数据库中填充的选项不会在更改时运行脚本

我从数据库（加载选项）加载我的选择选项，并试图在选定选项更改（不运行）时运行脚本脚本如下数据库中填充的选项不会在更改时运行脚本

<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>

脚本是在头元件，一旦我得到的脚本来运行，我会调整它的控制台只是一个peice的，看它是否运行

来源

2015-07-21 Jdoonan

什么是.selectmenu（）函数？ – Rob

@Rob它的一个jQuery UI功能 – Jdoonan

我明白了。我注意到你链接了两次。不知道这是否是问题。 – Rob

通过工作在我的代码进一步我发现答案是，我不能只是布局我想运行的功能，我需要围绕的情况它在另一个功能

<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>

来源

2015-07-23 07:01:35 Jdoonan

数据库中填充的选项不会在更改时运行脚本

回答

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