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我从我们的数据库(与GLua输入)与PHP的信息,然后使用json_decode将其更改为数组使用,但它返回NULL,而其他人是加工?json_decode返回NULL [{},{}]
// Get the RapSheet info from 'character_data_store'
$rap = mysql_query("SELECT * FROM `character_data_store` WHERE `character_id` = '$srp_uid' AND `key`='RapSheet'");
while($rapsheet=mysql_fetch_assoc($rap)){
$raps = $rapsheet['value'];
};
然后我用
// Deal with the rapsheet JSON
echo $raps;
$raps = json_decode($rapsheet, true);
echo var_dump($raps, TRUE);
回声的是,以检查它的工作,该信息被成功扳回一,因为它的回声,虽然后续代码var_dump返回
NULL布尔(真)
数据库内容:
[{"ArrestedBy":"James Scott","Date":1483732483,"Duration":60,"LeaveReason":"Jail time served.","ArrestReason":"test"}]
任何帮助将不胜感激!
试图达伦的响应后:
我试图
$rap = mysql_query("SELECT * FROM `character_data_store` WHERE `character_id` = '$srp_uid' AND `key`='RapSheet'");
$raps = array();
while($rapsheet=mysql_fetch_assoc($rap)){
$raps[] = $rapsheet['value'];
};
// encode
$rap = json_encode($raps, TRUE);
echo $rap;
这回:
[ “[{\” ArrestedBy \“: “詹姆斯·斯科特”,“日期”:148373248 3,\ “持续时间\”:60 \ “LeaveReason \”:\ “监狱的时间送达\”,\ “ArrestReason \”:\ “测试\”}]“] 所以,我想
echo $rap['ArrestedBy'];
,并返回
[
您好,我想这一点,但我仍然无法得到它的工作:( $ RAP =请求mysql_query(“SELECT * FROM'character_data_store' WHERE'character_id' = '$ srp_uid' 和'关键'= 'rapSheet'“); \t $斥责=阵列(); \t \t而($ rapsheet = mysql_fetch_assoc($ RAP)){ \t \t $斥责[] = $ rapsheet [ '值']; \t \t echo $ raps ['ArrestedBy']; \t}; – JamesScott
请将您的代码保留在您的答案中( *不是评论哈哈*)。什么没有用? – Darren
对不起达伦,通常不是一个stackoverflow的人!此外,它不会回应任何:/ – JamesScott