2014-10-10 111 views
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我有这个json编码数据使用'服务器端'版本的数据表。 但是,如何将这个值放入数据表中,以便我的数据可以以表格形式列出?json编码数据到数据表

{"sEcho":0,"iTotalRecords":"40","iTotalDisplayRecords":"40","aaData":[["[email protected]","5656565","656565"],["[email protected]","5656565","656565"],["[email protected]","5656565","656565"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","656","56"],["[email protected]","656","56"],["[email protected]","656","56"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","56565","56565"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Greater Noida","201308"],["[email protected]","Lucknow","226005"],["[email protected]","Lucknow","226005"],["[email protected]","ko","koko"],["[email protected]","ko","koko"],["[email protected]","Greater Noida","201308"],["[email protected]","Lucknow","226005"],["[email protected]","o","ok"],["[email protected]","TA","TA"],["[email protected]","Lucknow","226012"],["[email protected]","Lucknow","226012"],["[email protected]","Lucknow","226012"],["[email protected]","o","ok"],["[email protected]","o","ok"]]} 
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问题没有意义,不是表格构成'表格格式? – charlietfl 2014-10-10 19:39:30

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由于某种原因,它没有。没有表格,只有json被回显。 – user2619381 2014-10-10 19:41:40

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记录的插件方法的哪部分不起作用呢? – charlietfl 2014-10-10 19:42:38

回答

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$obj = json_decode($json); 

将创建与您的数据的对象

你也可以创建一个数组,但uusing真正的第二个参数

$array= json_decode($json, true);