_FILES['imagefile']['name'] != '') {
$extension = end(explode(".", $_FILES['imagefile']['name']));
$allowed_type = array("jpg", "jpeg", "png");
if (in_array($extension, $allowed_type)) {
$new_name = rand() . "." . $extension;
$path = "animals/" . $new_name;
if (move_uploaded_file($_FILES['imagefile']['tmp_name'], $path)) {
mysqli_query($connect, "INSERT INTO animals (animalname,image) VALUES ('".$animalname."','".$path."')");
$status = 'Successful!';
}
} else {
$status = 'This is not image file!';
}
} else {
$status = 'Please select image!';
}
echo $status;
?>
答案 0 :(得分:0)
经过试用和错误后,如果您更改此行,我发现该脚本有效:
data: new FormData($("#animal-upload")[0]),
因为它选择了表单对象。
您可以考虑一些安全提示:
要上传文件:
upload_tmp_dir
memmory_limit
指令