nik*_*hil 10 php ajax jquery json
我想在jqchart中构建一个图形,我需要获得两个数组
现在我想执行下面给出的操作.哪个是错误的.
html
$.ajax(
{
type: "GET",
url: "customer_coverage.php",
data: {id:id},
contentType: "application/json",
dataType: "json",
success: function (data21,data22) {
initChart2(data21,data22);
}
});
function initChart2(data21,data22) {
$('#jqChart2').jqChart({
series: [
{
type: 'column',
title: 'no of days ',
data:data21,
},
{
type: 'column',
title: 'no of days ',
data:data22,
},
]
});
}
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继承人PHP代码
echo json_encode($arr1);
echo json_encode($arr2);
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所以任何人都知道如何做到这一点?
bip*_*pen 11
无需回复json编码两次....合并数组并发送数据.......
echo json_encode(array('result1'=>$arr1,'result2'=>$arr2));
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并获取数据
initChart2(data.result1,data.result2);
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看看你是否能够生成两个 json 对象数组,然后你可以尝试这样做:
var data21,data22;
$.ajax({
type: "GET",
url: "customer_coverage.php",
data: {id:id},
contentType: "application/json",
dataType: "json",
success: function (data) {
$.each(data, function(i, item){
data21 = item.data21;
data22 = item.data22;
});
initChart2(data21,data22);
}
});
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我想如果你能够生产这个:
[
{
"data21": {
.........
},
"data22": {
........
}
}
]
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