I am posting data to PHP from jQuery using data from an HTML form.
Here is the jQuery line that sends the POST
$.post("InsertNewQuestion.php", $("Create_Question_Form").serialize());
Here is the PHP code
<?php
$con = mysql_connect("localhost","root","");
mysql_select_db("Quizzes",$con);
$Quiz_Name = $_POST['Question'];
echo $Quiz_Name;
$Option_1 = $_POST['Option1'];
echo $Option_1;
$Option_2 = $_POST['Option2'];
echo $Option_2;
$Option_3 = $_POST['Option3'];
echo $Option_3;
$Option_4 = $_POST['Option4'];
echo $Option_4;
$Option_5 = $_POST['Option5'];
echo $Option_5;
$rowIDList = mysql_query("SELECT rowID FROM TestQuiz");
$ColumnValues = array();
$CurrentGreatestRowID = -1;
$LCV = 1;
while($row1 = mysql_fetch_assoc($rowIDList)) {
if ($CurrentGreatestRowID < $row1['rowID']) {
$CurrentGreatestRowID = $row1['rowID'];
}
$LCV++;
}
$CurrentRowID = $CurrentGreatestRowID+1;
$sql = "INSERT INTO TestQuiz (rowID,Quiz_Name,Option_1,Option_2,Option_3,Option_4,Option_5,Option_1_Votes,Option_2_Votes,Option_3_Votes,Option_4_Votes,Option_5_Votes)
VALUES(".$CurrentRowID.",'".$Question."','".$Option1."','".$Option2."','".$Option3."','".$Option4."','".$Option5."',0,0,0,0,0);";
if (mysql_query($sql,$con)) {
echo "Inserted values";
}
else {
echo ("Could not insert values: ". mysql_error());
}
mysql_close($con);
?>
Here is the HTML form
<form id="Create_Question_Form" action="" method="POST">
Question Name: input id="Question" class="Create_Question_Text_Box" type="text" name="Question_Name"><span id="Invalid_1"></span><br>
Option 1: input id="Option1" class="Create_Question_Text_Box" type="text" name="Option_1"><span id="Invalid_2"></span><br>
Option 2: input id="Option2" class="Create_Question_Text_Box" type="text" name="Option_2"><span id="Invalid_3"></span><br>
Option 3: input id="Option3" class="Create_Question_Text_Box" type="text" name="Option_3"><span id="Invalid_4"></span><br>
Option 4: input id="Option4" class="Create_Question_Text_Box" type="text" name="Option_4"><span id="Invalid_5"></span><br>
Option 5: input id="Option5" class="Create_Question_Text_Box" type="text" name="Option_5"><span id="Invalid_6"></span><br>
input type="Submit" id="Question_Submit" value="Create Question"></input>
</form>
Your form selector should be #Create_Question_Form. Note the # indicating that this is an element ID.
$("#Create_Question_Form").serialize()
Update
You are accessing the $_POST values by ID instead of name. Try this:
$Option_1 = $_POST['Option_1'];
$Option_2 = $_POST['Option_2'];
// etc
input type="Submit" id="Question_Submit" value="Create Question"></input>
<input type="Submit" id="Question_Submit" value="Create Question" />
You seem t be missing a bracket
Related
i want to echo selected parent value. but i am getting error- Notice: Undefined index:
How can i echo selected parent value then? Whats wrong i am doing?
$q = mysql_query("SELECT * FROM menu");
echo '<form action="" method="post">
Menu name:<input type="text" name="mname"><br>
<select>';
while ($row = mysql_fetch_array($q)) {
$menu_name = $row['menu_name'];
echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select><br>
<input type="submit" name="submit" value="Add Menu">
</form>';
if (isset($_POST['submit'])) {
echo $mname = $_POST['mname'];
echo $parent = $_POST[$menu_name];
}
add name to the select box and get the value of select box by name.
Updated code:-
$q = mysql_query("SELECT * FROM menu");
echo '<form action="" method="post">
Menu name:<input type="text" name="mname"><br>
<select name="menu_name">';
while ($row = mysql_fetch_array($q)) {
$menu_name = $row['menu_name'];
echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select><br>
<input type="submit" name="submit" value="Add Menu">
</form>';
if (isset($_POST['submit'])) {
echo $mname = $_POST['mname'];
echo $parent = $_POST['menu_name'];
}
$_POST[$menu_name] probably doesn't exist, because only two elements in your form have name attributes. The text input and the submit input.
option elements aren't posted as part of the form, but rather the selected option's value for the select element. But your select element has no name, therefore no key to use in the key/value pair, so it isn't posted.
Give the element a name:
<select name="someName">
Then in the POST, you would be able to fetch the selected value just as you do for any other form element:
$_POST['someName']
You need to add name attribute to select tag.
echo '<form action="" method="post">
Menu name:<input type="text" name="mname"><br>
<select name="any_name">';
$q = mysql_query("SELECT * FROM menu");
while ($row = mysql_fetch_array($q)) {
$menu_name = $row['menu_name'];
echo '<option value="'.$menu_name.'">'.$menu_name.'</option>';
}
echo '</select><br>
<input type="submit" name="submit" value="Add Menu">
</form>';
if (isset($_POST['submit'])) {
echo $mname = $_POST['mname'];
echo $select_option_name = $_POST['any_name'];
}
Note: mysql_* functions are depricated, use mysqli_* functions
This is advertisement query
INSERT into advertise_management(advertisement_name,adv_code) values('".$adv_name."','".$adv_code."')
post management:
<input type="text" name="post_name" value="<?php if(isset($post_value)){ echo $post_value['post_name'];} else { echo ''; }?>" required="required">
<?php $res= mysql_query("select * from advertise_management ");
while($result= mysql_fetch_array($res)){
$adv=$result['advertisement_name'];
{ ?>
<input type="checkbox" value="<?php echo $adv?>" name="post_adv[]">
<?php } ?>
Post Management insert query :
$post_name = $_POST['post_name'];
$post_adv1 = $_POST['post_adv'];
$post_adv = implode(",", $post_adv1);
$post_id = $_POST['post_id'];
$res = mysql_query("INSERT into post_managment(post_name,post_adv) values('".$post_name."','".$post_adv."')") or die(mysql_error());
Post management edit :
$result = mysql_query("SELECT * FROM post_managment where post_id='$pid'");
$post_value = mysql_fetch_array($result);
><input type="text" name="post_name" value="<?php if(isset($post_value)){ echo $post_value['post_name'];} else { echo ''; }?>" required="required">
<?php $res= mysql_query("select * from advertise_management ");
while($result= mysql_fetch_array($res)){
$adv=$result['advertisement_name'];
{ ?>
<input type="checkbox" value="<?php echo $adv?>" name="post_adv[]">
<?php } ?>
<input type="submit" value="" id="create"><input type="reset" value="" id="cancel">
I want to return the values that is checked and the values that is not checked.
The browser normally won't send anything for checkboxes that aren't checked. You can work around this by also rendering a hidden input () with the same name and a different value. Make sure it's output right before the checkbox.
I am trying to display mysql records through hidden field values, but nothing displays.
A little help!
Here's the code;
Html:
<form name="form11" method="post" action="hpdata.php" enctype="multipart/form-data">
<input name="pro" id="pro" type="hidden" value= "CMS" />
<input name="piror" id="piror" type="hidden" value= "P1" />
<input name="stat" id="stat" type="hidden" value= "In Progress" />
<input type="submit" name="submit" id="submit" class="groovybutton" value="...">
</form>
PHP:
<?php
$project = $_POST["pro"];
$pirority = $_POST["piror"];
$status = $_POST["stat"];
mysql_connect ("one", "two", "three");
mysql_select_db ("wsms");
$rest = mysql_query("SELECT * FROM sheet WHERE project='$project' AND
pirority='$pirority' AND status='$status'");
while($row = mysql_fetch_array($rest))
{
echo $row['id'] . " " . $row['date']; echo "<br>";
}
?>
Put isset into your php code
Example
<?php
if(isset($_POST['submit'])){
echo $project = $_POST["pro"]."<br>";
echo $pirority = $_POST["piror"]."<br>";
echo $status = $_POST["stat"];
/* mysql_connect ("one", "two", "three");
mysql_select_db ("wsms");
$rest = mysql_query("SELECT * FROM sheet WHERE project='$project' AND
pirority='$pirority' AND status='$status'");
while($row = mysql_fetch_array($rest))
{
echo $row['id'] . " " . $row['date']; echo "<br>";
}*/
}
?>
<form name="form11" method="post" action="" enctype="multipart/form-data">
<input name="pro" id="pro" type="hidden" value= "CMS" />
<input name="piror" id="piror" type="hidden" value= "P1" />
<input name="stat" id="stat" type="hidden" value= "In Progress" />
<input type="submit" name="submit" id="submit" class="groovybutton" value="...">
</form>
Output
CMS
P1
In Progress
First of all check that if the data is coming in the post or not:
<?php
echo "<pre>";
print_r($_POST);
exit;
?>
If yes than remove the print code i provided , and use extract($_POST); at the top of your PHP code. You query will become like this:
$rest = mysql_query("SELECT * FROM sheet WHERE project='$pro' AND
pirority='$piror' AND status='$stat'");
I am trying to create a form that will edit rows from my db table. (Based on some code I got from a StackOverflow page.)
