Question

我正在尝试使用动态生成的下拉列表来填充表格。我有一个从我的数据库生成的下拉列表（它抓住了特定玩家可用的所有年份）。我希望能够从下拉列表中选择一年并让它更新我的表格。我生成了下拉列表，但我无法从下拉列表中获取所选值。我在这里找到了以下代码，但它似乎不起作用。这是我到目前为止的代码：

<input name="update" type="submit" value="Update" />
</form>
<p></p>
<form  action="player_login.html">
<input type="submit" value="Logout" />
</form>
</div>

<div style="float: left">
    <p></p>
<h1>Player Stats</h1>
<table width="300" border="1" cellpadding="2" cellspacing="2">

<?php
    // get "id" field from player table
    $login_id = $_COOKIE["DB"];
    $id = "select id from player where login_id='$login_id';";
    $result1=mysql_query($id)  or die('Select1 Query failed: ' . mysql_error());
    $row = mysql_fetch_array($result1);

    // create a dropdown from stats table in db
    echo "--Select Year--";
        $years_query = "select year from stats where player_id='$row[id]';";
        $years = mysql_query($years_query, $connect);

        // fill array with db info
        $var = array();
        while ($row2 = mysql_fetch_array($years))
        {
            $var[] = $row2['year'];
        }

        // create dropdown
    echo'<select name="years" id="years">'; 
    // For each value of the array assign variable name "city" 
    foreach($var as $year)
    { 
            echo'<option value="'.$year.'">'.$year.'</option>'; 
    }    
    echo'</select>';

    // get selected option from dropdown
    $selected_key = $_POST['years'];
    $selected_val = $var[$_POST['years']];
    echo "<p></p>selected key: " . $selected_val; // this wont print anything???

    $search_query="select * from stats where player_id='$row[id]' and year=2013;";
    $result=mysql_query($search_query)  or die('Select2 Query failed: ' . mysql_error());
    $num_cols = mysql_num_fields($result);
    $line = mysql_fetch_row($result);

    // create table with results
    echo "<tr>";
    echo "<td>Year</td>";  
        $j=1;
        echo "<td><input name='$j' type='text' value='$line[$j]' size=20/></td>";
    echo "</tr>";
    echo "<tr>";
    echo "<td>Total Points</td>";  
        $j=2;
        echo "<td><input name='$j' type='text' value='$line[$j]' size=20/></td>";
    echo "</tr>";
    echo "<tr>";
    echo "<td>PPG</td>";  
        $j=3;
        echo "<td><input name='$j' type='text' value='$line[$j]' size=20/></td>";
    echo "</tr>";


?> 

</table>
</div>

Answer 1

我看到你使用$ _POST，因为表单未提交，因此没有设置$ _POST的数据。我用来捕获事件并发送AJAX查询获取结果并更新它的最佳可用选项。

我在J Query的帮助下完成了这项工作，如

所示

$('#years').change(function() {
   $.ajax({
           //request of AJAX
    type : 'POST',
    url : 'players_data.php',
    dataType : 'json',
    data: {
        //Data with $_POST request
        years : $('#years').val();

    },
    success: function(data){
    //Things to be done with returned data
    }
}};

创建一个新文件players_data.php，然后编写用于从db获取数据的代码：

// get selected option from dropdown
$selected_key = $_POST['years'];
$selected_val = $var[$_POST['years']];
echo "<p></p>selected key: " . $selected_val; // this wont print anything???

$search_query="select * from stats where player_id='$row[id]' and year=2013;";
$result=mysql_query($search_query);
$num_cols = mysql_num_fields($result);
$line = mysql_fetch_row($result);
$return['year']=$line;
echo json_encode($return);

Answer 2

我发现您使用的是 $ _ POST ，为什么不使用表单？

//This is for get the form
    echo '<script type="text/javascript">
        //<![CDATA[
        function get_form( element )
        {
            while( element )
            {
                element = element.parentNode
                if( element.tagName.toLowerCase() == "form" )
                {
                    return element
                }
            }
            return 0; //error: no form found in ancestors
        }
        //]]>
    </script>';

//create a form
echo '<form action="'.$_SERVER['PHP_SELF'].'" method="post">';
// create dropdown; onchange will send the form when selected index changes...
echo '<select name="years" id="years" onchange="get_form(this).submit(); return false;">'; 
    // For each value of the array assign variable name "city" 
    foreach($var as $year)
    { 
            echo'<option value="'.$year.'">'.$year.'</option>'; 
    }    
    echo'</select></form>';

就是这样！：d

我也看到你正在使用一个独特的表单来更新所有页面......这是行不通的，因为你只有一个提交按钮，表单中没有更多的元素，请阅读：{{ 3}}

Answer 3

从你的代码我可以看到你想从选择框中获取值并立即填充表并显示结果。使用jquery获取所选对象的值并将javascript变量分配给php变量。并插入db ..

       <script type="text/javascript">
   $( "#years" ).change(function() {
     var value=document.getElementById("years").value;
         alert(value);
       </script>

将变量分配给php并执行php查询。

              <?php
             $data = "<script>document.write(value)</script>";
                  //execute  your query here..
        ?>

另外看看ajax ..就是那么好......

如何从下拉列表中获取所选值

3 个答案: