Im trying to send form data to my sql database but the database isn't receiving any of my values. The name of my database is taxibooking and table name is bookings.
I tried separating the php code in another file and using action on form to access the php code. on clicking submit I was redirected to a blank page with my php file name.
<form method="POST" action="">
Name of customer:<input type="text" name="fname"><br><br>
Enter pickup address:<textarea name="padd" rows="5" cols="10"></textarea><br><br>
Enter destination address:<textarea name="dadd" rows="5" cols="10"></textarea><br><br>
Select Taxi type:<select name="taxi"><option value="Viennese Fiaker">The Viennese Fiaker</option><option value="Indian Auto Rickshaw">Indian Auto Rickshaw</option><option value="Little Yellow">Little Yellow</option><option value="Mumbai Taxi Fiat">Mumbai Taxi Fiat</option><option value="Tricycles">Tricycles</option><option value="Water Taxi">Water Taxi</option><option value="Impeccable Taxi">Impeccable Taxi</option><option value="Red Taxi">Red Taxi</option></select><br><br>
<input type="submit" value="submit" name="sub">
</form>
<?php
$con=mysqli_connect("localhost","root","","taxibooking");
if(isset($_POST['sub']))
{
$n=$_POST['fname'];
$p=$_POST['padd'];
$d=$_POST['dadd'];
$type=$_POST['taxi'];
$sql="insert into bookings(name,pickup,destination,type) values ('$n','$p','$d','$type')";
mysqli_query($con,$sql);
}
?>
Try to debug and put query on if condition
if(mysqli_query($con,$sql)){
echo'done';
}else{
echo "error".$sql."</br>" . mysqli_error($con);
}
Here's how to add parameters on your insert script.
$sql="insert into bookings(name,pickup,destination,type) values ('".$n."','".$p."','".$d."','".$type."')";
I made a new database with a different name with the same table name and properties and now it sort of mysteriously works now. No changes done to the code.
Related
I have a form with few input fields and a update option ,suppose if i have 10 fields and i update only first two fields and not the rest who can i display a message that only the first to field A and B are updated using PHP and Mysqli
in the screenshot if i update the value of bill id and origin i should display message out in some other page that says that bill id and origin has been updated with "xyz " value
One way is to fetch fields with real and old name i.e.
Select name,name as old_name from table
In html
<input type="text" name="name" value="name from db">
<input type="hidden" name="old_name" value="old_name from db">
After submit form you can check it by.
And check after post.
If(post['name']==post['old_name'])
Something like this.
Another way is from js by checking onchange event
well it'll be something like this ..
html:
<html>
<body>
//action is defining the php file's name which the form will be sent to ;
//method is defining the method which the form will be sent with
<form action="updateCheck.php" method="post">
<input type="text" name="input-filed-1"/>
<input type="text" name="input-filed-2"/>
<input type="text" name="input-filed-3"/>
<input type="text" name="input-filed-4"/>
<input type"text" name="input-filed-5"/>
<input type="submit" value="submit/>
</form>
</body>
</html>
create a file named updateCheck.php in the same directory and enter the following code
<?php $updateMessage = "you have updated the following fields";
for($x = 0 ; $x<=5 ; $x++)
{
// trim for erasing the whitespaces
// that if clause checks if there was data entered in the update input fields
if(isset($_POST["input-field-".$x]) && trim($_POST["input-field-".$x]))
{
$updateMessage.= "input-field-".$x ;
}
}
echo "<script type='text/javascript'>alert('".$updateMessage."')</script>"
?>
I haven't tested the code .. but study it carefully and you'll see how it's done
Im trying to make like a refill code that will refill my table.
I have one table that have my refill code in it and other table that stores the balance of the account.
table1: card_credit (table that stores the balance of the account)
table2:card_refill (table that have me refill code)
I have created this code with session and PHP. Now I'm stuck and dont know how to move forward.
I want to make when i write in the refill code from table card_refill that its take the amount of credit into value in table card_refill
refill.php
<strong>Refill</strong>
<form action="refill.php" method"post">
<input type="text" name="refillcode"/>
<br>
<input type="submit" name="Submit" value="Refill" />
</form>
<?php
// starting the session
session_start();
if (isset($_POST['Submit'])) {
$_SESSION['refillcode'] = $_POST['refillcode'];
}
?>
Here is a possible solution, I am just don't know, where the card_id comes from.
This is inserting a new record into your card_credit table.
// starting the session before any output
session_start();
//include here the database connection file!
//for example:
//include('db_connection.php');
if (isset($_POST['Submit'])) {
//First do a validation here, is the refillcode number, exists, etc...
//Insert it into the table
$sql = "INSERT INTO card_credit (card_id, value) VALUES ('[YOUR_CARD_ID_HERE]', " . intval($_POST['refillcode']) . ")";
//Link is the resource variable when you created the mysqli_connect
mysqli_query($link, $sql);
//Redirect here if you want
}
?>
<!-- HTML CODE STARTS HERE -->
<strong>Refill</strong>
<form action="refill.php" method="post">
<input type="text" name="refillcode"/>
<br>
<input type="submit" name="Submit" value="Refill" />
</form>
Well I am pretty much trying to create database with some table, the values in the table and check them in phpMyAdmin. I am able to create the table and database, but not the values
2.) when I add the isset $_post['submit'] variable, when I click the submit button, nothing is getting created. Is there a syntax error I am not seeing?
