我正在寻找一种通过ajax将敏感变量传递给php文件的安全方法。目前,我一直在使用数据属性,但是可以使用类似于firebug的方法更改值。
HTML:
<div class="strip">
<?php
if($hide == 0) {
echo '<h2 class="action" data-type="1" data-id="<?php echo $id; ?>" data-action="0">Hide Business</h2>';
}
if($hide == 1) {
echo '<h2 class="action" data-type="1" data-id="<?php echo $id; ?>" data-action="1">Un-Hide Business</h2>';
}
?>
<h2 class="action" data-type="1" data-id="<?php echo $id; ?>" data-action="2">Delete Business</h2>
</div>JavaScript/JQuery:
$(".action").click(function() {
var type = $(this).data("type");
var id = $(this).data("id");
var action = $(this).data("action");
$.ajax({
url : 'assets/php/confirm.php',
type : 'POST',
data : "type="+type+"&action="+action+"&ids="+id,
success : function (result) {
alert(result);
}
});
});PHP:
if(isset($_POST['type'], $_POST['action'], $_POST['ids'])) {
$type = $_POST['type'];
$action = $_POST['action'];
$ids = explode(",", $_POST['ids']);
$count = count($ids);
if($type == 0) {
if($action == 1) {
$stmt = $mysqli->prepare("DELETE FROM deals WHERE id=?");
} else {
$stmt = $mysqli->prepare("UPDATE deals SET hide=0 WHERE id=?");
}
} else {
if($action == 1) {
$stmt = $mysqli->prepare("DELETE FROM businesses WHERE id=?");
} else {
$stmt = $mysqli->prepare("UPDATE businesses SET hide=0 WHERE id=?");
}
}
for($i = 0; $i < $count; $i++) {
$stmt->bind_param('s', $ids[$i]);
$stmt->execute();
$stmt->close();
}
echo 'Success updated '.$_POST['ids'];
}需要安全的变量是数据类型、数据id、数据操作值.原因是我不希望错误的数据库条目被删除。我不知道有别的选择,所以任何帮助都是很棒的。
https://stackoverflow.com/questions/23593764
复制相似问题