jQuery的閃存和動態變量

Question

我怎麼能包括這個varibale沒有它拍攝我一個錯誤

變量是在瓦爾線VID

$('.playVideo').live('click',function(z){ 
      z.preventDefault();//Disable Default Method 
      var vid = $(this).attr('data'); 

....

$('#video').empty().flash({ 
    "src":"video/videoplayer-loop.swf", 
    "width":322, 
    "height":275, 
    "vars":{"image":"img/video-posterframe-product.jpg","videoLink=player/video/"+vid+".flv","hideAudio1":"false","videoVolume":"50"}, 

}); 

我得到一個

Uncaught SyntaxError: Unexpected identifier

Answer 1

試圖聲明var vid任何功能

var vid; 
$('.playVideo').live('click',function(z){ 
    z.preventDefault();//Disable Default Method 
    vid = $(this).attr('data'); 
    .... 
} 

Answer 2

的vid變量超出範圍，則需要要麼把這些代碼兩位在同一個函數：

$('.playVideo').live('click',function(z){ 
    z.preventDefault();//Disable Default Method 
    var vid = $(this).attr('data'); 

    $('#video').empty().flash({ 
     "src":"video/videoplayer-loop.swf", 
     "width":322, 
     "height":275, 
     "vars":{"image":"img/video-posterframe-product.jpg", "videoLink=player/video/" + vid + ".flv", "hideAudio1": "false", "videoVolume": "50"}, 
    }); 
}); 

或將代碼設置閃光燈到它自己的功能，從您的單擊處理程序調用它，傳遞vid變量：

$('.playVideo').live('click',function(z){ 
    z.preventDefault();//Disable Default Method 
    var vid = $(this).attr('data'); 
    setUpFlash(vid); 
}); 

function setUpFlash(video) { 
    $('#video').empty().flash({ 
     "src":"video/videoplayer-loop.swf", 
     "width":322, 
     "height":275, 
     "vars":{"image":"img/video-posterframe-product.jpg", "videoLink=player/video/" + video + ".flv", "hideAudio1": "false", "videoVolume": "50"}, 
    }); 
}); 

你也可以去在全球範圍內變量爲vid，但這不是最佳實踐。