我试图将视频对象的图像属性作为源传递,以便在程序运行时可以在屏幕上显示图像。出于明显的原因,它找不到源,因为没有文件具有名称“ videoObj1.image”。我想知道是否有解决方法,也许是采用属性文本并将其作为源?甚至是直接使用videoObj1.image的方法。预先感谢。
Question2.html的一部分,我尝试使用image属性作为源:
function displayVideo(videoObj){
var html = "<h1>Search Result " + "</h1>" + "<b>";
html += "Search keyword: " + videoObj.result.searchKeyword;
html += "<table>";
for(var i=0; i < videoObj.result.video.length; i++){
var videoObj1 = videoObj.result.video[i];
html += "<tr>";
html += "<td>" + "<img src=videoObj1.image>" + "</td>";
html += "<td align='right'>" + videoObj1.channel + "</td>";
html += "<td style='color:green' align='right'>";
html += videoObj1.view;
html += "<img src='stockUp.png' />";
html += "</td>";
html += "<td align='right'>" + videoObj1.link + "%</td>";
html += "</tr>";
}
html += "</table>";
var displayDiv = document.getElementById("display");
displayDiv.innerHTML = html;
}
Question2.json:
{
"result": {
"searchKeyword": "Mathematics",
"video": [
{
"title": "Chaos Game",
"channel": "Numberphile",
"view": "428K",
"link": "http://www.youtube.com/watch?v=kbKtFN71Lfs",
"image": "http://i.ytimg.com/vi/kbKtFN71Lfs/0.jpg",
"length": "8:38"
},
{
"title": "Australian Story: Meet Eddie Woo, the maths teacher you wish you'd
had in high school",
"channel": "ABC News (Australia)",
"view": "223K",
"link": "http://www.youtube.com/watch?v=SjIHB8WzJek",
"image": "http://i.ytimg.com/vi/SjIHB8WzJek/0.jpg",
"length": "28:08"
},
{
"title": "Ham Sandwich Problem",
"channel": "Numberphile",
"view": "557K",
"link": "http://www.youtube.com/watch?v=YCXmUi56rao",
"image": "http://i.ytimg.com/vi/YCXmUi56rao/0.jpg",
"length": "5:53"
},
{
"title": "Magic Square Party Trick",
"channel": "Numberphile",
"view": "312K",
"link": "http://www.youtube.com/watch?v=aQxCnmhqZko",
"image": "http://i.ytimg.com/vi/aQxCnmhqZko/0.jpg",
"length": "3:57"
},
{
"title": "The 8 Queen Problem",
"channel": "Numberphile",
"view": "909K",
"link": "http://www.youtube.com/watch?v=jPcBU0Z2Hj8",
"image": "http://i.ytimg.com/vi/jPcBU0Z2Hj8/0.jpg",
"length": "7:03"
}
]
}
}
答案 0 :(得分:1)
问题是您正在将字符串react-native upgrade
传递给img src属性,该属性显然不起作用。
相反,您应该使用经典的字符串串联方法来传递变量,如下所示:
"videoObj1.image"
OR
使用推荐的现代template literals方法,如下所示:
"<td><img src=" + videoObj1.image + "></td>";