将外键自动插入表中的简单方法

Question

I have two table program(p_id(pk)AI, program_name and another table graduate_survey(id(pk)AI,total_PO1,total_PO2,session,p_id(fk) . In the program table values are already inserted as

p_id Program_name
    1    B.tech CSE
    2    B.tech IT.

..so on. I select from a drop down list B.tech CSE and then redirect to the survey form. I enter the total_PO1, total_PO2 there and submit it.Now i want to insert in the graduate_survey table where in the fk field the p_id of B.tech CSE should automatically enter so that i can know the survey is done for which program..Is there any query in MySQL to do that? The insert operation should be done in php code.Please suggest any query.

Answer 1

Your approach seems to be incorrect. When you generate your dropdown menu in HTML, then the display value (the value that the user actually sees and selects) should be the program name, but the underlying value should be the id of the same record.

When you send the data to the server, then you should send the underlying id value, not the display value. Using HTML's standard dropdown menu, you would create sg like this as an output:

<select name="programs">
    <option value="1">B.tech CSE</option>
    <option value="2">B.tech IT</option>
</select>

If this control is submitted via a form, then the browser will send the value of the selected option automatically. If you use ajax, then your code has to retrieve the selected value of the select element. In the php code you just insert the underlying value into the graduate survey table.