我正在使用onclick for myFunction(),它在按下"回复" - 按钮时执行。
该函数用于将parentId值设置为其子项(注释)。我的问题是,在用户第二次点击按钮之前,该值不会设置。
如果我按下回复按钮,一旦parentId为0,如果我在添加评论之前点击它两次,它将被赋予其父母ID。
获取评论,包含父ID为隐藏的输入字段
$query2 = "SELECT * FROM data WHERE `parent_id` IS NULL ORDER BY `id` DESC";
$result = mysqli_query($connect, $query2);
while($row = mysqli_fetch_array($result)){
$id = $row['id'];
$name = $row['name'];
$comment = $row['comment'];
$date = $row['date'];
$parent_id = $row['parent_id'];
printf("<div>");
printf("<div>");
printf("<div>");
printf("<strong> $name </strong><span style='float:right'class='text-muted'>$date + $id</span>");
printf("</div>");
printf("<div>%s",$comment);
printf("<div class='cmt'>");
printf("<button class='replybtn' onclick='return myFunction(%d);' style='float:right'>Reply</button>",$id);
printf("<div class='cmttext'></div>");
printf("<div class='replyform'></div>");
printf("</div>");
printf("</div>");
printf("</div>");
printf("</div>");
myFunction的
var myFunction = function(parentId){
document.getElementById('plopp').value = parentId;
};
回复表单
var varHtml = "\
</br><form method='post' name='form_child'>\n\
Namn: <br>\n\
<input type='text' name ='name_child'><br>\n\
Kommentar: <br>\n\
<textarea name='comment_child'></textarea> \n\
<input type='hidden' name='parent_child' id='plopp'><br>\n\
<input type='submit' name='submit_reply' value='Skicka'/> \n\
<button id='hide' type='submit' name='close'>Stäng</input>\n\
</form>";
var allElements = document.body.getElementsByClassName("replybtn");
var addCommentField = function() {
for (var i = 0; allElements.length > i; i++) {
if (allElements[i] === this) {
console.log("this "+ i);
if (document.getElementsByClassName("replyform")[i].innerHTML.length === 0) {
document.getElementsByClassName("replyform")[i].innerHTML = varHtml;
}
}
}
};
for (var i = 0; i < allElements.length; i++) {
allElements[i].addEventListener('click', addCommentField, false);
}
商店评论
if(isset($_POST['submit_reply'])){
if(isset($_POST['name_child']) && isset($_POST['comment_child'])){
if(!empty($_POST['name_child']) && !empty($_POST['comment_child'])){
$name = htmlentities($_POST['name_child']);
$comment = htmlentities($_POST['comment_child']);
$parent_id = $_POST['parent_child'];
$date = date("Y-m-d");
$connect = mysqli_connect(...);
if($connect){
mysqli_select_db($connect, "comments");
$query_child = "INSERT INTO data (...) VALUES (...)";
if(mysqli_query($connect, $query_child)){
} else {
die ("Failed: " . mysqli_error($connect));
}
} else {
die("Failed to connect to mysql: " . mysqli_errno($connect));
}
}else{
echo "";
}
}
//}
}
答案 0 :(得分:1)
将return false
添加到event handler
而不是功能:
onclick='myFunction(%d); return false;'
onclick
这里是事件处理程序。即使您在return false
中放置myFunction
,也会失败,因为返回的false
永远不会被事件处理程序捕获。
另外,要抓住它,您可以尝试使用:
onclick='return myFunction(%d);'
在函数中添加return false
。