I am using laravel 5.0. I want using a laravel validator where my MENU_NAME column in 'dbo.MS_MENU' is unique. I am using sql server database. I have made a validator code like in the below, but I still got an error
QueryException in Connection.php line 624: SQLSTATE[23000]: [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Violation of UNIQUE KEY constraint 'IX_MS_MENU'. Cannot insert duplicate key in object 'dbo.MS_MENU'. The duplicate key value is (Role). (SQL: EXEC dbo.M_INSERT_MENU_PARENT '0', '121', 'Role', 'Menu coba', '0', 'SDF')
$validator = Validator::make($request->all(), [
'MENU_NAME' => 'unique:dbo.MS_MENU']);
if ($validator->fails()) {
return redirect ('Menu')->withErrors($validator)->withInput();
}
pls try with this one
$validator = Validator::make($request->all(), [
'MENU_NAME' => 'unique:dbo.MS_MENU,MENU_NAME']);
You need to use table name and also column name
Hello check following code,
//Validation using ajax
public function checkvalidation(Request $request) {
$data['SUCC_FLAG'] = 0;
$formdata = $request->only('uname', 'umail', 'dob', 'cno', 'pwd', 'cpwd');
$rules = ['uname' => 'required|min:2|alpha', 'umail' => 'required|email', 'dob' => 'required|date', 'cno' => 'required|digits:10', 'pwd' => 'required|min:3|max:6', 'cpwd' => 'required|same:pwd'];
$validator = Validator::make($formdata, $rules);
if ($validator->fails()) {
$data['msg'] = $validator->errors();
} else {
$data['SUCC_FLAG'] = 1;
$data['msg'] = "success";
}
return json_encode($data);
}
this is sample code i use for my project you have to change according to your use or project.
for use validator you have to include use Validator; in header
i hope this works for you
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