I am trying update an id that has a foreign key to another table of names. I have a drop menu and in the drop menu I have name from table NAME_TEST. I need to select the name but the insert that I want is:
INSERT INTO (test) values (the value that i need is the ID of selected name)
Code:
<html> <h1>Update form</h1></html>
<?php
if (isset($_POST['submit'])) {
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = "UPDATE test SET location_name='".$_POST['new_location']."' WHERE id='".$_POST['location']."' LIMIT 1";
$res = mysqli_query($connect, $query);
if (!$res) {
die("Something went wrong");
}
}
// This is the code to load the select options
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = 'SELECT * FROM name_test';
$res = mysqli_query($connect, $query);
echo "Choose setup";
$options = array();
while($row = mysqli_fetch_assoc($res)) {
$options[] = $row;
}
?>
<form method="POST" action="<?= $_SERVER['PHP_SELF'] ?>"><BR><BR>
<select name="location">
<option value="0">--- Select an option ---</option>
<?php foreach ($options as $option): ?>
<option value="<?= $option['id'] ?>"><?= $option['name'] ?></option>
<?php endforeach; ?>
</select><br /><BR><BR>
<B> New name:</B> <BR> <input type="text" name="new_location"><br /><BR><BR>
<input type="submit" name="submit" value="Update" />
</form>
You can use a join for this. The idea is something like this:
INSERT INTO t(nameid)
select nameid
from names n
where n.name = ?;
t is the table you want to insert into. names is the table that has the name id and the name.
Related
I have select tag, and dropdown menu, I want to delete item, when it's selected, like, on change
Here is my code:
$db = mysqli_connect('localhost', 'root', 'root', 'php_wood');
$sql = "SELECT * FROM posts";
$post = $db->query($sql);
if(isset($_GET['postDelete'])){
$delete = $_GET['postDelete'];
$sql = "DELETE FROM posts WHERE id = `$delete`";
mysqli_query($db, $sql);
}
I don't know where is mistake, here is my form
<form action="delete.php" method="get">
<select id="" onchange="this.form.submit();" name="postDelete">
<?php
if($post->num_rows > 0){
while($row = $post->fetch_assoc()){
?>
<option value="<?php echo $row['id']?>"><?php echo $row['title']?></option>
<?php
}
}
?>
<option value="">
</option>
</select>
</form>
There are 02 tables called item and customer.
item(item_id, item_name)
customer(cus_id, iid, cus_name)
I just tried to store item_id from item to the iid in the customer.
but it always showing null values.
My database is item_sales.
Here is my PHP code
<html>
<title></title>
<head></head>
<body>
<?php
$hostname = "localhost";
$database = "item_sales";
$username = "root";
$password = "";
$con = mysql_pconnect($hostname, $username, $password);
error_reporting(0);
?>
<form action="index.php" method="post" enctype="multipart/form-data">
<p>Customer Name : <input type="text" name="cus_name" /><br/><br/> </p>
<p>Select an Item:
<select name="iid">
<?php
$sql = mysql_query("SELECT * FROM item");
mysql_select_db($database,$con);
while($sqlv = mysql_fetch_array($sql))
{ ?>
<option id="<?php echo $sqlv['item_id']; ?>"><?php echo $sqlv['item_name']; ?></option>
<?php } ?>
</select>
</p>
<?php
if(isset($_POST['submit']))
{
$sql2 = "SELECT * FROM item WHERE iid='%item_id%'";
mysql_select_db($database,$con);
$mydata = mysql_query($sql2);
$cus_name = $_POST['cus_name'];
$sql3 = "INSERT INTO customer (cus_id, iid, cus_name) VALUES ('', '$_POST[iid]', '$cus_name')";
mysql_query($sql3);
}
?>
<input type="submit" name="submit" value="Add Sale" />
</form>
</body>
</html>
The reason it is not working is that you are attempting to save the iid select into the iid field, and I'm guessing the iid field in customer is a numeric type field, like INT - using the POST variable like this, you are going to be saving the text of the SELECT rather than the val.
What you need to do to fix this particular problem is set a "value" on each of the select options. You've set an ID but thats no real help here.
<select name="iid">
<?php
$sql = mysql_query("SELECT * FROM item");
mysql_select_db($database,$con);
while($sqlv = mysql_fetch_array($sql))
{ ?>
<option value="<?php echo $sqlv['item_id']; ?>"><?php echo $sqlv['item_name']; ?></option>
<?php } ?>
</select>
This is besides the point your code is very dangerous. I would recommend you do not use the original mysql functions as, 1) they don't offer any real protection from malicious users, and 2) they will be removed from PHP support very soon.
