Post array from form using xmlhttprequest - php

Question

I have an php generated table/form with checkboxes like this:
if ($query) {
if (!mysqli_num_rows($query)) {
//Empty storage
echo "There are no items in '$storage'.";
exit();
} else {
//form
echo "<form name='send_parts_form' action='../includes/generatetab.php' method='post'>";
//Table header
echo "
<table>
<tr>
<th>Part</th>
<th>PN</th>
<th>Manufactured</th>
<th>Serial</th>
<th>Site</th>
<th>Date replaced</th>
<th>By user</th>
<th>Faulty</th>
<th>Send</th>
<th>Select</th>
</tr>";
//Retrieved data
while ($row = mysqli_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['part_type'] . "</td>";
echo "<td>" . $row['pn'] . "</td>";
echo "<td>" . $row['manufactured'] . "</td>";
echo "<td>" . $row['serial'] . "</td>";
echo "<td>" . $row['site_id'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['user'] . "</td>";
echo "<td>" . $row['faulty'] . "</td>";
echo "<td><input type='checkbox' name='send_parts[]' class='checkclass' value=" . $row['id'] . "></td>";
echo "</tr>";};
echo "</table>";
echo "</br>";
echo "</br>";
echo "<input type='button' onclick='sendToZG()' value='Send'/>";
echo "</br>";
echo "<input type='submit' name='submit' value='Generate tab' />";
echo "</form>";
exit();
}
} else {
die("Query failed");
}
User then checks option they want and upon submiting (Generate tab) they get tab delimited text with values they selected.
I now want when they click "Send" to have values posted to another php page and results returned on the same page (under SentList div). I have js like this:
//Browser Support Code
function sendToZG(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200){
var ajaxDisplay = document.getElementById('SentList');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from page and pass it to server script.
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
ajaxRequest.send("send_parts=" + formData);
}
Edited: ajaxRequest.send("send_part=" + formData); to ajaxRequest.send("send_parts=" + formData);
Now it returns:
Invalid argument supplied for foreach() on line 53 (That is where I fetch my data in sendtozg.php).
I'll add sendtozg.php at the end of the post.
If instead of:
<form name='send_parts_form' action='../includes/generatetab.php' method='post'>
I echo:
<form name='send_parts_form' action='../includes/sendtozg.php' method='post'>
Upon submit, script sendtozg.php gets executed fine but on a different page.
So basically what I'm trying to do is to have 2 options for the php generated form:
Generate tab delimited txt file
Execute sendtozg.php and return results on same page
I already have both scripts (generatetab.php and sendtozg.php) and they work fine.
sendtozg.php:
if (!empty($_POST['send_parts'])){
foreach ($_POST['send_parts'] as $send_parts){
$getchecked = mysqli_query($con, "SELECT * FROM $storage WHERE id=$send_parts");
while ($row = mysqli_fetch_array($getchecked)) {
$copypart = mysqli_query($con, "INSERT INTO sent_parts (part_type, pn, manufactured, serial, site_id, date, user, faulty, log)
SELECT part_type, pn, manufactured, serial, site_id, date, user, faulty, log
FROM $storage WHERE id=$send_parts");
// check to see if it copied
$getserial = mysqli_query($con, "SELECT serial FROM $storage WHERE id=$send_parts");
$getserial_row = mysqli_fetch_array($getserial, MYSQLI_NUM);
$foundserial = $getserial_row[0];
$checkcopy = mysqli_query($con, "SELECT id FROM sent_parts WHERE serial = '$foundserial'");
// add user info and date
$addinfo = mysqli_query($con, "UPDATE sent_parts SET sent2zg='$user', date2zg='$strdate' WHERE serial = '$foundserial'");
if (!$check1_res) {
printf("Error: %s\n", mysqli_error($con));
exit();
};
//delete from storage
if($checkcopy > 0) {
$getpart = mysqli_query($con, "SELECT part_type FROM sent_parts WHERE serial='$foundserial'");
$getpart_row = mysqli_fetch_array($getpart, MYSQLI_NUM);
$deletedpart = $getpart_row[0];
$removepart = mysqli_query($con, "DELETE FROM $storage WHERE id = '$send_parts'");
echo "Part " . $deletedpart . " has been transfered";
} else {
echo "Part " . $row['part_type'] . "was NOT transfered";
};
};
} exit ();
} else {
echo "Nothing was selected, please try again!";
}

Answer 1

Your <form> doesn't have an id attribute on it so you'll either need to add id="send_parts" to the <form> or you'll need to change your code from getElementById to getElementsByName like this:
// Now get the value from page and pass it to server script.
// Serialize the data
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.send(formData);
Then inside sendtozg.php you'll need to change the first two lines to:
if (!empty($_POST)){
foreach ($_POST as $send_parts){

Answer 2

This is the final code for the sendtozg.js:
// Get get the value from page and pass it to server script.
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.setRequestHeader("X-Requested-With", "XMLHttpRequest");
ajaxRequest.send(formData);
}
and sendtozg.php should be:
if (!empty($_POST)){
foreach ($_POST['send_parts'] as $send_parts){
$getchecked = mysqli_query($con, "SELECT * FROM $storage WHERE id=$send_parts");
By the way:
print_r ($some_array)
and
if (!$check1_res) {
printf("Error: %s\n", mysqli_error($con));
exit();
};
Are great tools for troubleshooting.