如何使用php ajax从mysqldb获取值到dropdownbox？

Question

如何使用php ajax从mysql数据库获取值到dropdownbox？

我需要的是

当我从dropdownbox中选择任何值时，相应的行值(从db中获取)将显示到另一个dropdownbox/textboxes？感谢Siva

Answer 1

假设这是我的select下拉列表，我想显示对应于一个月的数据，这是您的问题吗？

<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 2

您必须同时使用javascript和php来实现这一点。

<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方面...

<?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>';

Answer 3

你可以去w3得到php-ajax-sql的代码，如果你对php和sql知之甚少，你的工作就完成了。