I have a form that a user can insert his/her level of education either by picking a value from a drop down menu or just typing some text. My code is simple:
<label for="education">Education: <input type="text" name="education"/> </label>
<select name="education">
<option value=""></option>
<option value = "Primary education" name="Primary education">Primary education</option>
<option value = "Secondary education" name="Secondary education">Secondary education</option>
<option value = "Bachelor" name="Bachelor">Bachelor</option>
<option value = "Master" name="Master">Master </option>
<option value = "Doctoral" name="Doctoral">Doctoral </option>
</select>
So i want to insert that value in a column called education in database. Using code above the value is inserted only when someone is picking a value from menu. Input text is not stored in database.
In your textbox part;
Change
<label for="education">Hobby: <input type="text" name="education"/> </label>
to
<label for="education">Hobby: <input type="text" name="education_text"/> </label>
Duplicate name causes your problem(select and textbox name are same). I have given education_text
as an example, update it according to your needs. Do not forget to handle it backend with updated name.
On your backend;
$education = "";
if(!empty($_REQUEST["education_text"])) {
$education = $_REQUEST["education_text"];
} else if(!empty($_REQUEST["education"])) {
$education = $_REQUEST["education"];
} else {
die("Invalid education");
}
And use it in your query. You can change $_REQUEST
to $_POST
or $_GET
according to your needs