Jquery填充< table>基于< select> [英] Jquery to populate <table> based on <select>

问题描述

所以我有一个窗体，顶部有一个从mysql表中填入的值。基本上这种形式是允许将球员加入队伍。

So I've got a form with a drop down at the top that is populated with values from a mysql table. Basically this form is to allow the addition of players to a team.

    $seasonid = $_SESSION['SEASON_ID'];
$sql="SELECT TEAM_ID, TEAM_NAME FROM TEAMS WHERE SEASON_ID=$seasonid";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
        $tid=$row["TEAM_ID"];
        $tname=$row["TEAM_NAME"];
    $options.="<OPTION VALUE=\"$tid\">".$tname;
}

我想做的是从列表中选择一个团队使用TEAM_ID查询数据库，所有来自该团队的玩家都会显示在表单下方的列表中，以便填写表单的人员可以看到他们已经在他们所选择的团队中的谁。

What I'd like to do is when a team is selected from the list the database is queried with the TEAM_ID and all the players from that team are shown in a list below the form so that the person populating the form can see who is already on the team they've selected.

推荐答案

HTML

<select id = 'teamSelect'>....options....</select>
<div id = 'tableContainer'></div>

Javascript

$(function() {
    $("#teamSelect").bind("change", function() {
        $.ajax({
            type: "GET", 
            url: "path/to/server/script.php",
            data: "tid="+$("#teamSelect").val(),
            success: function(html) {
                $("#tableContainer").html(html);
            }
        });
    });

});

JavaScript代码将对服务器端的php脚本执行ajax请求（在代码中为 path / to / server / script.php ）此脚本将查询您的数据库，并根据需要输出表。所选择的小组将在 $ _ GET 变量'tid'中。

The javascript code will perform an ajax request to a server side php script (in the code it is the path/to/server/script.php) This script will query your database and simply output the table as you would like it. The team that is selected will be in the $_GET variable 'tid'.

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