【发布时间】:2014-11-10 13:42:23
【问题描述】:
Web 服务正在返回 jsonString。解析后,我必须将它的值绑定到表中。
最简单的方法是找到字符串中的每个对象。 例如:我的字符串以以下格式返回:
{
"Table1": [
{
"ProjectId": "VS200-001---",
"day1": "---",
"day2": "---",
"day3": "---",
"day4": "---",
"day5": "---",
"day6": "---",
"day7": "---",
"day8": "---",
"day9": "---",
"day10": "---",
"day11": "---",
"day12": "---",
"day13": "4.3",
"day14": "2",
"day15": "---",
"day16": "---",
"day17": "---",
"day18": "---",
"day19": "---",
"day20": "---",
"day21": "---",
"day22": "---",
"day23": "5",
"day24": "---",
"day25": "---",
"day26": "---",
"day27": "---",
"day28": "---",
"day29": "---",
"day30": "---"
},....
]
}
解析后我得到了对象的。
for(var x=0; x< _data.Length;x++)
{
for(var dys = 0; dys< dates.Length;dys++){
var val = "day" + (dys+1);
tbody += _data[0] .val + "</tr>";
///Here exception is generating incorrect format
//but if i use here _data[0].day1 it will show the value for day1
//how to convert string to object inorder to retrieve the value
}
}
【问题讨论】:
标签: javascript jquery json object