Select Data from Database with Id From URL - php

I've a database it is look like this
|ID | Name |
|081| John Davidson|
and i have "index.php" in my website, i've learnt about php form, using method get, and the url is change to index.php?id=081
<form action="index.php" method="get">
<input name="id" value="081"/>
<input type="submit" value="Submit"/></form>
and when the page is loaded i want to show the name of id 081 from my database, how to do that?

Try this,
//Your index.php file
if($_GET['id']){
$id = $_GET['id'];
$sql="SELECT * FROM tableName where id='$id'";
$data = mysql_query($sql);
$row = mysql_fetch_array($data);
echo $row['name'];
}
<form action="index.php" method="get">
<input name="id" value="081"/>
<input type="submit" value="Submit"/></form>

After submitting , you are sending the id value with GET to index.php by url.
You can catch it with $_GET['id'],and store it in database like this:
$sql="INSERT INTO table SET id='".$_GET['id']."'";
$query=mysqli_query($connection,$sql);
If you want to retrieve this value from the database,you can do it like this:
$sql="SELECT id FROM table";
$query=mysqli_query($connection,$sql);
$row=mysqli_fetch_array($query);
echo $row['id'];
UPDATE As Abhik mentioned, those statements are very vulnerable, you should probably learn about using prepared statements here
Another simple way of avoiding sql injection , since it's pretty obvious you are new to php , is to use POST method instead of GET , and check on user input with this little function:
function test_input($data){
$data=htmlentities($data);
$data=stripslashes($data);
$data=trim($data);
return $data;
}
$id=test_input($_POST['id']);
Of course,depending on the field type, there must be some validation like min,max length , character allowed , etc.

Related

button to call php to delete mysql row via ID

Hi I have a html form button called delete. Next to the button is an input area. The logged in user will enter a number corresponding to the ID then press delete. It should delete the row corresponding to that ID.
For some reason it doesn't work.
This is my html form:
<h3 class="block-head">Delete Job</h3>
<form action="deletejob.php" method="post" class="basic-grey">
<label>Please Enter Job ID</label><input type="text" name="idnum" id="id"/><br>
<input type="submit" name="delete" value="delete"/>
</form>
This is my php code:
mysql_connect("********", "root", "******")
or die("Connection Failed");
mysql_select_db("jobslist")or die("Connection Failed");
$idnum = $_POST['id'];
$query = "delete from jobs_list where idnum = '".$id."'";
if(mysql_query($query)){ echo "deleted";}
else{ echo "fail";}
I know about switching to mysqli and that...I just want to see if this method works first.
thanks.
the name of your input is idnum yet you are looking in the $_POST array for id and in your query using the variable $id when you declare $idnum in the line previous.
$idnum = intval($_POST['idnum']);
$query = "delete from jobs_list where idnum = '".$idnum."'";
PHP receive the data via $_POST as an array of input elements with the key as the NAME of the input field, so you have to change this line:
$idnum = $_POST['idnum'];
You can also change the input name and don't change the php code:
<input type="text" name="id" id="id"/>
It's a typical mistake, don't worry too much about that :P
You should change the POST value.. and then try
$idnum = $_POST['idnum'];
Change the HTML, update the input type id to:
id="idnum"
then, update the PHP query while fetching the post value to:
$idnum = $_POST['idnum'];
It should work.

Form session not working when multiple users use it

i need an help from you all.
i had created an form using PHP. it's a school application registration form. it has one page only.
i need to generate registration number(session id used as registration number here) for everyone who opens the form.
instead of creating a session i have used ID for all. that is when some one submits the form, it checks the DB and if the registration number is there, it will increment one value and add the current form to DB.
my code here
<td>Application No : <input type="hidden" name="disablusr_dummyid" autocomplete="off" style="background:#f0efed;" value="00<?php
include('config.php');
$q="select MAX(auto_gen_id) from application_form";
$result=mysql_query($q);
$data=mysql_fetch_array($result);
$max_val=$data[0];
echo $max_val+1;
?>"/>
<input type="hidden" name="applicant_id" autocomplete="off" value="00<?php
include('config.php');
$wer = "select MAX(auto_gen_id) from application_form";
$resultgh = mysql_query($wer);
$dates = mysql_fetch_array($resultgh);
$erfdqwe = $dates[0];
echo $erfdqwe+1;
?>" />
<input type="hidden" name="txt_applicant_id" style="display:none;" autocomplete="off" value="<?php
include('config.php');
$werqw = "select MAX(auto_gen_id) from application_form";
$resultghasw = mysql_query($werqw);
$dataqsax = mysql_fetch_array($resultghasw);
$erfdqweqti = $dataqsax[0];
echo $erfdqweqti+1;
?>" /></td>
but what is happening is when multiple users are using the form same session ID is generated and only one user is able to save the form and it reflects in DB. other forms are being not submitted and not added to DB.
help me in this error.. thanks in advance.
The solution here is not the use the hidden fields for the IDs :
At this moment u have something like this :
"INSERT INTO (id, ..., ...) VALUES('".addslashes($_POST['applicant_id']."', ...)
You should refactor your logic to calculate the id afterwards meaning that u drop the hidden fields and do this :
<?php
if (!empty($_POST)) {
$wer = "select MAX(auto_gen_id) from application_form";
$resultgh = mysql_query($wer);
$dates = mysql_fetch_array($resultgh);
$erfdqwe = $dates[0];
$new_applicant_id = $erfdqwe+1;
}
And then use the $new_applicant_id into the INSERT sql.
On a sidenote your code is vulnerable for SQL injection and is using outdated functions.
You should try to move to mysqli or PDO and preferable use prepared statements

