将 JSON 解析为 MySQL 表
我正在使用 Zend Framework (1.12),我想基于 JSON 文件创建一个表.我已经创建了表及其字段(现在它们都是长文本),它所要做的就是将它们插入正确的列中.我遵循了以下示例:
I'm using Zend Framework (1.12) and I'd like to create a table based on a JSON file. I've already created the table and its fields (they're all longtext for now), all it has to do is insert them into the right columns. I followed these examples:
http://www.daniweb.com/web-development/php/threads/381669/json-to-mysql-with-php(第二篇)将 JSON 解析为 mySQL
问题是我的 JSON 构造不同(我的有一个名为 Actie 的根元素",不知道正确的术语)它包含一个包含所有对象的数组.目前,我正在使用此代码:
The problem is that my JSON is constructed differently (mine has a "root element" called Actie, don't really know the right term) which contains an array with all the objects. Currently, I'm using this code:
$actieurl = "http://creative3s.com/thomas/nmdad/actie.json";
$my_arr = json_decode(file_get_contents($actieurl));
$db = new Zend_Db_Adapter_Pdo_Mysql(array(
'host' => 'localhost',
'username' => 'root',
'password' => NULL,
'dbname' => 'zf-tutorial'
));
foreach($my_arr as $key => $value){
$sql[] = (is_numeric($value)) ? "`$key` = $value" : "`$key` = '" . mysql_real_escape_string($value) . "'";
}
$sqlclause = implode(",",$sql);
$query = "INSERT INTO `testerdetest` SET $sqlclause";
$db->query($query);
但是我收到一个错误,说我正在传递一个数组:
But I'm getting an error saying that I'm passing an array:
Warning: mysql_real_escape_string() expects parameter 1 to be string, array given in C:UsersThomasDocumentsGitHubNMDAD-testingapplicationcontrollersIndexController.php on line 29
有谁知道如何用这种格式的 JSON 解决这个问题?请记住,我无法以任何方式更改 JSON.额外链接:
Does anyone know how to solve this with a JSON of this format? Keep in mind that I cannot alter the JSON in any way. Extra links:
JSON:http://creative3s.com/thomas/nmdad/actie.json一>表结构:http://i.imgur.com/KtXeEuw.png
推荐答案
您的 json 数据具有顶级键 'Actie',因此您需要循环遍历 $my_arr->Actie
.
Your json data has the top level key 'Actie', so you need to be looping through $my_arr->Actie
.
您可以将代码简化为:
$actieurl = "http://creative3s.com/thomas/nmdad/actie.json";
$my_arr = json_decode(file_get_contents($actieurl));
$db = new Zend_Db_Adapter_Pdo_Mysql(array(
'host' => 'localhost',
'username' => 'root',
'password' => NULL,
'dbname' => 'zf-tutorial'
));
foreach($my_arr->Actie as $row){
$db->insert('testerdetest', (array)$row);
}
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