![]() Using (SqlCommand cmd = conn.CreateCommand())Ĭmd.CommandText = SqlDbType.Date).Value = SqlDbType.Char, 40).Value = SqlDbType.Char, 40).Value = SqlDbType.Char, 40).Value = SqlDbType.VarChar, 40).Value = txtChip. RadioButton rb1 = ().FirstOrDefault(r => r.IsChecked = true) Ĭonst string SqlString = "insert into PetItem(DuoDate,PetName,PetRace, PetColor, PetChip,PetGender,PetSterile) values (SqlConnection conn = new SqlConnection(connectionString)) Misalnya untuk memilih jenis kelamin, pengguna hanya dapat memilih salah satunya saja dan tidak boleh lebih dari. The way I know its not getting through isset() is because it is not echoing out 'in isset ' therefore can not reach the rest of my code. I was wondering what would be the best way to do this < form action''. I have tried this method before but with text fields, so I'm assuming I am doing something wrong with the configuration of the radio buttons. ![]() ![]() If you mean you want the same radio button selected as the one they previously selected then make a variable for each one, and if that was the previously selected one set it as "CHECKED".Īnd put the relevant variable in the end of each radio button.RadioButton rb = ().FirstOrDefault(r => r.IsChecked = true) Cara Membuat Radio Button di PHP Radio Button adalah salah satu komponen input didalam tag html yang berfungsi untuk membuat inputan dari pengguna untuk memilih salah satu jawaban yang tersedia. I have 4 different MySQL query's and I want to be able to choose which one is used based on which radio button is clicked. On your diycake.php you should check, isset(POST'submit'), if the form was sent.The value from the radio button is also in the POST variable, accessible by it's name. If they didn't put in a value then the box will be empty. There is no need for an isset, as the input will just be blank if it isn't set. How I imagine a solution: I've read up online (particularly here on stackoverflow), and it seems this. However, it doesn't, because it creates an undefined variable issue. However, it doesnt, because it creates an undefined variable issue. Marie, I'm not sure what your createforminput function does (and I'm confused about some of your HTML), but the following is a very simple works-out-of-the-box demonstration of how to handle radio button logic. If the radio button is left unchecked, I'd still like the rest of the search to run. If the radio button is left unchecked, Id still like the rest of the search to run. Stack Overflow is leveraging AI to summarize the most relevant questions and answers from the community, with the option to ask follow-up questions in a conversational format. The first is 'question1', the second is 'question2' and so on. ![]() ![]() Keep all the previously submitted info in variables and set them as the inputs values so the info is not lost. Problem: I have a search-form with a radio button, which uses PHP to post to a database. The radio buttons do belong to a radio group with each pass of the while statement. To show the values in the input fields after the user hits the submit button, we add a little PHP script inside the value attribute of the following input fields: name, email, and website. I'm building a quiz for a server-side scripting class. I like how you acted like we were morons yet you were wrong.ĭBookatay rather than have an error page, just have the errors appear on the same page as the form. How can i recieve the value of the radio button once the form is posted (in PHP) POST'radio' Once it is posted on the same page, how can I remember the selected radio button and keep that checked Add a checked attribute if the value is equal to POST'radio'. I've searched through a lot of previously asked questions about php and radio buttons but I haven't yet found something helping in my situation. He's trying to get the value of his radio choice into a text field.īookatay, what you have should work if POST is set. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |