Html代码
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"> </script>
<script>
$(function() {
var dmJSON = "three.json";
$.getJSON( dmJSON, function(data) {
$.each(data.records, function(i, f) {
var tblRow = "<tr>" + "<td>" + f.Clue + "</td>" + "<td>" + f.Answer + "</td>" + "<td>" + f.Status + "</td>" + "<td> " + f.Views + "</td>" + "</tr>"
$(tblRow).appendTo("#entrydata tbody");
});
});
});
</script>
</head>
<body>
<div class="wrapper">
<div class="profile">
<table id= "entrydata" border="1">
<thead>
</thead>
<tbody>
</tbody>
</table>
</div>
</div>
</body>
</html>
JSON数据
{
"records": [
{
"Clue" : "First Clue",
"Answer" : "Answer to the first clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Second Clue",
"Answer" : "Answer to the second clue",
"Status" : "Rejected",
"Views" : "15"
},
{
"Clue" : "Third Clue",
"Answer" : "Answer to the third clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Fourth Clue",
"Answer" : "Answer to the fourth clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Fifth Clue",
"Answer" : "Answer to the fifth clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Sixth Clue",
"Answer" : "Answer to the sixth clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Seventh Clue",
"Answer" : "Answer to the seventh clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Eigth Clue",
"Answer" : "Answer to the eigth clue",
"Status" : "Rejected",
"Views" : "10"
},
{
"Clue" : "Nintht Clue",
"Answer" : "Answer to the ninth clue",
"Status" : "Rejected",
"Views" : "10"
}
]
}
从上面的代码中,JSON数据存储在一个表中。我想将每条记录存储在不同的表中。有8条记录,所以我想要8个表,每个表有1条记录。 例如:在第一张表中,我希望存储第一条记录(即第一条线索,第一条线索的答案,拒绝10条) 同样在第二个表中我想存储第二个记录。怎么做? 任何解决方案将不胜感激。提前致谢
答案 0 :(得分:1)
如果要在单独的表中呈现每个记录,那么最好的办法是动态创建表。例如:
$(function() {
var dmJSON = "three.json";
$.getJSON(dmJSON, function(data) {
console.log(data)
$.each(data.records, function(i, f) {
var table = '<table>';
table += "<tr><td>" + f.Clue + "</td><td>" + f.Answer + "</td><td>" + f.Status + "</td><td> " + f.Views + "</td></tr>";
table += "</table>";
$('.profile').append(table);
});
});
});
但是使用多个表格时,更难以正确对齐列,但不确定这是否是您需要的。
答案 1 :(得分:1)
这可以帮到你试试这个:
HTML:
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"> </script>
<script>
$(function() {
var dmJSON = "three.json";
$.getJSON( dmJSON, function(data) {
$.each(data.records, function(i, f) {
var $table="<table border=1><tbody><tr>" + "<td>" + f.Clue + "</td>" + "<td>" + f.Answer + "</td>" + "<td>" + f.Status + "</td>" + "<td> " + f.Views + "</td>" + "</tr></tbody></table>"
$("#entrydata").append($table)
});
});
});
</script>
</head>
<body>
<div class="wrapper">
<div class="profile">
<div id='entrydata'></div>
</div>
</div>
</body>
</html>
答案 2 :(得分:0)
将JSON数据存储在多个表Demo
中$(function() {
var data = {
"records": [{
"Clue": "First Clue",
"Answer": "Answer to the first clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Second Clue",
"Answer": "Answer to the second clue",
"Status": "Rejected",
"Views": "15"
}, {
"Clue": "Third Clue",
"Answer": "Answer to the third clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Fourth Clue",
"Answer": "Answer to the fourth clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Fifth Clue",
"Answer": "Answer to the fifth clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Sixth Clue",
"Answer": "Answer to the sixth clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Seventh Clue",
"Answer": "Answer to the seventh clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Eigth Clue",
"Answer": "Answer to the eigth clue",
"Status": "Rejected",
"Views": "10"
}, {
"Clue": "Nintht Clue",
"Answer": "Answer to the ninth clue",
"Status": "Rejected",
"Views": "10"
}]
}
$.each(data.records, function(i, f) {
var tblRow = "<table border='1'>" + "<tr>" + "<td>" + f.Clue + "</td>" + "<td>" + f.Answer + "</td>" + "<td>" + f.Status + "</td>" + "<td> " + f.Views + "</td>" + "</tr>" + "</table>"
$(tblRow).appendTo("body");
});
});