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发送卷曲-i -X POST -d 'JSON = { “金额”:1000, “idPlayers”:1}' http://www.myurl.com/updateamount.phpPHP不获取JSON值I从卷曲
我的PHP代码:
<?php
$json = urldecode($_GET['jsonSendData']);
$json = str_replace("\\", "", $json);
$data = json_decode($json, true);
// if i hard code it here it works fine, but need from a post from curl
//$json = '{"Amount":1000,"idPlayers":1}';
$data = json_decode($json, true);
echo "json = ", $json;
$amount = mysql_real_escape_string($data['Amount']);
$id = mysql_real_escape_string($data['idPlayers']);
echo "Amount = ",$amount;
return;
奏效的感谢!我尝试过,但不正确! – 2012-02-25 22:12:31
我的Xcode人认为我可以像下面这样做,所以我们不必发送'json'值。但它没有任何想法? $ json = urldecode($ _ POST ['json']); \t $ json = str_replace(“\\”,“”,$ json); \t $ data = json_decode($ json,true); // $ amount = json_decode($ _ POST ['Amount']); // $ player_id = json_decode($ _ POST ['idPlayers']); – 2012-02-26 23:55:57