Question

我有一个3步链式选择序列game -> battle -> winning side，它从MySQL数据库中提取所有数据。

在互联网上徘徊之后，我发现了一个非常棒的jQuery脚本。但是，我不知道如何使用此脚本来允许现有数据：<option selected="selected"></option>。

链接选择javascript：

<script>
var ajax = new Array();

function getScenNumList(sel)
{
    var game = sel.options[sel.selectedIndex].value;
    document.getElementById('scenarioNumber').options.length = 0;   // Empty scenario number select box
    if(game.length>0){
        var index = ajax.length;
        ajax[index] = new sack();

        ajax[index].requestFile = 'js/getPlayData.php?gameName='+game;  // Specifying which file to get
        ajax[index].onCompletion = function(){ createScenarioNumbers(index) };  // Specify function that will be executed after file has been found
        ajax[index].runAJAX();      // Execute AJAX function
    }
}

function createScenarioNumbers(index)
{
    var obj = document.getElementById('scenarioNumber');
    eval(ajax[index].response); // Executing the response from Ajax as Javascript code
}

function getNations(sel)
{
    var scenNum = sel.options[sel.selectedIndex].value;
    document.getElementById('victor').options.length = 0;   // Empty nation select box
    if(scenNum.length>0){
        var index = ajax.length;
        ajax[index] = new sack();

        ajax[index].requestFile = 'js/getPlayData.php?scenID='+scenNum; // Specifying which file to get
        ajax[index].onCompletion = function(){ createNations(index) };  // Specify function that will be executed after file has been found
        ajax[index].runAJAX();      // Execute AJAX function
    }
}

function createNations(index)
{
    var obj = document.getElementById('victor');
    eval(ajax[index].response); // Executing the response from Ajax as Javascript code
}

</script>

摘自PHP数据库检索脚本（getPlayData.php）：

$gameName = mysql_real_escape_string($_GET['gameName']);

$q = "SELECT a, b, c FROM table WHERE game='$gameName' ORDER BY num ASC";

$r = mysql_query($q);

echo "obj.options[obj.options.length] = new Option('#','');\n";

while ($row = mysql_fetch_row($r)) {

    $string = mysql_real_escape_string(($row[0].' - '.$row[1])); // needed so quotes ' " don't break the javascript

    echo "obj.options[obj.options.length] = new Option('$string','$row[2]');\n";

}

回显obj.options是此脚本使用的stock方法。这对我来说似乎很难看，但我不知道任何javascript，所以我不想乱用它。

HTML很简单，只是一个包含几个空<select>个对象的表，其ID与javascript和onchange="getXXX(this)"调用中的ID相匹配。

我的问题是：一切都适用于新记录，但我不知道如何改变它以支持从每个选择标记一个选项作为选择，假设我有手头的数据（例如：用户正在编辑现有记录）？

非常感谢！

Answer 1

您可以看到本教程，用于创建默认选项。 http://www.javascriptkit.com/javatutors/selectcontent.shtml选项构造函数中的一个参数指示是否选择了该选项。
在PHP文件中，您将编辑如下：

$gameName = mysql_real_escape_string($_GET['gameName']);

$q = "SELECT a, b, c FROM table WHERE game='$gameName' ORDER BY num ASC";

$r = mysql_query($q);

echo "obj.options[obj.options.length] = new Option('#','');\n";

while ($row = mysql_fetch_row($r)) {

    $string = mysql_real_escape_string(($row[0].' - '.$row[1])); // needed so quotes ' " don't break the javascript
    if ($string ......)
    echo "obj.options[obj.options.length] = new Option('$string','$row[2]', false, true);\n";
    else
     ......

}

帮助链式选择和现有表单值，例如：添加<option selected =“selected”> </option>

1 个答案: