Php 从包含多个值的sql单元格中选择一个值

我有一个sql表，其中包含用于为诊所预订客户的事件。表值是从Joomla扩展中收集的。 我的问题是，有关客户端的所有信息都收集在一个表单元格中，如下面的示例所示

我如何取出这些值

"I_WANT_CLIENT_FIRST_NAME", "I_WANT_CLIENT_LAST_NAME", "I_WANT_CLIENT_EMAIL", "I_WANT_CLIENT_PHONE" and "I_WANT_CLIENT_GENDER"

所以我可以和他们分开工作

我不太熟悉sql，所以我以前从未见过这种技术。它叫什么

以下是我专栏的内容：

[{  "id":"1",
    "fieldname":"First Name",
    "fieldtype":"text",
    "varname":"firstname",
    "size":"60",
    "is_email":null,
    "is_required":"1",
    "is_first_name":"1",
    "is_last_name":"0",
    "values":null,
    "ordering":"0",
    "data":"I_WANT_CLIENT_FIRST_NAME"
},

{   "id":"2",
    "fieldname":"Last Name",
    "fieldtype":"text",
    "varname":"lastname",
    "size":"60",
    "is_email":null,
    "is_required":"1",
    "is_first_name":"0",
    "is_last_name":"1",
    "values":null,"ordering":"0",
    "data":"I_WANT_CLIENT_LAST_NAME"
},

{   "id":"3",
    "fieldname":"Email",
    "fieldtype":"text",
    "varname":"email",
    "size":"60",
    "is_email":"1",
    "is_required":"1",
    "is_first_name":"0",
    "is_last_name":"0",
    "values":null,
    "ordering":"0",
    "data":"I_WANT_CLIENT_EMAIL"},

    {"id":"4","fieldname":"Mobile","fieldtype":"text","varname":"mobile","size":"60","is_email":null,"is_required":"1","is_first_name":"0","is_last_name":"0","values":null,"ordering":"0","data":"I_WANT_CLIENT_PHONE"},{"id":"5","fieldname":"Gender","fieldtype":"radio","varname":"gender","size":"60","is_email":"0","is_required":"1","is_first_name":"0","is_last_name":"0","values":"Male|Female","ordering":"0","data":"I_WANT_CLIENT_GENDER"
}]

使用不同的：

SELECT DISTINCT first_name, last_name,email, phone, gender FROM your_table WHERE email = 'specified_mail';

---

您可以为WHERE子句使用任何字段，特别是如果它是唯一的。

使用json_decode（），因此，您的意思是表中的单行中有一列包含所有json字符串，对吗？每个客户端预订都会创建一个新行。“customfields\u data”列包含上述信息。我想取出5个要处理的值-可能在一个数组中。希望它有意义…

它叫什么

。这称为违反，这是将关系数据库与面向文档的数据库混淆的人所犯的一个常见错误。@RandomSeed可能是OP使用JSON列类型