你好我无法弄清楚如何遍历这个json编码数组,并为每个对象,获取它的所有值。我需要将每个值作为自己的变量。json_encode获取值foreach对象
echo json_encode($formulars);
这是我所得到的,当我赞同它
[{"project_name":"polle","type":"support","title":"vi","reason":"prover","solution":"igen","comments":"okay ","date_stamp":"2013-08-20 14:06:37","used_time":132},{"project_name":"dolla","type":"support","title":"lolol","reason":"skl","solution":"dskal","comments":"kflafda ","date_stamp":"2013-08-20 14:11:36","used_time":210},{"project_name":"polle","type":"fejl","title":"lol","reason":"aksdl","solution":"fdjks","comments":"djsks ","date_stamp":"2013-08-20 14:13:27","used_time":1230}]
我已经试过这段代码,我设法脱身,从第一个对象的PROJECT_NAME,这就是它:
foreach ($formulars as $current => $project_name) {
$project_name['project_name'];
}
那么,有什么办法可以让每个对象中的所有变量我的数组而不仅仅是PROJECT_NAME的?
像这样:
foreach ($formulars as $current){
$projectName = $current['project_name'];
$type = $current['type'];
$reason = $current['reason'];
}
在此先感谢
尝试json_decode(),那么你可以访问它。 – Parixit
“像这样:”代码似乎就足够了。为什么它不工作?发生了什么? – mario
哦,上帝马里奥我竟然没有测试出“这样”的代码,我只是快速输入了它。我只是试了一下,它的工作。我觉得很愚蠢,呵呵。谢谢你! –