php–将JSON解析为mySQL

获取我的PHP脚本JSON字符串,看起来像这样(包含任何对象的数组)：

[
{
"source":"symbols/2/2.png",
"ypos":133,
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28",
"rotation":0,
"type":"MyImage",
"width":252,
"depth":5,
"height":159,
"xpos":581
},
{
"source":"symbols/2/2.png",
"ypos":175,
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28",
"rotation":0,
"type":"MyImage",
"width":258,
"depth":3,
"height":163,
"xpos":214
},
{
"color":"0",
"ypos":468.38,
"fontSize":28,
"xpos":156.95,
"rotation":0,
"type":"MyTextArea",
"width":268.05,
"depth":7,
"height":244.62,
"fontFamily":"Verdana Bold",
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28"
}
]


如何在mySQL中使用记录保存此数组中的每个JSON对象？

解决方法:

尝试这个：

$json = '[
{
"source":"symbols/2/2.png",
"ypos":133,
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28",
"rotation":0,
"type":"MyImage",
"width":252,
"depth":5,
"height":159,
"xpos":581
},
{
"source":"symbols/2/2.png",
"ypos":175,
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28",
"rotation":0,
"type":"MyImage",
"width":258,
"depth":3,
"height":163,
"xpos":214
},
{
"color":"0",
"ypos":468.38,
"fontSize":28,
"xpos":156.95,
"rotation":0,
"type":"MyTextArea",
"width":268.05,
"depth":7,
"height":244.62,
"fontFamily":"Verdana Bold",
"template":"8B82CA47-41D2-D624-D6A2-37177CD82F28"
}
]';
//create a DB connection
con = mysql_connect("localhost","username","password");
mysql_connect _db('your_database',$con);
$result = json_decode($json);
foreach($result as $key => $value) {
if($value) {
//how to use json array to insert data in Database
mysql_query("INSERT INTO tablename (source, ypos, template) VALUES ($value->source, $value->ypos,$value->template)");
}
mysql_close($con);
}


注意：但建议使用PHP数据对象(PDO)来执行数据库操作.
检查here