我检查了其他类似的问题，但我找不到帮助我的解决方案。
我正在尝试从数据库获取数据并将其插入到我的 dataTable 复选框类型） >。我希望这两个输入包含在我的表单中，并将信息发送到.php文件并将其放在我的数据库中的另一个表中。 我可以将我的“原始”表单的数据发送到.php文件，但是我找不到复选框输入的值时遇到问题。
$(document).ready(function(){
$.ajax ({
url: "dati_education.php",
metod: "POST",
dataType: "json",
success: function(data) {
$("#education").dataTable({
data: data,
columns: [
{ 'data' : 'Title' },
{ 'data' : 'Year' },
{ 'data' : "Place" },
{ 'data' : ' ',
'render': function(data, type, row){
return "<input class='check' id = '"+row.education_id+"'name='edu_id' type='checkbox' onclick= 'return "+ row.education_id + "'/>";
}
}
],
paging: true
})
}
});
$.ajax ({
url: "dati_work.php",
metod: "POST",
dataType: "json",
success: function(data) {
$("#work").dataTable({
data: data,
columns: [
{ 'data' : 'Company' },
{ 'data' : 'Role' },
{ 'data' : 'Year' },
{ 'data' : 'Place'},
{ 'data' : ' ',
'render': function(data, type, row){
return "<input class='chk'name='work_id' type='checkbox' onclick='return "+ row.work_id + "'/>";
}
}
],
paging: true
})
}
});
});
var id1 = 0;
$('#edu_id').on('click', function(){
id1 = $(".check:checked").each(function(){
$(this).find('input[type="checkbox"]').val();
});
});
//I add some console.log() to do some tests.
var id2= 0;
var work =[];
console.log('ciao');
$('#work_id').on('click', function(){
console.log('ciao');
$(".chk:checked").each(function(){
var id2 = $(this).attr("id");
//var id =$(this).find('input[type="checkbox"]').val();
console.log(id2);
work = JSON.stringify(id2);
work[0] = id2;
});
});
var data = $("#Form").serializeArray();
console.log(id1);
console.log(work);
data.push({
name : "edu_id",
value: id1
});
data.push({
name : "work_id",
value: work
});
console.log(data);
$("#Form").on("submit", function(){
$.ajax({
type: "POST",
url: "create_cards.php",
data: data,
success: function(data){
alert(data);
//alert('Finished! Starting over!');
}
});
});
正如您所看到的，我尝试使用两种不同的方法来尝试获取复选框的值，但两者都不起作用。我认为使用“serializeArray”方法是正确的，因为从控制台我可以看到它添加了字段。但我仍然无法从复选框中获取值
这是我的HTML代码：
<div class="container">
<div class="row">
<form id="Form" class="form-horizontal" action="create_cards.php" method="POST">
<div class="col-sm-6">
<br> <br> <br> <br> <br>
<div class="form-group">
<label class="control-label col-sm-2" for="title">Title:</label>
<div class="col-sm-10">
<input type="text" class="form-control" id="title" placeholder="Enter the Title" name="title">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="name">Name:</label>
<div class="col-sm-10">
<input type="text" class="form-control" id="name" placeholder="Enter your Name" name="name">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="surname">Surname:</label>
<div class="col-sm-10">
<input type="text" class="form-control" id="surname" placeholder="Enter your Surname" name="surname">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="email">Email:</label>
<div class="col-sm-10">
<input type="email" class="form-control" id="email" placeholder="Enter your Email" name="email">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="phone">Phone:</label>
<div class="col-sm-10">
<input type="phone" class="form-control" id="phone" placeholder="Enter your Phone" name="phone">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="photo">Photo:</label>
<div class="col-sm-10">
<input type="select" class="form-control" id="photo" placeholder="Enter your Photo" name="photo">
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="note">Note:</label>
<div class="col-sm-10">
<input type="text" class="form-control" id="note" placeholder="Enter some Note" name="note">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" class="btn btn-default">Submit</button>
</div>
</div>
</div>
<div class="col-sm-6">
<table class="table table-hover table-border" id="education">
<thead>
<tr>
<th>Title</th>
<th>Year</th>
<th>Place</th>
<th> </th>
</tr>
</thead>
<tbody>
<td></td>
</tbody>
</table>
<table class="table table-hover" id="work">
<thead>
<tr>
<th>Company</th>
<th>Role</th>
<th>Year</th>
<th>Place</th>
<th> </th>
</tr>
</thead>
</table>
</div>
</form>
</div>
这是我从中获取数据以插入dataTable的文件之一（另一个是完全相同的）：
<?php
include("database.php");
$conn = mysqli_connect($db_host, $db_user, $db_password);
mysqli_select_db($conn, $db_database);
session_start();
$query = "select * from education_experience where user_id = '"
.$_SESSION["user_id"] ."'";
$result = mysqli_query($conn, $query);
$dati = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$dati[] = array(
"Title" => $row['title'],
"Year" => $row['year'],
"Place" => $row['place'],
"education_id" => $row['education_experience_id']
);
}
echo json_encode($dati);
?>
这是我发送数据的文件。我还没有编写查询，因为我想先解决问题。它无法识别索引'edu_id'和'work_id'。
<?php
include("database.php");
$conn = mysqli_connect($db_host, $db_user, $db_password);
mysqli_select_db($conn, $db_database);
session_start();
echo "CIAO";
if(isset($_POST['title']) && isset($_POST['name']) &&
isset($_POST['surname']) && isset($_POST['email']) && isset($_POST['phone']) ) {
$Title = mysqli_escape_string($conn, $_POST['title']);
$Name = mysqli_escape_string($conn, $_POST['name']);
$Surname = mysqli_escape_string($conn, $_POST['surname']);
$Email = mysqli_escape_string($conn, $_POST['email']);
$Phone = mysqli_escape_string($conn, $_POST['phone']);
$Photo = mysqli_escape_string($conn, $_POST['photo']);
$Note = mysqli_escape_string($conn, $_POST['note']);
$Edu_id = mysqli_escape_string($conn, $_POST['edu_id']);
$Work_id = mysqli_escape_string($conn, $_POST['work_id']);
echo $Title, $Name, $Edu_id, $Work_id;
}
?>
我仍然对这个主题很新，所以请你好。 很抱歉可能存在语法错误，但英语不是我的母语。