我不知道如何将博客标题放入 ajax 调用的数据字段中。我一直在寻找关于初学者 SQL/PHP/AJAX 的教程,但我正在努力,所以如果有人有任何关于此类内容的好的教程,我会很高兴听到。我了解 jquery 和 php 还好,但我正在尝试建立一个简单的博客系统,您可以使用 ajax 请求编辑和删除帖子,但我很挣扎。谢谢!
函数.js
$(document).ready(function(){
$('#deletePost').click(function(){
$.ajax({
url:"post_action.php",
data: { action: "deletePost", postTitle: title of blog post },
success: function(result){
$('ul.left').html(result);
}
});
});
});
index.php
<?php
include 'scripts/db_connect.php';
include 'scripts/functions.php';
sec_session_start();
$sql = "SELECT * FROM blog";
$result = mysqli_query($mysqli, $sql);
while($row = mysqli_fetch_array($result))
{
echo'<div class="blog"><h3 class="blog">' . $row['Title'] .
"</h3><h3>" . $row['Date'] . "</h3><h3>" . $row['Tag'] .
"</h3><hr>";
echo'<p class="blog">' . $row['Body'] . '</p><form name="postForm"
method="post" action="process_post.php">
<input type="radio" name="postAction"
value="editPost" class="postButton" type="button">Edit</input>
<input type="radio" name="postAction" value="deletePost"
class="postButton" type="button">Delete</input>
<input type="radio" name="postAction" value="commentPost"
class="postButton" type="button">Comment</input>
</form></div>';
}
?>
最佳答案
您不能在每篇博文中重复 id="deletePost"
。你应该使用一个类来代替。然后你可以写:
$('.deletePost').click(function(){
$.ajax({
url:"post_action.php",
data: { action: "deletePost",
postTitle: $(this).siblings("h3.blog").text()
},
success: function(result){
$('ul.left').html(result);
}
});
});
关于javascript - 如何将PHP Post数据放入jquery ajax请求中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20258451/