jQuery Ajax 通過單擊按鈕但有兩種不同的操作將數據發送到 php

Question

我想要實現的是一個用戶無需刷新頁面即可關注另一個用戶。 到目前為止，我已經玩過並且在 mysql 表中插入和刪除行沒有問題，但是現在當我嘗試使用 AJAX 時我無法讓它工作。

查詢

$(document).ready(function(){
$("#followbutton").click(function(e) {
e.preventDefault();
var theuserid = $('#theuserid').val();
var thefollower = $('#thefollower').val();

$.ajax({
url: 'includes/followuser.inc.php',
type: 'post',
data: {'theuserid': theuserid, 'thefollower': thefollower, 'submitFollow': true},
success: function(response){
$('#followmessage').html(response);   
$("#followmessage").show().delay(3000).fadeOut();
$('#followbutton').hide();
$('#unfollowbutton').show();

// $("#unfollowbutton").hover(function(){
// $(this).text("Unfollow");
// }, function(){
// $(this).text("Unfollow");
// });
    
}
});
});
});

$(document).ready(function(){
$("#unfollowbutton").click(function(e) {
e.preventDefault();
var theuserid = $('#theuserid').val();
var thefollower = $('#thefollower').val();

$.ajax({
url: 'includes/followuser.inc.php',
type: 'post',
data: {'theuserid': theuserid, 'thefollower': thefollower, 'submitUnfollow': true},
success: function(response){
$('#followmessage').html(response);   
$("#followmessage").show().delay(3000).fadeOut();
$('#unfollowbutton').hide();
$('#followbutton').show();

//I want the button to change its text to Following and when hovering it should say unfollow if user is followed

}
});
});
});

followuser.inc.php

<?php
require_once 'dbh.inc.php';
require_once 'functions.inc.php';
if (isset($_POST["submitFollow"])){
$userthatisfollowed = $_POST["thefollower"];
$theuserid = $_POST["theuserid"];

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$stmt = $conn->prepare('INSERT INTO userfollow (thefollower, theuserid, followstatus) VALUES (?,?,?)');
$followstatus = 1;
$stmt->bind_param('sss', $userthatisfollowed, $theuserid, $followstatus);
$stmt->execute();
echo $response = "<span>Followed!</span>";
$stmt->close();

} else if(isset($_POST["submitUnfollow"])){
$userthatisfollowed = $_POST["thefollower"];
$theuserid = $_POST["theuserid"];
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$stmt = $conn->prepare('DELETE userfollow FROM userfollow WHERE thefollower = ? AND theuserid = ?');

$stmt->bind_param('ss', $userthatisfollowed, $theuserid);
$stmt->execute();
echo $response = "<span>Unfollowed!</span>";
$stmt->close();
} else {
    echo "DID NOT WORK";
}

配置文件.php

if(isset($_SESSION["userid"]) && $_SESSION["userid"] != $userthatisfollowed) {
?>
<form action="<?php echo htmlspecialchars("includes/followuser.inc.php");?>" id="followform" method="post">
<?php

if ($resulted->num_rows > 0){
$subscribe_status = "Unfollow";
$subscribe_text = "Following";

} else {
$subscribe_status = "Follow";
$subscribe_text = "Follow";
}


echo "<button name='submit".$subscribe_status."' id ='unfollowbutton' type='submit' style='display:none'>";
echo "<span>".$subscribe_text."</span>";
echo "</button>";

echo "<button name='submit".$subscribe_status."' id ='followbutton' type='submit'>";
echo "<span>".$subscribe_text."</span>";
echo "</button>";


// echo "<button name='submit".$subscribe_status."' id ='notificationbell' type='submit' style='display:none'>";
// echo "<i class='fa fa-bell'></i>";
// echo "</button>";

echo "<div id='followmessage'></div>";
?>
<input type="hidden" name="theuserid" id="theuserid" value="<?php echo $_SESSION["userid"] ?>">
<input type="hidden" name="thefollower" id="thefollower" value="<?php echo $userthatisfollowed; ?>">
</form>
<?php
}

值得注意的是，我收到的響應DID NOT WORK告訴我if(isset($_POST["submitUnfollow"]))未設置。 但是，如果我嘗試使用if(isset($_POST["theuserid"]) && (isset($_POST["thefollower"]))那么它實際上適用於插入查詢但不適用於刪除查詢。

Answer 1

您缺少data:對象中的submitFollow參數。 相反，您有followbutton: true ，PHP 代碼未使用它。 所以將其更改為：

data: {'theuserid': theuserid, 'thefollower': thefollower, 'submitFolow': 'true'},

對於取消關注按鈕，請改用submitUnfollow 。