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[英]How to insert the selected data from drop down menu into multiple table in database
[英]How to obtain the option selected from the drop down menu and insert it into a mysql table
我一直在嘗試使用帶有ajax的php / mysql做一個動態下拉菜單。 但是我是php和ajax的新手,所以我不知道如何將所選選項插入mysql表。 我創建了一個表單,其中的操作導致了一個insertsql.php文件,但它不起作用。我正在使用wamp服務器。 任何幫助將不勝感激。
index1.php
<?php
//index.php
$connect = mysqli_connect("localhost", "root", "", "projects");
$pname = '';
$query = "SELECT pname FROM project_details GROUP BY pname ORDER BY pname ASC";
$result = mysqli_query($connect, $query);
while($row = mysqli_fetch_array($result))
{
$pname .= '<option value="'.$row["pname"].'">'.$row["pname"].'</option>';
}
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>zz
</head>
<body>
<br /><br />
<div class="container" style="width:600px;">
<h2 align="center">Dynamic Dependent Select Box using JQuery Ajax with PHP</h2><br /><br />
<form method = "POST" action = "insertsql.php" >
<select name="pname" id="pname" class="form-control action">
<option value="">Select Project</option>
<?php echo $pname; ?>
</select>
<br />
<select name="user" id="user" class="form-control action">
<option value="">Select User Name</option>
</select>
<br />
<input type="submit" name="update" value="Update">
<p id="dem"></p>
<p id="demo"></p>
</form>
</div>
</body>
</html>
<script>
$(document).ready(function(){
$('.action').change(function(){
if($(this).val() != '')
{
var action = $(this).attr("id");
var query = $(this).val();
var result = '';
if(action == "pname")
{
result = 'user';
}
$.ajax({
url:"fetch.php",
method:"POST",
data:{action:action, query:query},
success:function(data){
$('#'+result).html(data);
}
})
}
});
});
</script>
fetch.php
<?php
//fetch.php
if(isset($_POST["action"]))
{
$connect = mysqli_connect("localhost", "root", "", "projects");
$output = '';
if($_POST["action"] == "pname")
{
$query = "SELECT fname,lname FROM users WHERE pname = '".$_POST["query"]."' GROUP BY fname";
$result = mysqli_query($connect, $query);
$output .= '<option value="">Select User</option>';
while($row = mysqli_fetch_array($result))
{
$output .= '<option value="'.$row["fname"].' '.$row["lname"].'">'.$row["fname"]." ".$row["lname"].'</option>';
}
}
echo $output;
}
?>
insertsql.php
<?php
include('sqlconfig.php');
$pname= $_POST['pname'];
$username=$_POST['user'];
$date=$_POST['wdate'];
$hours=$_POST['hours'];
echo "$pname";
echo "<br>$username<br>";
$sql="INSERT INTO worklogging(pname,username,wdate,wkhours) VALUES ('$pname','$username','$date','$hours')";
if(!mysqli_query($con,$sql))
{echo 'not inserted';}
else
{echo 'Inserted';
}
?>
這是下拉菜單中的數據插入mysql表的部分。 謝謝大家的幫助:)
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