使用PHP从Mysql获取Json
嗨,我正在用php抓取我的表并得到这个输出使用PHP从Mysql获取Json,php,mysql,json,Php,Mysql,Json,嗨,我正在用php抓取我的表并得到这个输出 {"id":"1","player":"123","name":"his_name","reason":"reason text","instance":"1","lastupdated":"datetime"} 然而,我希望得到这样的东西 text{"id":"1","player":"123","name":"his_name","reason":"reason text","instance":"1","lastupdated":"dateti
{"id":"1","player":"123","name":"his_name","reason":"reason text","instance":"1","lastupdated":"datetime"}
然而,我希望得到这样的东西
text{"id":"1","player":"123","name":"his_name","reason":"reason text","instance":"1","lastupdated":"datetime"}
当前php:
mysql_connect($dbhost.':'.$dbport, $dbuser, $dbpass) or die (mysql_error());
mysql_select_db("dbname") or die (mysql_error());
$sth = mysql_query("SELECT *"."FROM `table_name` Order By id Desc;");
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
$rows[] = $r;
}
print json_encode($rows);
数据库结构:
player | name | reason | instance | lastupdated
如果您想在文字文本前加前缀,请使用字符串连接运算符打印:
print "text" . json_encode($rows);
这对你有用吗?但是,如果只想在从数据库获得的完整结果集之前添加一个键,那么
文本的内容应该是什么。正如我从您的示例链接中了解到的
然后您可以像这样更新代码:
// Assuming $rows is already populated with desired data.
$res = array();
$res['text'] = $rows;
print_r(json_encode($res));
输出:
{"text":{"id":1,"player":123,"name":"his_name","reason":"reason text","instance":1,"lastupdated":"datetime"}}
希望有帮助。为什么要添加文本
?我轮询它的方式是我需要文本在那里,我在各种API上见过它,比如twitch的kraken API。添加这样的文本会导致json无效。除非您是作为jsonp类型响应之王来执行此操作,否则在json字符串之外添加文本是没有意义的;他们的方式-频道-名称-描述-等等…啊。。。所以,从技术上讲,你是在这个数组之前附加了一个数组,对吗?正是我想要的,只是我不知道如何解释它。