【发布时间】:2013-09-14 01:23:27
【问题描述】:
如果我使用线:
var colorsone = ["#F0F8FF", "#FAEBD7", "#00FFFF", #7FFFD4", #F0FFFF","#F5F5DC", "#FFE4C4", "#000000", "#FFEBCD", "#0000FF", "#8A2BE2", "#A52A2A", "#DEB887", "#5F9EA0", "#7FFF00", "#D2691E", "#FF7F50", "#6495ED", "#FFF8DC", "#DC143C", "#00FFFF", "#00008B", "#008B8B", "#B8860B", "#A9A9A9", "#006400", "#BDB76B", "#8B008B"];
我在 firebug 中收到此错误
SyntaxError: illegal character
...one = ["#F0F8FF", "#FAEBD7", "#00FFFF", #7FFFD4", #F0FFFF","#F5F5DC", "#FFE4C4",...
jquery....min.js (строка 6, столбец 50)
我正在使用 jquery 1.10.2
请告诉我为什么会出错?
【问题讨论】:
-
缺少
"s ->"#00FFFF", (here)#7FFFD4", (and here)#F0FFFF" -
строка 6, столбец 50都在这里说 -
@FDL 将此作为答案发布;)
标签: javascript jquery illegal-characters