I am able to populate the form with relevant data, but when I submit the form, the row isn't updated. In fact, some of my columns are deleted.
What did I do wrong?
edit.php
<?php
$UID = (int)$_GET['f'];
$query = mysql_query("SELECT * FROM user_feeds WHERE feed_id = '$UID'") or die(mysql_error());
if(mysql_num_rows($query)>=1){
while($row = mysql_fetch_array($query)) {
$feedtitle = $row['feed_title'];
$feedurl = $row['feed_url'];
$feedorder = $row['feed_order'];
$feedowner = $row['feed_owner'];
}
?>
<form action="update.php" method="post">
<input type="hidden" name="ID" value="<?=$UID;?>">
Title:<br /> <input type="text" name="ud_feedtitle" value="<?=$feedtitle?>"><br>
URL: <br /> <input type="text" name="ud_feedurl" value="<?=$feedurl?>"><br>
Order: <br /> <input type="text" name="ud_feedorder" value="<?=$feedorder?>"><br>
Owner:<br /> <input type="text" name="ud_feedowner" value="<?=$feedowner;?>"><br>
<input type="Submit">
</form>
<?php
}else{
echo 'No entry found. Go back';
}
?>
</div>
</body>
</html>
update.php
<?php
$ud_ID = $_REQUEST["ID"];
$ud_feedtitle = $_POST["feed_title"];
$ud_feedurl = $_POST["feed_url"];
$ud_feedorder = $_POST["feed_order"];
$ud_feedowner = $_POST["feed_owner"];
$query = "UPDATE user_feeds SET feed_title = '$ud_feedtitle', feed_url = '$ud_feedurl', feed_order = '$ud_feedorder', feed_owner = '$ud_feedowner', WHERE feed_id = '$ud_ID'";
$res = mysql_query($query);
if ($res)
echo "<p>Record Updated<p>";
else
echo "Problem updating record. MySQL Error: " . mysql_error();
?>
Reason:
The name of the input field is the same name by which $_POST is populated. The variables you are currently requesting :
$_POST["feed_title"];, $_POST["feed_url"];, $_POST["feed_order"];, $_POST["feed_owner"];
are all empty as they don't exist. When updating, you are replacing the values in your table with blank values.
Solution:
In your update.php, the following should be there instead.
$ud_ID = $_POST["ID"];
$ud_feedtitle = $_POST["ud_feedtitle"]; //corresponding to <input type="text" name="ud_feedtitle" ...
$ud_feedurl = $_POST["ud_feedurl"]; //corresponding to <input type="text" name="ud_feedurl" ...
$ud_feedorder = $_POST["ud_feedorder"]; //corresponding to <input type="text" name="ud_feedorder" ...
$ud_feedowner = $_POST["ud_feedowner"]; //corresponding to <input type="text" name="ud_feedowner" ...
I am working on a project. What I need to do is basically enter some info into a form, have that form save it into a database, display the data, and then be able to edit the data. So far, I am able to do everything except edit the data. I've tried using $_GET to get the ID of the particular "bug" I need to edit, and I am able to do that, and get all of the information but I am not sure how to edit that particular ID in my database. Here is my handler: http://pastebin.com/mR6QWpJ7 and my form:
<form action="week10handle.php" method="POST">
<fieldset width="300px">
<legend width="300px"><b>Add a bug report</b></legend>
Product Name:<br/><input type="text" name="product_name"><br/>
Product Version: <br/><input type="text" name="product_version"><br/>
Hardware Type: <br/><input type="text" name="hardware"><br/>
Operating System: <br/><input type="text" name="os"><br/>
Frequency: <br/><input type="text" name="frequency"><br/>
Proposed Solutions: <br/><textarea name="solutions"></textarea><br/>
<input type="submit" value="Submit">
</fieldset>
</form>
Here is where I obtain the get data in my edit form page so far, but as of right now, I am not sure how to edit a particular ID in the database.
