Replicate the same data request query, but use a JavaScript Ajax post so that it doesn't refresh the page on button press, just requests another page and display in a section on the page. i need some help changing this to a ajax post
my code
*<?php
if(isset($_POST['submit']))
{
$success = $_POST['success'];
$dates = $_POST['dates'];
$datee = $_POST['datee'];
/*** mysql hostname ***/
$hostname = 'localhost';
$dbname = '*******';
/*** mysql username ***/
$username = 'root';
/*** mysql password ***/
$password = '*******';
try {
$dbh = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password);
$tablename = 'login_attempts';
$sql = 'SHOW COLUMNS FROM `'.$tablename.'`';
$fields = array();
$csv = array();
$stmt = $dbh->query($sql);
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
array_push($fields, $row['Field']);
}
array_push($csv, $fields);
$success = mysql_real_escape_string($success);
$sql = "SELECT * FROM $tablename WHERE success = '".$success."' AND attempted >='".$dates."' AND attempted <='".$datee."'";
$stmt = $dbh->query($sql);
$stmt->execute();
$csv = array();
while($row = $stmt->fetch(PDO::FETCH_NUM))
{
array_push($csv, $row);
}
$fp = fopen('file.csv', 'w');
foreach ($csv as $row) {
fputcsv($fp, $row);
}
fclose($fp);
header("Content-type: application/csv");
header("Content-Disposition: attachment; filename=export.csv");
header("Pragma: no-cache");
header("Expires: 0");
readfile('file.csv');
$dbh = null;
} catch(PDOException $e) {
echo $e->getMessage();
}
exit();}
?>
<html>
<head>
<title>csv with criteria</title>
</head>
<body>
<form action="csv2.php" method="post" enctype="multipart/form-data">
Select data range
<br>
<input type="date" name="dates" id="dates"> Starting date
<br>
<input type="date" name="datee" id="datee"> Ending date
<br>
Select what data you'd like
<br>
<input type="radio" name="success" value="1" checked> Yes<br>
<input type="radio" name="success" value="0"> No<br>
<input type="submit" value="show" name="submit">
<br>
</form>
</body>
</html>*
You can see this post where it's explained how to make a good ajax call to POST parameters. But you will need to separate this portion of code :
<?php
if(isset($_POST['submit']))
{
$success = $_POST['success'];
$dates = $_POST['dates'];
$datee = $_POST['datee'];
/*** mysql hostname ***/
$hostname = 'localhost';
$dbname = '*******';
/*** mysql username ***/
$username = 'root';
/*** mysql password ***/
$password = '*******';
try {
$dbh = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password);
$tablename = 'login_attempts';
$sql = 'SHOW COLUMNS FROM `'.$tablename.'`';
$fields = array();
$csv = array();
$stmt = $dbh->query($sql);
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
array_push($fields, $row['Field']);
}
array_push($csv, $fields);
$success = mysql_real_escape_string($success);
$sql = "SELECT * FROM $tablename WHERE success = '".$success."' AND attempted >='".$dates."' AND attempted <='".$datee."'";
$stmt = $dbh->query($sql);
$stmt->execute();
$csv = array();
while($row = $stmt->fetch(PDO::FETCH_NUM))
{
array_push($csv, $row);
}
$fp = fopen('file.csv', 'w');
foreach ($csv as $row) {
fputcsv($fp, $row);
}
fclose($fp);
header("Content-type: application/csv");
header("Content-Disposition: attachment; filename=export.csv");
header("Pragma: no-cache");
header("Expires: 0");
readfile('file.csv');
$dbh = null;
} catch(PDOException $e) {
echo $e->getMessage();
}
exit();}
?>
into one another php file.
If you want to use AJAX request on form submit you should:
<script src="//code.jquery.com/jquery-1.12.0.min.js"></script>
Create a new JavaScrpit file where you will catch the form submit event, send jQuery request and analize the response, eg:
$(document).ready(function(){ $("#myForm").submit(function(e){ e.preventDefault(); //stop sending the form //here you should validate your form //submit it via AJAX: $.ajax({ url: "csv2.php", data: { dates: $("#dates").val(), datee: $("#datee").val(), $('input[name='success']:checked', '#myForm').val() } }) }) });
You can use another functions of AJAX object such as success or error callback functions: jQuery.ajax() . Don't forget to add ID to your form.
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