从HTML表单插入到＆从数据库表插入

Question

-- UPDATED --

I am trying to INSERT data from a HTML FORM and a database TABLE(1) INTO TABLE(2). The below code works partially. So far data from table1 does insert into table2, as well as incrementing (having the same user enter multiple form data). But the UPDATE data is not inserting. When I check the table values, after submitting the form, room_temp shows the same 1 of 3 options (even if I've not selected it) and stage1 - stage10 shows the value of 0, even if I've entered 1,2,3 etc....

           $sql = "
            INSERT INTO table2 (user_id, user_fname, user_lname, user_dob)
            SELECT user_id, user_fname, user_lname, user_dob FROM table1;

            UPDATE table2 SET 

            room_temp='$rtemp',
            stage1='$ustage1',
            stage2='$ustage2',
            stage3='$ustage3',
            stage4='$ustage4',
            stage5='$ustage5',
            stage6='$ustage6',
            stage7='$ustage7',
            stage8='$ustage8',
            stage9='$ustage9',
            stage10='$ustage10'

            WHERE test_id=1";

Please tell me how this query should be.

Form Code - examples of my 2 input fields

        <label id="radio-inline"><input type="radio" name="roomTemp" value="warm" <?php if (isset($rtemp) && $rtemp=="warm") echo "checked";?>>&nbsp;Warm</label>
        <label id="radio-inline"><input type="radio" name="roomTemp" value="cool" <?php if (isset($rtemp) && $rtemp=="cool") echo "checked";?>>&nbsp;Cool</label>
        <label id="radio-inline"><input type="radio" name="roomTemp" value="cold" <?php if (isset($rtemp) && $rtemp=="cold") echo "checked";?>>&nbsp;Cold</label>

        <div class="form-group">
        <label class="rateLabel" for="0-3mins">Stage 1<br />5 / 8 (miles/km) ph</label>
        <input type="text" class="form-control" name="txt_stage1" placeholder="HR after 0 - 3 mins" value="<?php if(isset($error)){echo $ustage1;}?>">
        </div>

Answer 1

What is the primary key of the table? If it has one Ie userid then

Replace the second statement by an update statement

update table2 
set value1=...
from table1
where userid=1