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php - 我想要以下结果 : $name = 'Doron' ; $email = 'doron@myemail.com' ; $phone = '0501234567' ;

这个问题在这里已经有了答案:HowtoextractandaccessdatafromJSONwithPHP?(1个回答)关闭3年前。我想从我的JSON中提取以下数据:$name='Doron';$email='doron@myemail.com';$phone='0501234567';它是一个多维数组。电子邮件有一个键type:email,电话有type:tel,名字有id:name$myJson='[{"name":{"id":"name","type":"text","title":"","value":"Doron","raw_value":"Doron","required

MySQL GROUP_CONCAT 和 DISTINCT

我有这样的tb1表:nameemaillinkjohnmyemail@gmail.comgooglejohnmyemail@gmail.comfacebookjohnmyemail2@gmail.comtwitter........andmore当我用查询调用数据时看起来像SELECTname,email,group_concat(DISTINCTlinkSEPARATOR'/')assourceFROMtb1groupbyemail结果是这样的:NAMEEMAILSOURCEjohnmyemail2@gmail.comtwitterjohnmyemail@gmail.comfaceb