在this React tutorial中，React演示组件内的按钮单击事件回调函数明确绑定到if(isset($_POST["upload_report"])) {
echo $_POST['title']; // giving title index undefined
$max_report_id += 1;
$target_dir = "Reports/";
$file_title = mysqli_real_escape_string($connection,trim($_POST['title']));
$file_name = mysqli_real_escape_string($connection,trim($_FILES['report']["name"]));
//$file_size = mysqli_real_escape_string($connection,trim($_FILES['report']["size"]));
$target_file = $target_dir . basename($file_name); //basename gets filename from path
$uploadOk = 1;
$file_type = pathinfo($target_file,PATHINFO_EXTENSION);
$target_file = $target_dir . $max_report_id . "." . $file_type;
// Check if file already exists
if (file_exists($target_file)) {
//echo "Sorry, file already exists.";
$uploadOk = 0;
}
/* Check file size
if ($file_size > 5000000) {
echo "Sorry, your file is too large.";
$uploadOk = 0;
}
*/
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
$status_fail = true;
$fail_message = "Sorry, there was an error uploading your file.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["report"]["tmp_name"], $target_file)) {
$status_success = true;
$success_message = "The file ". $file_title. " has been uploaded.";
$query = "insert into reports(title,fname,path,platform,added_by,added_on) values ('$file_title','$file_name','$target_file','$platform','$user_id',now())";
$result = mysqli_query($connection, $query) or die($query);
}
else {
$status_fail = true;
$fail_message = "Sorry, there was an error uploading your file.";
}
}
}
：
null但是，在回调函数中，<button onClick={_this.props.toggleActive.bind(null, user.id)}>Toggle Active</button>
关键字用于访问React Container Component的状态：
this我不明白为什么当toggleActive: function(userId) {
...
var newState = Object.assign({}, this.state)
...
}
应该等于this时这个例子有效，或者在非严格模式下，全局null对象。任何人都可以告诉我为什么这个例子仍在运作？
可以找到相应的Codepen here。
答案 0 :(得分：0)
<button onClick={_this.props.toggleActive.bind(null, user.id)}>Toggle Active</button>
这里你是
null的绑定方法，而不是实例。你可以 不能在方法中访问this，因此this将无效。
答案 1 :(得分：0)
我之前错了，并没有理解这个问题。请看下面的代码： -
function a(){
this.x = 5;
this.y = function(){
console.log(this.x)
}
}
var x = 20;
var c = new a();
var d = c.y.bind(null);
d(); // prints 20 and not null or undefined
当null作为绑定函数中的上下文传递时，该函数保留其原始上下文，即此处的全局范围（定义函数d的上下文）。但是，如果在绑定函数中传递了某个上下文，则该函数将打印＆＃39; X＆＃39;该背景的财产。
同样，在
中传递null时<button onClick={_this.props.toggleActive.bind(null, user.id)}>Toggle Active</button>
在触发器上，toggleActive函数内的上下文是定义它的上下文，即UserListContainer.Hence，它能够访问其状态