use*_*551 8 php mysql arrays json
嗨,我正在尝试将json数组插入我的MySQL数据库.我将数据从我的iphone传递到那里我将数据转换为json格式,并且我使用url将数据传递到我的服务器,而不是插入我的服务器.
这是我的json数据.
[{ "Name": "0", "手机": "DSF", "城市": "sdfsdf", "电子邮件": "DSF"},{ "名": "13123123", "手机":"sdfsdfdsfsd ", "城市": "sdfsf", "电子邮件": "13123123"}]
这是我的Php代码.
<?php
$json = file_get_contents('php://input');
$obj = json_decode($data,true);
//Database Connection
require_once 'db.php';
/* insert data into DB */
foreach($obj as $item) {
mysql_query("INSERT INTO `database name`.`table name` (name, phone, city, email)
VALUES ('".$item['name']."', '".$item['phone']."', '".$item['city']."', '".$item['email']."')");
}
//database connection close
mysql_close($con);
//}
?>
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我的数据库连接代码.
<?php
//ENTER YOUR DATABASE CONNECTION INFO BELOW:
$hostname="localhost";
$database="dbname";
$username="username";
$password="password";
//DO NOT EDIT BELOW THIS LINE
$link = mysql_connect($hostname, $username, $password);
mysql_select_db($database) or die('Could not select database');
?>
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请告诉我在上面的代码中我做错了基本上我不是一个php开发人员我是移动应用程序开发人员所以我使用php作为服务器端脚本请告诉我如何解决这个问题.
Ami*_*mit 10
$json = file_get_contents('php://input');
$obj = json_decode($json,true);
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我认为你传递了错误的变量.你应该通过$json在json_decode如上图所示.
小智 5
您缺少 JSON 源文件。创建一个 JSON 文件,然后将其分配给 var 数据:
<?php
require_once('dbconnect.php');
// reading json file
$json = file_get_contents('userdata.json');
//converting json object to php associative array
$data = json_decode($json, true);
// processing the array of objects
foreach ($data as $user) {
$firstname = $user['firstname'];
$lastname = $user['lastname'];
$gender = $user['firstname'];
$username = $user['username'];
// preparing statement for insert query
$st = mysqli_prepare($connection, 'INSERT INTO users(firstname, lastname, gender, username) VALUES (?, ?, ?, ?)');
// bind variables to insert query params
mysqli_stmt_bind_param($st, 'ssss', $firstname, $lastname, $gender, $username);
// executing insert query
mysqli_stmt_execute($st);
}
?>
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