我有这个:
PHP(Wordpress页面模板):
//grab $projects from database then...
echo '<table class="project-list">';
foreach($projects as $project) {
echo '<tr class="alternate-rows"><td>';
echo '<span class="project" id="' . $project->PID . '">'; //PID is $projectid
echo $project->project_name;
echo '</span>';
echo '</td></tr>';
}
echo '</table>';
<div id="project-files">
</div>
PHP(Wordpress functions.php):
function load_projectfiles($projectid) {
global $wpdb;
$displayprojectfiles = $wpdb->prepare("select * from projectfiles where project_id=%s ", $projectid );
$projectfiles = $wpdb->get_results($displayprojectfiles);
return $projectfiles;
wp_die();
}
add_action('wp_ajax_get_data', 'load_projectfiles');
JS:
jQuery(document).ready(function($){
$('.project').on('click', function(){
var projectid = $(this).attr('id');
//I want to put $projectfiles into $('#project-files') using AJAX so I can make a list using foreach($projectfiles->$projectfile)
});
})
基本上,我想在用户通过jQuery AJAX使用<div id="project-files">
单击<span class="project">
来填充load_projectfiles()
时,不刷新整个页面。
我不太清楚如何编写此代码(尤其是jQuery部分),或者我的逻辑是否正确(例如,jQuery应该是第一个将projectid传递给函数的函数吗?)>