2016-11-14 37 views
1

試圖讓選擇下拉菜單中的值,如下葛亭選擇下拉

<select id="utype" class="text" style="width:100%;padding:10px !important" name="user-type"> 
    <option value="default">Select </option> 
    <option value="passenger">Passenger</option> 
    <option value="driver">Driver</option> 
</select> 

但始終司機值inseted到數據庫,甚至乘客由用戶選擇

if (isset($_POST['submit'])) { 

    $date = date('Y-m-d'); 
    $full_name = mysql_real_escape_string(addslashes($_REQUEST['fname'])); 
    $email_address = mysql_real_escape_string(addslashes($_REQUEST['email'])); 
    $pass = mysql_real_escape_string(addslashes($_REQUEST['password'])); 
    $contact_no = mysql_real_escape_string(addslashes($_REQUEST['mobile'])); 
    $usertype = mysql_real_escape_string(addslashes($_REQUEST['user-type'])); 
} 

同樣的價值我試圖以3種方式獲得usertype值,但仍然只有驅動程序值被插入到數據庫中。

1.$usertype = mysql_real_escape_string($_POST['user-type']); 
2.$usertype= isset($_POST['user-type']) ? mysql_real_escape_string($_POST['user-type']) : ''; 
3.$usertype=$_POST['user-type']; 

即使當我修復usertype像$ usertype ='passenger';並插入到分貝值$用戶類型.driver插入DB 這裏是compelet php文件

<?php 
    include ('../includes/include_files.php'); 
    $msg = ""; 


/* ----ON SUBMIT------ */ 


if (isset($_POST['submit'])) { 

$date = date('Y-m-d'); 

$full_name = mysql_real_escape_string(addslashes($_REQUEST['fname'])); 
$email_address = mysql_real_escape_string(addslashes($_REQUEST['email'])); 
$pass = mysql_real_escape_string(addslashes($_REQUEST['password'])); 
$contact_no = mysql_real_escape_string(addslashes($_REQUEST['mobile'])); 
$usertype = 'passenger'; 
// $usertype = mysql_real_escape_string($_POST['user-type']); 
//$usertype= isset($_POST['user-type']) ? mysql_real_escape_string($_POST['user-type']) : ''; 
//$usertype=$_POST['user-type']; 
// $("#utype").val(); 
// $("#elementId").val(); 
if($usertype='driver'){ 
    $status='Inactive'; 
} 
else 
{ 
    $status="Active"; 
} 

if ($full_name != "" && $email_address != "" && $pass != "" && $contact_no != "" && $usertype != "") { 

    $checkinfo = mysql_query("select * from tbl_user where email='" . $_POST['email'] . "' "); 
    $count_email = mysql_num_rows($checkinfo); 
    if ($count_email == 0) { 

     $reg_ins = mysql_query("insert into tbl_user(`fullname`, `email`, `password`, `mobile`, `usertype`,`add_date`, `status`) values ('$full_name','$email_address','$pass','$contact_no','$usertype','$date', '$status')"); 

     if ($reg_ins) { 

      $uinfo = mysql_query("select * from tbl_user where email='" . $_POST['email'] . "' and password='" . $_POST['password'] . "' "); 
      $res = mysql_fetch_object($uinfo); 

      echo "<script>window.android.clickOnRegister();</script>"; 
     } 
    } else { 

     $msg = " Email ID Already Exist."; 
    } 
} else { 

    $msg = "Please fill all required fields *"; 
} 
} 
?> 
+0

你是如何將數據傳遞到PHP – guradio

+0

你讓通過AJAX或一些特殊的職位? –

+0

你的代碼似乎很好。可能是你可以嘗試通過更改您的選擇框的名稱user_type –

回答

0

的問題在你的代碼在這條線上:

if($usertype='driver'){ 

由於您只使用一個=,因此您需要使用這種==這樣的雙倍來進行比較。有一個等於$usertype的值driver。 (也將永遠免費)。

因此改變該行:

if($usertype=='driver'){ 
+0

什麼錯誤! thx哥們 –

0

看起來你背後有一些特殊的邏輯。

如果您將使用標準方式發送您的表單,您將獲得user-type參數的正確值。

檢查您的表單中是否有唯一的name="user-type"

請檢查在firebug/chrome開發人員工具中發送到服務器的值是什麼,以確定問題出在後端還是前端。

+0

;並插入到db值$ usertype .driver插入到數據庫中!!!!!!!!!!!!! –

+0

顯示你的mysql查詢然後 – luke

+0

完整的php文件添加到問題 –

0

嘗試: -

<select size="1" id="slt_country" name="country"> 
<option value=""> - Select - </option> 
<option value="AF">Afghanistan</option> 
<option value="AL">Albania</option> 
<option value="DZ">Algeria</option> 

... 

<option value="ZM">Zambia</option> 
<option value="ZW">Zimbabwe</option> 
</select> 

,可以用來獲得用戶選擇的國家的JavaScript代碼:

var selected_index = oForm.elements["country"].selectedIndex; 

if(selected_index > 0) 
{ 
    var selected_option_value = oForm.elements["country"].options[selected_index].value; 
    var selected_option_text = oForm.elements["country"].options[selected_index].text; 
} 
else 
{ 
    alert('Please select a country from the drop down list'); 
} 

鏈接: - http://www.javascript-coder.com/javascript-form/javascript-get-select.phtml