如何使用php ajax從mysqldb到dropdownbox的值？

Question

如何使用php ajax從mysql數據庫獲取值到dropdownbox？

我需要的是

當我從dropdownbox選擇任何值，則相應的行值（從數據庫獲取）將顯示另一個dropdownboxes /文本框？ 謝謝 Siva

Answer 1

你可以轉到w3並獲得php-ajax-sql的代碼，如果你對php和sql知之甚少，那麼你的工作就會完成。

Answer 2

假設這是我的選擇下拉菜單，我想要顯示對應一個月的數據， 這是你的問題嗎？

<select name="monthview" onchange="showUser(this.value)"> 
      <option><b>Change Month</b> 
               <option value="1">January 
               <option value="2">February 
               <option value="3">March 
               <option value="4">April 
               <option value="5">May 
               <option value="6">June 
               <option value="7">July 
               <option value="8">August 
               <option value="9">September 
               <option value="10">October 
               <option value="11">November 
               <option value="12">December 

     </select></td> 

<div id="txtHint">YOUR NEW DROP DOWN WITH ROWS FROM DB WILL BE HERE </div > 

    *** AJAX CALL ***: 
<script> 
    function showUser(str) 
    { 
    if (str=="") 
    { 
    document.getElementById("txtHint").innerHTML=""; 
    return; 
    } 
    if (window.XMLHttpRequest) 
    {// code for IE7+, Firefox, Chrome, Opera, Safari 
    xmlhttp=new XMLHttpRequest(); 
    } 
    else 
    {// code for IE6, IE5 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    xmlhttp.onreadystatechange=function() 
    { 
    if (xmlhttp.readyState==4 && xmlhttp.status==200) 
    { 
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText; 
    } 
    } 
    var x=document.myForm.eid.value; 
    xmlhttp.open("GET","databasefetch.php?q="+str,true); 
    xmlhttp.send(); 
    } 
    </script> 

PHP SIDE YOU NEED TO RECEIVE THE MONTH VALUE AND THEN SEND IN THE CORRESPONDING ROWS IN A DROP DOWN AS A RESPONSE TO AJAX CALL , LIKE THIS . 

$month_selected = $_GET['q']; 
//connect to database 
$result ="your sql query" 
//create a drop down now 
echo "<select>"; 
echo "<select name='projectassignedto'>"; 
             echo "<option>"; 
             while ($row = mysql_fetch_array($result)) { 
             echo "<option value='" . $row['employeeid'] ."'>" . $row['employeeid'] ."</option>"; 
              } 
             echo "</select>"; 
echo ""; 

Answer 3

，你將不得不使用JavaScript和PHP來實現這個

<select id="month" onchange="change()"> 
<option value="1">January</option value="2"> 
<option value="1">January</option value="2"> 
</select> 
<div id="result"></div> 
<script> 
function change(){ 
var sel_month = $('#month').val(); 
$.post("url_to_your_php_file", {month:sel_month}) 
       .done(function(data) { 
        $('#result').html(data); 
        }); 
} 

在PHP端

...

<?php 
$month = $_POST['month']; 
//run your query here 
//result should be an array... 
//populate a select box with the array... 
echo '<select>'; 
foreach($result as $data){ 
echo '<option value="'.$data['id'].'">'.$data['name/title'].'</option>'; 
} 
echo '</select>';