2014-02-10 10:29
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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>

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.

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1条回答 默认 最新

  • douxian6086 2014-02-10 10:34

    In your textbox part;


    <label for="education">Hobby: <input type="text" name="education"/> </label>


    <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

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