使用jQuery從MySQL數據庫更改第二個下拉選擇框

我想獲得第二個下拉框來改變和填充時，第一個框有一輛汽車選擇。它連接到數據庫就好了，如果我在第二個下拉列表中鍵入'Honda'而不是'carmake3'，那麼它會給我列出所有本田汽車。我只需要根據用戶選擇的內容來工作。任何幫助將非常感謝！使用jQuery從MySQL數據庫更改第二個下拉選擇框

<script src="http://code.jquery.com/jquery-1.9.1.js"></script> 
<form action="step-3" method="post"> 
Car 3 
</div> 

<?php 
mysql_connect('localhost', '**********', '**********'); 
mysql_select_db('**********'); 
$sql = "SELECT Make FROM CarMakes"; 
$result = mysql_query($sql); 
echo "<select name='carmake3'>"; 
echo "<option value='Make'>Make</option>"; 
while ($row = mysql_fetch_array($result)) { 
    echo "<option value='" . $row['Make'] . "'>" . $row['Make'] . "</option>";} 
echo "</select>"; 
?> 

<script type="text/javascript"> 
jQuery(document).ready(function() { 
    jQuery('carmodel3').change(function() { 
     jQuery('carmake3').change(); 
    }); 
}); 
</script> 

<?php 
mysql_connect('localhost', '**********', '**********'); 
mysql_select_db('**********'); 
$sql1 = "SELECT * FROM myTable WHERE Make='carmake3'"; 
$result1 = mysql_query($sql1); 
echo "<select name='carmodel3'>"; 
    echo "<option value='Model'>Model</option>"; 
while ($row = mysql_fetch_array($result1)) { 
    echo "<option value='" . $row['Model'] . "'>" . $row['Model'] . "</option>";} 
echo "</select>"; 
?> 
<input type="text" maxlength="4" name="car3year" placeholder="year" class="WriteInBox"/> 

<input type="text" maxlength="6" name="car3miles" placeholder="miles" class="WriteInBox"/> 

</div> 

</form>

來源

2013-08-27 Mitchell White

這麼多東西..子請 –

幫我出。一些技巧。或者如果你想爲我重做它，我將不勝感激 –

好吧，我很想去，但是SO並不是爲了那個......你需要聯繫我的'elance'。 –

做一個HTML下面讓

$sql = "SELECT Make FROM CarMakes"; 
$result = mysql_query($sql); 
<select id="carmake3" name='carmake3' onchange="get_makes();"> 
<option value='Make'>Make</option> 
<?php while ($row = mysql_fetch_array($result)) { 
    echo "<option value='" . $row['Make'] . "'>" . $row['Make'] . "</option>";} 
?> 
</select> 
<div id="get_car_make"></div> // Sub will be appended here using ajax

寫的AJAX功能get_makes給出（）;

<script type="text/javascript"> 
function get_makes() { // Call to ajax function 
    var carmake3 = $('#carmake3').val(); 
    var dataString = "carmake3="+carmake3+"&redirecturl="+redirecturl; 
    $.ajax({ 
     type: "POST", 
     url: "getcarmakenames.php", // Name of the php files 
     data: dataString, 
     success: function(html) 
     { 
      $("#get_car_make").html(html); 
     } 
    }); 
} 
</script>

文件getcarmakenames.php - 會得到下面的文件，該文件將被追加到div需要加以糾正

if ($_POST) { 
    $carmake3 = $_POST['carmake3']; 
    if ($carmake3 != '') { 
     $sql1 = "SELECT * FROM myTable WHERE Make='carmake3'"; 
     $result1 = mysql_query($sql1); 
     echo "<select name='carmodel3'>"; 
     echo "<option value='Model'>Model</option>"; 
     while ($row = mysql_fetch_array($result1)) { 
      echo "<option value='" . $row['Model'] . "'>" . $row['Model'] . "</option>";} 
     echo "</select>"; 
    } 
    else 
    { 
     echo ''; 
    } 
}

來源

2013-08-27 04:38:51

'if（$ _POST）{'將始終返回true .... –

使用jQuery從MySQL數據庫更改第二個下拉選擇框

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