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  1. #1
    SitePoint Member
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    May 2011
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    PHP survey form + DB

    Hello,

    I have a problem displaying the results of a form.
    Right now when I complete and send the form to DB I get results ok, but I want to know when someone is checking a value from the first question what checked in other questions dependig on the fist question .
    The first question I want to be about domain of activity so I want to know and to display the result dependig on the fist question, so I want to know what are the answers to other questions depending of the answer of the first question !
    So basically to have results displayed differt depending on the first question answers !

    This is a part of my html form :

    HTML Code:
    Question1?
    
    <br>Productie<input type="radio" name="domeniu_activitate" VALUE="0" CHECKED /> 
    <br>Materiale<input type="radio" name="domeniu_activitate" VALUE="1" />
    <br>Administratie<input type="radio" name="domeniu_activitate" VALUE="2" />
    <br>Management<input type="radio" name="domeniu_activitate" VALUE="3" />
    <br>Rework<input type="radio" name="domeniu_activitate" VALUE="4" />
    <br>Engineering<input type="radio" name="domeniu_activitate" VALUE="5" />
    
    
    
    <br><br>
    Question2?
    
    
    <br>mai putin de 6 luni <input type="radio" name="de_cat_timp_lucrati" VALUE="0" CHECKED />
    <br>intre 6 luni si 1 an<input type="radio" name="de_cat_timp_lucrati" VALUE="1"  />
    <br>intre 1 - 2 ani<input type="radio" name="de_cat_timp_lucrati" VALUE="2"  />
    <br>intre 2 - 3 ani <input type="radio" name="de_cat_timp_lucrati" VALUE="3"  />
    <br>mai mult de 3 ani<input type="radio" name="de_cat_timp_lucrati" VALUE="4"  />

    This is my connection for sending results to DB :

    PHP Code:
    <?php
    $con 
    mysql_connect("127.0.0.1","root","");
    if (!
    $con)
      {
      die(
    'Could not connect: ' mysql_error());
      }

    mysql_select_db("Baza de date"$con);

    $sql="INSERT INTO Tabel_bzd (domeniu_activitate,de_cat_timp_lucrati)

    VALUES

    ('
    $_POST[domeniu_activitate]','$_POST[de_cat_timp_lucrati]')";

    if (!
    mysql_query($sql,$con))
      {
      die(
    'Error: ' mysql_error());
      }
    echo 
    "1 record added";

    mysql_close($con)
    ?>

    This is my query for displaying results :


    PHP Code:
    <?php
    $con 
    mysql_connect("127.0.0.1","root","");
    if (!
    $con)
      {
      die(
    'Could not connect: ' mysql_error());
      }

    mysql_select_db("Baza de date"$con);

    //$result = mysql_query("SELECT * FROM Tabel_bzd");
    echo "<b><p class='one'>Question1? </p></b><br>";
    $i=0;
    while(
    $i<6)
    {
        
    $result mysql_query("SELECT COUNT(*) as suma FROM Tabel_bzd WHERE domeniu_activitate = "$i ." ");
        while(
    $row mysql_fetch_array($result))
        {
          switch (
    $i)
            {
                case 
    0:
                        echo 
    "Productie: ";
                        echo 
    $row['suma'];
                        break;
                case 
    1:
                        echo 
    "<br>Materiale: ";
                        echo 
    $row['suma'];
                        break;
                case 
    2:
                        echo 
    "<br>Administratie: ";
                        echo 
    $row['suma'];
                        break;
                case 
    3:
                        echo 
    "<br>Management: ";
                        echo 
    $row['suma'];
                        break;
                case 
    4:
                        echo 
    "<br>Rework: ";
                        echo 
    $row['suma'];
                        break;            
                default:
                        echo 
    "<br>Engineering: ";
                        echo 
    $row['suma'];
            }
        }
        
    $i++;
    }

    echo 
    "<br><br><b><p class='one'> Question2?</p></b>";
    $i=0;
    while(
    $i<5)
    {
        
    $result mysql_query("SELECT COUNT(*) as suma FROM Tabel_bzd WHERE de_cat_timp_lucrati = "$i ." ");
        while(
    $row mysql_fetch_array($result))
        {
          switch (
    $i)
            {
                case 
    0:
                        echo 
    "<br>mai putin de 6 luni: ";
                        echo 
    $row['suma'];
                        break;
                case 
    1:
                        echo 
    "<br>intre 6 luni si 1 an: ";
                        echo 
    $row['suma'];
                        break;
                case 
    2:
                        echo 
    "<br>intre 1 2 ani: ";
                        echo 
    $row['suma'];
                        break;
                case 
    3:
                        echo 
    "<br>intre 2 3 ani: ";
                        echo 
    $row['suma'];
                        break;        
                default:
                        echo 
    "<br>mai mult de 3 ani: ";
                        echo 
    $row['suma'];
            }
        }
        
    $i++;
    }
    mysql_close($con);
    ?>

  2. #2
    SitePoint Member
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    May 2011
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    I found a way defining a variable for the first question and displaying it permanently of the resuts page and when I select an option is showing me different result !


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