<html>
<body>
<p> welcome to my Page
Please insert the data below
<br>
<br>
<form action="input.php" method="post">
<p> Name: <input type="text" name="name">
<p> Age: <input type="text" name="age">
<p> Address: <input type="text" name="address">
<p> Email: <input type="text" name="email">
<input type="submit" value="Send!" name="submit">
</form>
<?php
if(isset($_POST['submit'])){
//connects to the sql database
$con = mysql_connect("localhost", "willc86", "tigers330");
if (!$con) {
echo 'can not connect to Database' . "<br>";
}
//creates the database in mySQL
$db = mysql_query("CREATE DATABASE form");
if (!$db) {
echo 'Did not create database';
}
//select the database and connect to it. on where you want to create tables.
mysql_select_db("form", $con);
//create the table once "form database" is selected
$table = "CREATE TABLE users (
Name varchar(30),
Age varchar(30),
Address varchar(30),
Email varchar(30)
)";
mysql_query($table,$con);
//insert the data from the form
$value= "INSERT INTO users (Name, Age, Address, Email)
VALUES ('$_POST[name]','$_POST[age]','$_POST[address]','$_POST[email]')";
mysql_query($value,$con);
mysql_close();
}//end of submit
?>
</body>
</html>
Your form action is input.php, is your file called input.php as well? Otherwise you'd be executing input.php when you're submitting the form instead of executing the PHP on your page.
I think user willc86 don't have access rights for create databases.
In second your script is incorrect, because it run for each "user add" operation and tried create database and table.
You can create it once in phpadmin and use in your script only insert.
No point in highlighting particular errors here as others are unlikely to have the exact same issue. Better to just give you the tools/means to debug the issue for yourself.
Instead of:
mysql_query($table,$con);
Try:
$res = mysql_query($table,$con);
if($res === false){
throw new Exception(mysql_error($conn));
}
Better yet, use PDO.
First of all your code is fine if this file name is input.php. There can be few reasons, one that you have incorrect credentials second that the user does not have a right to create table, for that go to Phpmyadmin and give user all rights.
Secondly and most importantly, use Prepared statements or mysqli otherwise you are vulnerable to SQL Injection
For that do go through this post
How can I prevent SQL injection in PHP?
I want to create a project that save the fill up forms in previous form and insert into database using one button only. For example answer1.php and answer2.php the save button is in the answer2.php i want to fetch data from answer1.php and save to databse same as in answer2.php
this code below insert data in one form only
$query = mysql_query("INSERT into holiday (holiday_no,holiday_name, status,campaign_name,holiday_type, createdBy, holiday_date, createdDate)
VALUES('$holiday_no', '$id','$status','$campaign_name','$hol', 'System','$date','$createdDate')") or die(mysql_error());
echo "Data has been saved with holiday name";
Not quite sure what you're asking for ...but giving it a try ;-) :
You can put the key-value pairs the first script receives into the next form so they get transmitted once again to the second script. E.g. if in the first step something like category=foo and country=bar gets transmitted write out a form that looks like
<form method="POST" action="answer2.php">
<p>
<input type="hidden" name="category" value="foo" />
<input type="text" readonly="readonly" name="country" value="bar" />
<!-- all the other things you want to add to the form -->
<input type="submit" />
</p>
</form>
But keep in mind that a) you need to encode the values properly for html output, otherwise your scripts are vulnerable for injection attacks, see http://docs.php.net/htmlspecialchars
and b) your second script can't "be sure" that the values haven't been altered or even transmitted to the first script at all; if you need that (e.g. for some transaction mechanism) you need something else like e.g. http://docs.php.net/features.sessions
Use hidden fields, a session, or a temporary table.
<?php
if (empty($_REQUEST)) {
echo '<form action="', $_SERVER['PHP_SELF'], '">
</form>';
} elseif (empty($_REQUEST['some_field_from_your_second_form']) {
// Do the second part of your form
} else {
// Do the final submission
// Sanitize the values
// Insert in the database
}
having a bit of trouble adding some data to a database. I have the file new_entry.php which is a form, which posts the data added to insert_new.php.
Every time the fields are filled in and submitted the data does not go to the database with the error message "Could not add the data to table" appearing..any ideas?
NEW_ENTRY.PHP
<body>
<form method="post" action="insert_new.php"><!-- form sent to insert_new.php-->
Section: <input type="text" name="section"/><br />
Food: <input type="text" name="food"/><br />
Description: <input type="text" name="description"/><br />
Price: <input type="text" name="price"/><br />
<br />
<input type="submit" value="submit"/>
</form>
</body>
INSERT_NEW.PHP
<?php
include 'library/connect.php';//connect to databse
$section = $_REQUEST["section"]; // get data from the HTML form on new student form
$food = $_REQUEST["food"];
$description = $_REQUEST["description"];
$price = $_REQUEST["price"];
mysql_query ("INSERT INTO food_menu (section, food, description, price) VALUES ('$section', '$food', '$description', $price)")/* insert the data to the food_menu table*/
or die ("Could not add the data to table");//error message
header('Location:index.php');//auto redirect to view page
include 'library/closedb.php';
?>
It seems that you have a mistake at the end of your MySQL query near price.
Please replace the code below with existing line:
mysql_query ("INSERT INTO food_menu (section, food, description, price) VALUES ('$section', '$food', '$description', '$price')")
Tell me the result please.
First: Don't do this. You really need to research SQL Injection or you will be very sorry.
Secondly, your price has no numeric validation (assuming it's going into a numeric column)... this is also bad... what if someone put in a dollar sign or something?
Next, please post your table definition and connection code (not the connection values).
You can also get more feedback if you do something like:
or die (mysql_error());//error message