从json api到mysql

Question

我需要帮助解析json数据到mysql。我的代码不工作，这是我的代码。

<?php 
$server = "localhost"; 
$username = "root"; 
$password = "12345"; 
$database = "json"; 
mysql_connect($server,$username,$password) or die("Failed"); 
mysql_select_db($database) or die("Database Failed"); 

$url = "http://demo.miliarta.com/cityapi/all/?dealerid=TEN000005&user=dealer&passwd=dealer&cityid=316"; 
$string = file_get_contents($url); 
$arr = json_decode($string, true); 

//array instances specific to json items 
$id = $arr["cityid"]; 
$id2 = $arr["stateid"]; 
$id3 = $arr["cityname"]; 
$id4 = $arr["statename"]; 
$s=0; 
//Enumerating Array index 
foreach($arr as $item=> $value){ 
    $s=count($value); // WIN 
} 
echo $s; 
//suck the array for loop 
for($i=0;$i<$s;$i++){ 
    $cityid= $id[0]; 
    $stateid = $id2[$i]; 
    $cityname = $id3[$i]; 
    $statename = $id4[0]; 
    mysql_query("INSERT INTO city (cityid, stateid, cityname, statename) VALUES('$cityid', '$stateid', '$cityname', '$statename')") or die (mysql_error()); 
} 
?> 

问题是上线22 说明：未定义指数：cityid在C：\ XAMPP \ htdocs中\ JSON \ jsontosql.php上线22 说明：用C STATEID：：未定义指数\ XAMPP \第23行的htdocs \ json \ jsontosql.php 注意：未定义的索引：在C：\ xampp \ htdocs \ json \ jsontosql.php中的cityname在第24行 注意：未定义索引：statename在C：\ xampp \ htdocs \ json \ jsontosql.php 25行 4表'json.city'不存在

tq寻求帮助。

Answer 1

为你修复它。你的循环是错误的，你需要先看$arr来获取每一行数据。

试试这个：

<?php 
// Your database configs here 

$url = "http://demo.miliarta.com/cityapi/all/?dealerid=TEN000005&user=dealer&passwd=dealer&cityid=316"; 
$string = file_get_contents($url); 
$arr = json_decode($string, true); 

foreach($arr as $item){ 
    $cityid = $item['cityid']; 
    $stateid = $item['stateid']; 
    $cityname = $item['cityname']; 
    $statename = $item['statename']; 

    mysql_query("INSERT INTO city (cityid, stateid, cityname, statename) VALUES('$cityid', '$stateid', '$cityname', '$statename')") or die (mysql_error()); 
}