數據庫中填充的選項不會在更改時運行腳本

我從數據庫（加載選項）加載我的選擇選項，並試圖在選定選項更改（不運行）時運行腳本腳本如下數據庫中填充的選項不會在更改時運行腳本

<script> 
$(function() 
{ 

    function subrace() 
    { 
     var race = $("#race").val(); 

     console.log(race + "!"); 
    } 

    $("#race") 
     .selectmenu() 
     .selectmenu(
     { 
      change: subrace() 
     }) 
     .selectmenu("menuWidget") 
     .addClass("overflow"); 
}); 
</script> 


<select name="race" id="race"> 
<?php 

$dbhost = "localhost"; 
$dbuser = "DanD_user"; 
$dbpass = "****************"; 
$dbname = "dand_user"; 

$mysqli = new mysqli($dbhost, $dbuser, $dbpass, $dbname); 

$query = "SELECT * FROM races ORDER BY name ASC"; 

$result = mysqli_query($mysqli, $query); 
$x = 0; 
while($row = $result->fetch_array(MYSQLI_BOTH)) 
{ 
    echo" <option value=\"" . $row['name'] . "\">" . $row['name'] . "</option>\n"; 
    $x++; 
} 

if($x === 0) 
{ 
    echo"<option disabled selected>No Races Available</option>"; 
} 
?> 
</select>

腳本是在頭元件，一旦我得到的腳本來運行，我會調整它的控制檯只是一個peice的，看它是否運行

來源

2015-07-21 Jdoonan

什麼是.selectmenu（）函數？ – Rob

@Rob它的一個jQuery UI功能 – Jdoonan

我明白了。我注意到你鏈接了兩次。不知道這是否是問題。 – Rob

通過工作在我的代碼進一步我發現答案是，我不能只是佈局我想運行的功能，我需要圍繞的情況它在另一個功能

<script> 
$(function() 
{ 

    function subrace() 
    { 
     var race = $("#race").val(); 

     console.log(race + "!"); 
    } 

    $("#race") 
     .selectmenu() 
     .selectmenu(
     { 
      change: function(event, ui) 
      { 
       subrace(); 
      } 
     }) 
     .selectmenu("menuWidget") 
     .addClass("overflow"); 
}); 
</script> 


<select name="race" id="race"> 
<?php 

$dbhost = "localhost"; 
$dbuser = "DanD_user"; 
$dbpass = "****************"; 
$dbname = "dand_user"; 

$mysqli = new mysqli($dbhost, $dbuser, $dbpass, $dbname); 

$query = "SELECT * FROM races ORDER BY name ASC"; 

$result = mysqli_query($mysqli, $query); 
$x = 0; 
while($row = $result->fetch_array(MYSQLI_BOTH)) 
{ 
    echo" <option value=\"" . $row['name'] . "\">" . $row['name'] . "</option>\n"; 
    $x++; 
} 

if($x === 0) 
{ 
    echo"<option disabled selected>No Races Available</option>"; 
} 
?> 
</select>

來源

2015-07-23 07:01:35 Jdoonan

數據庫中填充的選項不會在更改時運行腳本

回答

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