从php中获取json的值 [英] get value from json in php
本文介绍了从php中获取json的值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在PHP中有一个json,想要从中获取名称和数字并将其存储在数据库中
I have a below json in PHP and want to fetch name and number from it and store it in database
{
"student": [{
"Name": "John",
"Number": "1234567890"
}, {
"Name": "Jonny",
"Number": "0987654321"
}, {
"Name": "Tom",
"Number": "0212365895"
}]
}
我尝试过以下代码但无法获取代码:
I have tried below code but not able to fetch it:
<?php
//http://localhost:8080/sample1/webservice2.php?json={%22UserName%22:1,%22FullName%22:2}
//$json=$_GET ['json'];
$json = file_get_contents('php://input');
$obj = json_decode($json);
echo $json;
//Save
$con=mysql_connect("localhost","USERNAME","PASSWORD");
mysql_select_db("DATASASE",$con);
/* grab the posts from the db */
//$query = "SELECT post_title, guid FROM wp_posts WHERE post_author = $user_id AND post_status = 'publish' ORDER BY ID DESC LIMIT $number_of_posts";
//for($i=0;i<count($obj);$i++){
mysql_query("INSERT INTO Contact (Name, Number) VALUES ('".$obj->{'student'}{'Name'}."', '".$item->{'Number'}."')");
//}
mysql_close($con);
//
//$posts = array($json);
$posts = array(1);
header('Content-type: application/json');
echo json_encode(array('posts'=>$posts));
?>
推荐答案
json =
_GET ['json'];
_GET ['json'];
json = file_get_contents(' php:// input');
json = file_get_contents('php://input');
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