如何解决使用 AJAX 和 PHP,创建的表单提交错误的错误值
***更新 我已经按照建议取消了表格元素,并且正在使用 CSS。我还看到有一个“表单”属性,我也试过了。当我提交时,它仍然像以前一样运行 - 发送错误的值,因为它正在发送整个表。我已经用更新的 HTML 输出和 PHP 代码更新了下面的内容。看起来是正确的,这是我最近的尝试。我错过了什么?
我正在使用 PHP 为每一行数据创建一个表单。我通过 AJAX 调用 PHP。表单正确构建。每一行都正确地列出了它的值并且有它自己的形式。在这个例子中,有三行,因此是三种形式。当我在第一行提交用户名时,发送的 ID 来自第三行。不知道发生了什么。
AJAX 调用 PHP FORM - 主页
<script>
window.onload = function signupForm() {
var xmlHttp = new XMLHttpRequest();
xmlHttp.onreadystatechange = function() {
if(xmlHttp.readyState == 4 && xmlHttp.status == 200) {
document.getElementById("signupForm").innerHTML = this.responseText;
}
}
xmlHttp.open("GET","ajaxInput.php",true);
xmlHttp.send();
}
</script>
PHP 表单 - signupForm.php
<?php
$con=mysqli_connect("localhost","xxxxxx","xxxxxxx","xxxxxxxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT id,DATE_FORMAT(startTime,'%b-%d-%Y') as eventDate,endTime FROM events
WHERE now() < endTime");
echo "
<style>
.table { display: table; }
.table>* { display: table-row; }
.table>*>* { display: table-cell; padding: 5px; border-style: inset;}
</style>
<div class='table'>
<div>
<div><b>Event Id</b></div>
<div><b>Date</b></div>
<div><b>@username</b></div>
<div><b>Sign Up!</b></div>
</div>";
while($row = mysqli_fetch_array($result))
{
//echo "<form action='ajaxSignup.php' method='post'>";
echo "<div>";
echo "<div>" . $row['id'] . "</div>";
echo "<div>" . $row['eventDate'] . "</div>";
echo "<div><form id='form" .$row['id']. "' method='post'><input class='formSignup' type='text' name='pi_username' id='pi_username' maxlength='20' placeholder='@username' form='form" .$row['id']. "'></div>";
echo "<div><input class='formSignup' type='hidden' name='event_id' id='event_id' value='" . $row['id'] . "' form='form" .$row['id']. "'>
<input name='submit". $row['id'] . "' type='submit' value='Sign up!' onclick='signup(); return false;'></form></div>";
echo "</div>";
}
//echo "</div>";
echo "</div>";
mysqli_close($con);
?>
表格绘制正确。我也在不同的地方放置了表单标签。下面,我在输入标签中使用了 form 属性。表格是用 CSS 而不是表格元素绘制的。
<html>
<head></head>
<body>
<p>Something here</p>
<div id="signupForm">
<style>
.table { display: table; }
.table>* { display: table-row; }
.table>*>* { display: table-cell; padding: 5px; border-style: inset;}
</style>
<div class="table">
<div>
<div><b>Event Id</b></div>
<div><b>Date</b></div>
<div><b>@username</b></div>
<div><b>Sign Up!</b></div>
</div>
<div>
<div>11</div>
<div>Feb-25-2021</div>
<div><input class="formSignup" type="text" name="pi_username" id="pi_username" maxlength="20" placeholder="@username" form="form11"></div>
<div>
<input class="formSignup" type="hidden" name="event_id" id="event_id" value="11" form="form11">
<form id="form11" method="post"><input name="submit11" type="submit" value="Sign up!" onclick="signup(); return false;"></form>
</div>
</div>
<div>
<div>12</div>
<div>Feb-26-2021</div>
<div><input class="formSignup" type="text" name="pi_username" id="pi_username" maxlength="20" placeholder="@username" form="form12"></div>
<div>
<input class="formSignup" type="hidden" name="event_id" id="event_id" value="12" form="form12">
<form id="form12" method="post"><input name="submit12" type="submit" value="Sign up!" onclick="signup(); return false;"></form>
</div>
</div>
<div>
<div>13</div>
<div>Feb-27-2021</div>
<div><input class="formSignup" type="text" name="pi_username" id="pi_username" maxlength="20" placeholder="@username" form="form13"></div>
<div>
<input class="formSignup" type="hidden" name="event_id" id="event_id" value="13" form="form13">
<form id="form13" method="post"><input name="submit13" type="submit" value="Sign up!" onclick="signup(); return false;"></form>
</div>
</div>
</div>
</div>
正如在 PHP 中所见,每一行都是它自己的形式。但是查看开发人员工具中的元素,表单很早就关闭了。我认为这与问题有关。 Elements
当我在第一行(顶行)输入用户名时,该用户名似乎没有出现,而产生它的 ID 是 13 而不是 11。
处理提交按钮 onclick(主页)的 AJAX 脚本...
<script>
function signup() {
var elements = document.getElementsByClassName("formSignup");
var formData = new FormData();
for(var i=0; i<elements.length; i++) {
formData.append(elements[i].name,elements[i].value);
}
var xmlHttp = new XMLHttpRequest();
xmlHttp.onreadystatechange = function() {
if(xmlHttp.readyState == 4 && xmlHttp.status == 200) {
document.getElementById("signupSuccess").innerHTML = this.responseText;
}
}
xmlHttp.open("post","ajaxSignup.php");
xmlHttp.send(formData);
}
</script>
注册页面上的 PHP 代码。早些时候我收到了一些回声,显示 id 是 13,而不是 11,并且没有用户名。
<?php
$pi_username = $high_score = $attempts = $event_id = "";
echo $event_id;
echo $pi_username;
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$pi_username = test_input($_POST["pi_username"]);
$event_id = test_input($_POST["event_id"]);
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
echo $event_id;
echo $pi_username;
if($_SERVER["REQUEST_METHOD"] == "POST") {
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost","xxxxxx");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
//check if user has already signed up for event
$alreadySignedUp = mysqli_query($link,"SELECT count(*) AS total FROM signup WHERE event_id = $event_id AND pi_username = '$pi_username'");
while ($worm = mysqli_fetch_array($alreadySignedUp)){
//echo $bird['total'];
if($worm['total'] >= 1 ){
echo "You have already signed up for this event.";
echo $event_id;
echo $pi_username;
echo $worm['total'];
mysqli_close($link);
return;
}
}
// Attempt insert query execution
$sql = "INSERT INTO signup (event_id,pi_username) VALUES ('$event_id','$pi_username')";
if(mysqli_query($link,$sql)){
echo "You have been successfully added to event ".$event_id."!";
mysqli_close($link);
return;
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
else {
echo "Don't forget to submit your high scores before you leave!";
}
?>
我哪里出错了?
解决方法
signup() 函数正在寻找具有相同类 (formSignup) 的元素。所有输入字段都具有相同的类名,并且都被发送到 signup() 函数。我已将类名更新为每行唯一。现在只发送一行。 signup() 函数更新为:
window.onload = function signupForm(getClass) {
var xmlHttp = new XMLHttpRequest(getClass);
具有唯一类名的输入字段示例:
<input class='formSignup" .$row['id']. "' type='hidden' name='event_id' id='event_id' value='" . $row['id'] . "' form='form" .$row['id']. "'>
删除了表格元素并按照建议创建了一个 CSS 样式的“表格”。现在都在工作。问题已更新为 CSS 样式的表格。
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