See this SO article on how to replace the mysql functionality from your PHP code : How can I prevent SQL injection in PHP?
That article also might help you understand the dangers your code offers.
The correct code is following :
<html>
<title></title>
<head></head>
<body>
<?php
$hostname = "localhost";
$database = "item_sales";
$username = "root";
$password = "";
$con = mysql_pconnect($hostname, $username, $password);
error_reporting(0);
?>
<form action="index.php" method="post" enctype="multipart/form-data">
<p>Customer Name : <input type="text" name="cus_name" /><br/><br/> </p>
<p>Select an Item:
<select name="iid">
<?php
$sql = mysql_query("SELECT * FROM item");
mysql_select_db($database,$con);
while($sqlv = mysql_fetch_array($sql))
{ ?>
<option value="<?php echo $sqlv['item_id']; ?>"><?php echo $sqlv['item_name']; ?></option>
<?php } ?>
</select>
</p>
<?php
if(isset($_POST['submit']))
{
$sql2 = "SELECT * FROM item";
mysql_select_db($database,$con);
$mydata = mysql_query($sql2);
$cus_name = $_POST['cus_name'];
$iid = $_GET['item_id'];
$sql3 = "INSERT INTO customer (cus_id, iid, cus_name) VALUES ('', '$_POST[iid]', '$cus_name')";
mysql_query($sql3);
}
?>
<input type="submit" name="submit" value="Add Sale" />
</form>
</body>
</html>
i need insert new location to row with the name i choose from my drop down.
How can i connect between the name and the location?
<?php
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = 'SELECT * FROM test';
$res = mysqli_query($connect, $query);
echo "<select name='testform'>";
while($row=mysqli_fetch_assoc($res)){
echo "<option value=>$row[name]</option>";
}
echo "</select>";
?>
<html>
<form action="indexx.php" method="POST">
<br>Locatio name:<input type= "text" method="POST"><BR>
<input type="submit" value="Insert" method="POST">
</form>
</html>
Check if the form was submitted by giving it a name and checking if it is set in your PHP code. If it is set, it means that the user submitted your data entry form. Perform an INSERT statement with the value the user entered.
Sidenote: input-tags don't have an attribute "method". You can remove those.
You might want to read about the lifecycle of a PHP script and the usage of the $_POST array in PHP. The syntax of the INSERT statement can be found in numerous language specs or tutorials.
I think you mixed things up a bit and wrote the following code for you. Please tell me if this explains it a bit more.
The code you require
<?php
if (isset($_POST['submit'])) {
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = "UPDATE test SET location_name='".$_POST['new_location']."' WHERE id='".$_POST['location']."' LIMIT 1";
$res = mysqli_query($connect, $query);
if (!$res) {
die("Something went wrong");
}
}
// This is the code to load the select options
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = 'SELECT * FROM test';
$res = mysqli_query($connect, $query);
$options = array();
while($row = mysqli_fetch_assoc($res)) {
$options[] = $row;
}
?>
<form method="POST" action="<?= $_SERVER['PHP_SELF'] ?>">
<select name="location">
<option value="0">--- Select an option ---</option>
<?php foreach ($options as $option): ?>
<option value="<?= $option['id'] ?>"><?= $option['name'] ?></option>
<?php endforeach; ?>
</select><br />
New name: <input type="text" name="new_location"><br />
<input type="submit" name="submit" value="Update" />
</form>
<?php
if (isset($_POST['submit'])) {
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = "UPDATE test SET location_name='".$_POST['new_location']."' WHERE id='".$_POST['location']."' LIMIT 1";
$res = mysqli_query($connect, $query);
if (!$res) {
die("Something went wrong");
}
}
// This is the code to load the select options
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = 'SELECT * FROM test';
$res = mysqli_query($connect, $query);
$options = array();
while($row = mysqli_fetch_assoc($res)) {
$options[] = $row;
}
?>
<form method="POST" action="<?= $_SERVER['PHP_SELF'] ?>">
<select name="location">
<option value="0">--- Select an option ---</option>
<?php foreach ($options as $option): ?>
<option value="<?= $option['id'] ?>"><?= $option['name'] ?></option>
<?php endforeach; ?>
</select><br />
New name: <input type="text" name="new_location"><br />
<input type="submit" name="submit" value="Update" />
</form>
I'm creating a task management list for a project. I've created a form to add each task and echo it into a list.
