JSON DUMP: {"id":1,"name":"demo","pixeldata":"[{ \"time\" : \"1\", \"colour\" : \"#ff5357\"},{ \"time\" : \"2\", \"colour\" : \"#2424ff\"},{ \"time\" : \"3\", \"colour\" : \"#ff0d13\"},{ \"time\" : \"4\", \"colour\" : \"#f7ff4a\"},{ \"time\" : \"5\", \"colour\" : \"#fa24ff\"},{ \"time\" : \"6\", \"colour\" : \"#ff3e43\"}]"}
[0:]

var days = [
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday",
"Sunday",
];
document.getElementById("demo2").innerHTML = days[0] + "<br>" +
days[1] + "<br>" +
days[2] + "<br>" +
days[3] + "<br>" +
days[4] + "<br>" +
days[5] + "<br>" + days[6];
&#13;
body {
margin: 0;
padding: 0;
}
table {
border: 1px solid #000;
border-collapse: collapse;
width: 500px;
height: 150px;
margin: 5px;
}
table tr td {
border:1px solid #000;
padding-left:10px;
}
&#13;
嗨,这听起来像一个奇怪的问题。有没有办法从[0]到[6]显示而不提及中间数字(1-5)?我尝试过循环,但我不认为我做得对。 for(var i = 0; i&gt; days.length; i ++);
答案 0 :(得分:5)
您可以使用.join()
:
document.getElementById("demo2").innerHTML = days.join("<br>");
<强>演示:强>
var days = [
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday",
"Sunday",
];
document.getElementById("demo2").innerHTML = days.join("<br>");
body {
margin:0;
padding:0;
}
table {
border:1px solid #000;
border-collapse: collapse;
width:500px;
height:150px;
margin:5px;
}
table tr td{
border:1px solid #000;
padding-left:10px;
}
<h1>Concatenation Challenge</h1>
<table>
<tr>
<td>All days of the week</td>
<td id="demo2"></td>
</tr>
</table>
<强>文档:强>
答案 1 :(得分:1)
var days = [
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday",
"Sunday",
]
// From HERE
var myString = "";
for (var i = 0; i < days.length; i++) {
myString = myString + days[i] + "<br>";
}
document.getElementById("demo2").innerHTML = myString;
// TO HERE
body {
margin:0;
padding:0;
}
table {
border:1px solid #000;
border-collapse: collapse;
width:500px;
height:150px;
margin:5px;
}
table tr td{
border:1px solid #000;
padding-left:10px;
}
<h1>Concatenation Challenge</h1>
<table>
<tr>
<td>All days of the week</td>
<td id="demo2"></td>
</tr>
</table>