【发布时间】:2021-03-01 05:24:31
【问题描述】:
我使用 jQuery 将一组用户帖子发布到他们的个人资料页面。在数组中,我为每个帖子添加了一个删除按钮。我在命名每个按钮以使其与用户的特定帖子协调时遇到问题,因此我可以在单击时删除正确的帖子。
$(document).ready(show_user_tweets);
function show_user_tweets() {
var url = "controller.php";
var query = { page: 'Profile', command: 'ShowUserTweets'};
$.post(url, query,
function(data) {
var rows = JSON.parse(data);
if(rows.length > 0) {
var t = "<table class='table table-borderless'>";
for (var i = 0; i < rows.length; i++) {
t += "<tr><td>"
t += "<div class='modal-content'>";
t += "<div class='modal-header'>";
t += "<h2 class='modal-title'>" + rows[i]['Username'] + "</h2></div>";
t += "<div class='modal-body'>";
t += "<div class='form-group'>" + rows[i]['Tweet'] + "</div></div>";
t += "<div class='modal-footer justify-content-between'>";
t += "<div class='form-group'>" + "<input type='button' id='deleteBtn' name='rows[i]['UserId']' value='Delete' class='btn btn-danger'></input>" + "</div>";
t += "<div class='form-group'>" + rows[i]['Date'] + "</div></div>";
t += "</div></div></td></tr>";
}
t += '</table>';
$('#pane-result').html(t);
}
});
}
我目前有每个按钮,例如:<input type='button' id='deleteBtn' name='rows[i]['UserId']' value='Delete' class='btn btn-danger'></input>
如何将每个按钮绑定到每个特定帖子,以便我可以在控制器中访问它以从 SQL 数据库中删除?
干杯
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