我对 Laravel 完全陌生。我有一张 table
+---------------------+
| id | Content |
+---------------------+
| 1 | {"_token":"yAzxyLrdQfVscyc3fNB87PU4e1p6sS4JwX6AfmUQ","datefrom":"2017-07-07","dateto":"2017-07-31","Productivity":"Productivity","Productivityrating":"2","Technical_Skills":"Technical skill","Technical_Skillsrating":"3","Work_Consistency":"Work consistency","Work_Consistencyrating":"4","Presentation_skills":"Presentaion skills","Presentation_skillsrating":"3","test":"test","testrating":"5","cycle_id":"1","save":"proceed"} |
|
|
|
+------+--------------+
字段名称内容为json。喜欢
{"_token":"yAzxyLrdQfVscyc3fNB87PU4e1p6sS4JwX6AfmUQ",
"datefrom":"2017-07-07",
"dateto":"2017-07-31",
"Productivity":"Productivity",
"Productivityrating":"2",
"Technical_Skills":"Technical skill",
"Technical_Skillsrating":"3",
"Work_Consistency":"Work consistency",
"Work_Consistencyrating":"4",
"Presentation_skills":"Presentaion skills",
"Presentation_skillsrating":"3",
"test":"test",
"testrating":"5",
"cycle_id":"1",
"save":"proceed"}
我想从cycle_id=1的表中搜索 我有一个 mysql 查询
$sql="SELECT *from table where `Content`->>'$.cycle_id'=1";
如何将其转换为 Laravel？
喜欢
$user = DB::table('table')->where('Content', '$.cycle_id'=1)->first();
请帮助我。任何帮助将不胜感激
最佳答案
从 Laravel 5.6.24 开始，您可以使用WhereJsonContains():
User::whereJsonContains('content', ['cycle_id' => 1])->get()
关于php - 如何将查询转换为 Laravel，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/45025812/