我用下面的代碼按鈕添加到頁面如何用函數創建一個按鈕,然後使用jQuery提交它?
$("#myDiv").html("<button id='fileUpload'>Upload</button>");
我然後創建按鈕的Ajax Upload實例。
var button = $('#fileUpload'), interval;
new AjaxUpload(button, {
action: '/upload.ashx',
name: 'myfile',
onSubmit: function(file, ext) {
button.text('Uploading');
this.disable();
// Uploding -> Uploading. -> Uploading...
interval = window.setInterval(function() {
var text = button.text();
if (text.length < 13) {
button.text(text + '.');
} else {
button.text('Uploading');
}
}, 200);
},
onComplete: function(file, response) {
button.text('Upload');
window.clearInterval(interval);
}
});
我想要做的是追加按鈕到頁面,然後模擬自動點擊它。我會如何去做這件事?
更新
現在,該代碼如下:
$("#myDiv").html("<button id='fileUpload'>Upload</button>");
var button = $('#fileUpload'), interval;
new AjaxUpload(button, {
action: '/upload.ashx',
name: 'myfile',
onSubmit: function(file, ext) {
button.text('Uploading');
this.disable();
// Uploding -> Uploading. -> Uploading...
interval = window.setInterval(function() {
var text = button.text();
if (text.length < 13) {
button.text(text + '.');
} else {
button.text('Uploading');
}
}, 200);
},
onComplete: function(file, response) {
button.text('Upload');
window.clearInterval(interval);
}
});
$('#fileUpload').click();
的。點擊事件似乎並不火。它達到了代碼,但什麼都不做......
** **更新
$('#fileUpload').click();
需求是
$('input').click();
請檢查爲什麼公認的答案。只要
你得到的錯誤是什麼?你是否在開始時觸發這段代碼? 如果記得在dom被加載時等待,等等。 – 2009-12-21 16:00:17