jQuery和Auto-populat选择 [英] jQuery and Auto-populat selects

问题描述

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。

I have a simple page where I can select a client, then once I chose that autopopulate to the projects that belong to the client. I am using PHP/MySQL to pull the results.

我看了一眼： 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。

I took at a look at this: http://remysharp.com/2007/01/20/auto-populating-select-boxes-using-jquery-ajax/ but I think that starts with both fields on the page. I tried to rework the code but didn't come out that well.

 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;


?>

推荐答案

如果你有这样的HTML：

If you have HTML like this:

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

你可以拥有这样的jQuery代码：

You can have jQuery code like this:

$(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：

And then have getProjects.php return something like:

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

我还没有测试过所有这些，但它或多或少都是你想要的，我想。

I haven't tested all of this, but it is more or less what you want, I think.

编辑 - 经过测试和工作，显然它可能不是您需要的，但它可以让您了解如何去做。希望它有所帮助。

edit - Tested and works, obviously it might not be exactly what you need but it gives you an idea of how to go about it. Hope it helps.

编辑2 - 您的代码存在问题：

edit 2 - The problem with your code is here:

$.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);
    });
});

您正在覆盖传入此处的项目在第二行使用项目。我在我的代码中使用 $ 来区分这两者，这可能不是最好的方法。要解决此问题，请将代码更改为：

You are overwriting the projects passed into the function with the projects in the 2nd line. I used a $ in my code to differentiate the two, which was admittedly perhaps not the best of ways. To fix, change the code to this:

$.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);
    });
});

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