$getbug = htmlspecialchars($_GET["bugid"]);
if (!empty($getbug)){
$getbuginfo = mysql_query("SELECT * FROM `bugs` WHERE `id`= '$getbug'");
if ($getbuginfo = mysql_fetch_assoc($getbuginfo)){
$edit_product_name = $getbuginfo['product_name'];
$edit_prod_version = $getbuginfo['product_version'];
$edit_hardware = $getbuginfo['hardware_type'];
$edit_os = $getbuginfo['os'];
$edit_frequency = $getbuginfo['frequency'];
$edit_solutions = $getbuginfo['solutions'];
?>
<form action="week10handle.php" method="POST">
<fieldset width="300px">
<legend width="300px"><b>Edit bug <?php echo $getbug;?></b></legend>
Product Name:<br/><input type="edit" name="product_name" value="<?php echo $edit_product_name;?>"><br/>
Product Version: <br/><input type="edit" name="product_version" value="<?php echo $edit_prod_version;?>"><br/>
Hardware Type: <br/><input type="edit" name="hardware" value="<?php echo $edit_hardware;?>"><br/>
Operating System: <br/><input type="edit" name="os"value="<?php echo $edit_os;?>"><br/>
Frequency: <br/><input type="edit" name="frequency"value="<?php echo $edit_frequency;?>"><br/>
Proposed Solutions: <br/><textarea name="solutions"><?php echo $edit_product_name;?></textarea><br/>
<input type="submit" value="Submit">
</fieldset>
</form>
EDIT: Here is my update php code, but it is still not working, when I submit my form, it refreshes the page, but it doesn't update the database:
<?php
if (mysql_connect('localhost','root','') && mysql_select_db('bug_reports')){
$errors = array();
if (isset($_POST['product_name'], $_POST['product_version'],$_POST['hardware'],$_POST['os'],$_POST['frequency'], $_POST['solutions'])){
$product_name = mysql_real_escape_string(htmlentities($_POST['product_name']));
$product_version = mysql_real_escape_string(htmlentities($_POST['product_version']));
$hardware = mysql_real_escape_string(htmlentities($_POST['hardware']));
$os = mysql_real_escape_string(htmlentities($_POST['os']));
$frequency = mysql_real_escape_string(htmlentities($_POST['frequency']));
$solutions = mysql_real_escape_string(htmlentities($_POST['solutions']));
$getbug = mysql_real_escape_string(htmlentities($_POST['bugid']));
if (empty($product_name) || empty($product_version) || empty($hardware) || empty($os) || empty($frequency) || empty($solutions)){
$errors[] = 'All fields are required.';
}
if (!is_numeric($product_version) || !is_numeric($frequency)){
$errors[] = 'Product version and frequency must both be numbers';
}
if (empty($errors)){
$update = "UPDATE `bugs` SET `product_name` = '$product_name', `product_version = '$product_version', `hardware_type = '$hardware', `os` = '$os', `frequency` = '$frequency', `solutions` = '$solutions' WHERE `id` = $getbug";
if ($update = mysql_query($update)){
header('Location: week10handle.php');
} else{
$errors[] = 'Something went wrong, please try again.';
}
} else{
foreach($errors as $error){
echo '<p><strong>'.$error.'</strong></p>';
}
}
}else{
$getbug = htmlspecialchars($_GET["bugid"]);
}
if (!empty($getbug)){
$getbuginfo = mysql_query("SELECT * FROM `bugs` WHERE `id`= '$getbug'");
if ($getbuginfo = mysql_fetch_assoc($getbuginfo)){
$bugid = $getbuginfo['id'];
$edit_product_name = $getbuginfo['product_name'];
$edit_prod_version = $getbuginfo['product_version'];
$edit_hardware = $getbuginfo['hardware_type'];
$edit_os = $getbuginfo['os'];
$edit_frequency = $getbuginfo['frequency'];
$edit_solutions = $getbuginfo['solutions'];
?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST">
<fieldset width="300px">
<legend width="300px"><b>Edit bug <?php echo $getbug;?></b></legend>
Product Name:<br/><input type="edit" name="product_name" value="<?php echo $edit_product_name;?>"><br/>
Product Version: <br/><input type="edit" name="product_version" value="<?php echo $edit_prod_version;?>"><br/>
Hardware Type: <br/><input type="edit" name="hardware" value="<?php echo $edit_hardware;?>"><br/>
Operating System: <br/><input type="edit" name="os"value="<?php echo $edit_os;?>"><br/>
Frequency: <br/><input type="edit" name="frequency"value="<?php echo $edit_frequency;?>"><br/>
Proposed Solutions: <br/><textarea name="solutions"><?php echo $edit_product_name;?></textarea><br/>
<input type="hidden" name="bugid" value="<?php echo $bugid;?>" >
<input type="submit" value="Update">
</fieldset>
</form>
<?
}else{
echo "something went wrong";
}
}else{
echo "No bug found.";
}
}else
echo 'Could not connect at this time.';
?>
A typical way to detect an update, as opposed to an insert, would be to check for a value for id. So, in your edit form add a hidden field to pass the id to the handler and then in your handler you can decide whether to process it as insert or update based on the presence of the id field.
if (isset($_GET['id']) {
// do update
$sql = 'UPDATE `bugs` SET ... WHERE id = ' . intval($_GET['id']);
} else {
// do insert
$sql = 'INSERT INTO `bugs` VALUES ....';
}
UPDATE `bugs` SET `product_name` = '...', `product_version` = '...', ... WHERE `id` = $bugid;
Where the "..." will be replaced with newly $_POST-ed values for each column