2013-02-03 16 views
0

有人可以請告訴我我將如何張貼選擇下拉列表值沿着我的文字輸入值到MySQL數據庫。將下拉值發佈到mysql?

我有以下,不知道應該如何處理下拉值。

感謝

<form action=\"includes/welcomestats.php\" method=\"post\" id=\"form1\">    

    <input type=\"text\" name=\"location\" id=\"location\" maxlength=\"30\" placeholder=\"Location: e.g. London\"> 
    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Enter Your Location</label><br/> 
    <br/> 

    <input type=\"text\" name=\"local_station\" id=\"local_station\" maxlength=\"30\" placeholder=\"Local Train Station\"><label>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Enter a Local Train Station</label><br/><br/><br/> 

    <h5>About You</h5><br/> 
    <select name=\"formGender\" style=\"min-width:125px;\"> 
     <option value=\"\">Select...</option> 
     <option value=\"M\">Male</option> 
     <option value=\"T\">Female</option> 
    </select> 
    <label>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Enter Your Gender</label> 
    <br/><br/> 

    <input type=\"submit\" class=\"welcome-submit2\" name=\"submit\" value=\"Next ->\" id=\"submit\"/> 
    </form> 


    <?php 
    require_once("session.php"); 
    require_once("functions.php"); 
    require('_config/connection.php'); 
    ?> 
    <?php 
    session_start(); 
    include '_config/connection.php'; 
    $location = $_POST['location']; 
    $result = mysql_query("SELECT location FROM ptb_profiles WHERE id=".$_SESSION['user_id'].""); 
    if(!$result) 
    { 
    echo "The username you entered does not exist"; 
    } 
    else 
    if($location!= mysql_result($result, 0)) 
    { 
    echo ""; 
     $sql=mysql_query("UPDATE ptb_profiles SET location ='".addslashes($display_name)."' WHERE id=".$_SESSION['user_id'].""); 

    } 

回答

0

選擇字段值的處理方式相同任何其他領域。所以,如果你有選擇字段一樣 -

<select name=\"formGender\" style=\"min-width:125px;\"> 
     <option value=\"\">Select...</option> 
     <option value=\"M\">Male</option> 
     <option value=\"T\">Female</option> 
    </select> 

您可以使用

 $gender = $_POST['formGender'] 

更新 這會給你M或F.

+0

它會給你的男或女,不男性或女性 – bhttoan

+0

是的。我忽略**值** – SachinGutte