jQuery和Auto-populat选择

Question

我有一个简单的页面,我可以选择一个客户端,然后我选择autopopulate到属于客户端的项目.我使用PHP/MySQL来提取结果.

我看了看这个:http://remysharp.com/2007/01/20/auto-populating-select-boxes-using-jquery-ajax/但我认为从页面上的两个字段开始.我试图重做代码,但没有那么好.

 var client_id = $('#c_id').val();
    $.getJSON("../inc/get-projects.php", {id: client_id}, function(data){
        projects = $('#p_id');
        projects.empty();
        $.each(data, function() {
            var option = $('<option/>').attr('value', this.id).text(this.name);
            projects.append(option);
        });
    });

PHP:

<?php
    include "config.inc.php";
    $sth = mysql_query(
        sprintf(
        "SELECT c_id,p_id,p_title FROM projects WHERE c_id = %s",
        mysql_real_escape_string($_GET['id'])
        )
    );
    $projects = array();
    while($r = mysql_fetch_assoc($sth)) {
        $projects[] = array('id' => $r['p_id'], 'name' => $r['p_title']);
    }
    print json_encode($projects);
    exit;


?>

Answer 1

如果您有这样的HTML:

<select id='clients'>...</select>
<select id='projects'>...</select>

你可以有这样的jQuery代码:

$(document).ready(function() {
    var $clients = $('#clients');
    $clients.change(function() {
        var client_id = $clients.val();
        $.getJSON("getProjects.php", {id: client_id}, function(projects) {
            $projects = $('#projects');
            $projects.empty();
            $.each(projects, function() {
                var option = $('<option/>').attr('value', this.id).text(this.name);
                $projects.append(option);
            });
        });
    });
});

然后让getProjects.php返回如下内容:

$sth = mysql_query(
    sprintf(
    "SELECT * FROM projects WHERE client_id = %s",
    mysql_real_escape_string($_GET['id'])
    )
);
$projects = array();
while($r = mysql_fetch_assoc($sth)) {
    $projects[] = array('id' => $r['id'], 'name' => $r['name']);
}
print json_encode($projects);
exit;

我没有测试过所有这些,但我认为它或多或少都是你想要的.

编辑 - 测试和工作,显然它可能不是你需要的,但它让你知道如何去做.希望能帮助到你.

编辑2 - 您的代码存在问题:

$.getJSON("../inc/get-projects.php", {id: client_id}, function(projects){
    projects = $('#p_id');
    projects.empty();
    $.each(projects, function() {
        var option = $('<option/>').attr('value', this.id).text(this.name);
        projects.append(option);
    });
});

您将projects使用projects第二行覆盖传入函数的内容.我$在我的代码中使用了一个来区分这两者,这可能不是最好的方法.要修复,请将代码更改为:

$.getJSON("../inc/get-projects.php", {id: client_id}, function(data){
    projects = $('#p_id');
    projects.empty();
    $.each(data, function() {
        var option = $('<option/>').attr('value', this.id).text(this.name);
        projects.append(option);
    });
});