I want to add a select tag to the form that will allow the user to select the size of the task (small, medium, large or extra large).
My database is set up with 3 columns a unique key that auto increments, description and status.
I want to manipulate the value of the the status at the same time I enter the description. For example selecting "small" from the drop down will give the task entered a value of "1" and a different appearance in the list than a description with a status of "2".
Any feedback is greatly appreciated!!
Here is the code I have:
<?php
mysql_connect('localhost', 'root', 'root');
$query = "INSERT INTO `project1`.`tasklist` (
`key` ,
`description` ,
`status`
)
VALUES (
NULL , '".$_GET["newToDo"]."', '".$_GET["status"]."'
);";
mysql_query($query);
header('Location: index.php');
?>
<?php
mysql_connect('localhost', 'root', 'root');
foreach($_GET['toDone'] as $toDoKey) {
$query = "UPDATE `project1`.`practice` SET `status` = '0' WHERE `tasklist`.`key` =".$toDoKey.";";
mysql_query($query);
}
header('Location: index.php');
?>
<h1>Get stuff done!</h1>
<form action = "secondary.php" method="GET">
<label for="newToDo">New To Do:</label>
<input type="text" name="newToDo" id="newToDo" />
<select name="taskSize">
<option value="small" selected="selected">Small</option>
<option value="medium">Medium</option>
<option value="large">Large</option>
<option value="extraLarge">X-Large</option>
</select>
<input type="submit" />
</form>
<h2>to do</h2>
<form action="toDone.php" method="GET">
<ul>
<?php
mysql_connect('localhost', 'root', 'root');
$query = "SELECT * FROM `project1`.`tasklist` WHERE `status` = 1";
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
echo $row['description'];
echo "</li>";
}
?>
</ul>
<input type="submit" />
<h2>done</h2>
<ul id="dunzo">
<?php
$query = "SELECT * FROM `project1`.`tasklist` WHERE `status` = 0";
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
echo "<li>";
echo $row['description'];
echo "</li>";
}
?>
</ul>
What I want is to change how each new row in the list looks based on what is selected in the dropdown.
Dude just one thing to add to your code..
Deprecation of mysql_ functions
I have this code and I'm trying to put the selected state in a subcat table.
So far it returns an empty value. I'm not sure if this is clear or not, but all I want is: select a state from the select option and submit it. I want to get the selected state name into my table subcat.
enter <?php
include("connect.php");
$state = $row['states']; //Select name
if (isset($_POST[submit])){
$query = "INSERT INTO subcat (sub_name) VALUES ('$state')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo "<select name='states'>
<option value=''>Select a state</option>";
while ($row = mysql_fetch_assoc($sql)) {
echo "<option value='$row[id]'>$row[name]</option>";
}
echo "</select>";
?>
<input type="submit" name="submit" value="Continue" />
</form> here
Thanks
Change $state = $row['states'] to $state = $_POST['states']
<?php
include("connect.php");
if (isset($_POST[submit]))
{
$state = $_POST['states']; //Select name
$query = "INSERT INTO subcat (sub_name) VALUES ('$state')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo "<select name='states'>
<option value=''>Select a state</option>";
while ($row = mysql_fetch_assoc($sql)) {
echo "<option value='$row[id]'>$row[name]</option>"; // if you want to
//get the name into table, then use like this
//echo "<option value='$row[name]'>$row[name]</option>"; or
//echo "<option>$row[name]</option>";
}
echo "</select>";
?>
<input type="submit" name="submit" value="Continue" />
</form>
Try this:
enter <?php
include("connect.php");
if (isset($_POST[submit])){
$state = $_POST['states'];
$query = "INSERT INTO subcat (sub_name) VALUES ('".mysql_real_escape_string($state)."')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo '<select name="states" id="states">
<option value="">Select a state</option>';
while ($row = mysql_fetch_assoc($sql)) {
echo '<option value="'.$row['name'].'">'.$row['name'].'</option>';
}
echo '</select>';
?>
<input type="submit" name="submit" value="Continue" />
</form> here
Dont forget to use mysql_real_escape_string to prevent SQL injections. I have replaced $state = $row['states']; with $state = $_POST['states'];
I dont know where u got $row from...
The above will insert the states name into the database.