After form submit get the id from databse of last record

I need help for my web service project where user create an order when submit the order form, Driver get notification with accept or reject the order and when he accept order details list display to him. these all process done with order form submission. my all coding working fine. I don't understand where from get id of current order list. plz give solution. thx
<?php
if(isset($_POST['order_form']))
{
mysql_connect('localhost','root','');
mysql_select_db('live_help');
$user = $_POST['user'];
$password = $_POST['mobile'];
$password = $_POST['address'];
$password = $_POST['order_item'];
$password = $_POST['price'];
$sql= mysql_query("insert into `json_web`(user,mobile,address,order_item,price) values('$user','$mobile','$address','$order_item','$price')");
if($sql)
{
echo 'Value Saved';
}
}
?>
<form name="order_form" action="" method="post">
<input type="text" name="user">
<input type="text" name="mobile">
<input type="text" name="address">
<input type="text" name="order_item">
<input type="text" name="price">
<input type="submit" name="field1" value="Submit" />
</form>
use mysql_insert_id(); to grap the last insert id. Take a look for more information
mysql_insert_id()
Warning:
mysql_* is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:
After your mysql_query(); you can do this:
$id = mysql_insert_id();
mysql_insert_id(); retrieves the ID of the last inserted element to the database.
Use mysqli_insert_id() as from documentation
mysqli::$insert_id -- mysqli_insert_id — Returns the auto generated id used in the last query
In order to use mysql_insert_id() you need to add another field into the database table so you can have distinct order id for each order.
Perhaps you already have it, if not, add it:
ALTER TABLE `json_web` ADD order_id INT(10) NOT NULL AUTOINCREMENT;
Then just select from database, use mysql_insert_id()
Cheers

Retrieve all record and insert to another table form another db

I have troubles about php & mysql. I've to retrieve all records from DB1's table and then I have to insert them again to DB2's table.
<?php
require_once 'includes/config.php';
include 'includes/header.php';
if(isset($_POST['go'])){
$query = mysql_query("SELECT id,username,password FROM $db_database1.account")
or die(mysql_error());
echo "Record ".mysql_num_rows($query)." retrieve";
while($result_row = mysql_fetch_array($query, MYSQL_ASSOC)){
$account_ID = $result_row['id'];
$username = $result_row['username'];
$password = $result_row['password'];
$query = mysql_query("INSERT INTO $db_database2.account(uid,username,password) VALUES('$account_ID','$username','$password')")
or die(mysql_error());
$selectId = mysql_insert_id();
}
}
mysql_close($conn);
?>
<div class="wrapper">
<div class="content">
<form method="post" action="<?PHP $_SERVER['PHP_SELF'];?>">
<input type="submit" name="go" value="Go" />
</form>
</div>
<?php include 'includes/footer.php';?>
According to this code just one record was inserted. How can I insert all retrieved records?
To insert records into another table you need one single query, run from mysql console without PHP:
INSERT INTO db_database2.account SELECT id,username,password FROM db_database1.account
Notes on your code
you have to escape strings you are adding to the query
for some reason you are inserting into the same database
asking for the mysql_insert_id() makes no sense as you are apparently inserting a_i id already
there is no use for storing second mysql_query result into variable
yet this variable gets overwritten <- here is the reason your code runs once.
there is no use for echoing $_SERVER['PHP_SELF'] here. just leave form action blank.
yet you are actually leaving form action blank as you just forgot to echo this variable
I see no use for all the form and HTML here. Can't you just run this code without forms?
as it seems that whole mess is just to hash passwords, you need no extra tables then
just simple
UPDATE account SET password = md5(concat(id,username,password));
always have a database backup before such manipulations

Editting data by ADMIN

I am new to PHP and working on small local webpage and database that takes user information and database stores the same user information .If i Login with ADMIN it shows all data. My requirement is that the loggined user is an admin, then he has a right to edit all the informtion of the users that i stored in the database.And this is to be done using GET method . How it will be working?
Heres some example code purely to demonstrate how to update a table using a GET method form. The code doesn't have any kind of error checking and assumes you already know how to connect to your database (and that its MySQL).
Assuming you've landed on a page which invites you to edit data, which record you're editing is referenced by an 'id' variable on the URL which matches a numerical primary key in your database table.
<?php
$SQL = "SELECT myField1,myField2 FROM myTable WHERE myKeyField = '".intval($_GET['id'])."'";
$QRY = mysql_query($SQL);
$DATA = mysql_fetch_assoc($QRY);
?>
<form method='get' action='pageThatStoresData.php'>
<input type='hidden' name='key' value='<?php echo $_GET['id']; ?>' />
<input type='text' name='myField1' value="<?php echo $DATA['myField1']; ?>" />
<input type='text' name='myField2' value="<?php echo $DATA['myField2']; ?>" />
<button type='submit'>Submit</button>
</form>
So, this will give you a page that takes the data out of your table, displays it in a form with pre-filled values and on submit, will go to a URL like:
http://mydomain.com/pageThatStoresData.php?key=1&myField1=someData&myField2=someMoreData
In that page, you can access variables 'key', 'myField1', 'myField2' via the $_GET method.
Then you just need to update your table within that page:
$SQL = "UPDATE myTable
SET myField1 = '".mysql_real_escape_string($_GET['myField1'])."',
myField2 = ".mysql_real_escape_string($_GET['myField1'])."
WHERE key = '".intval($_GET['key'])."'
";
$QRY = mysql_query($SQL);
PLEASE NOTE: The code above is unsuitable for a straight copy/paste as it doesn't do any error checking etc, its purely a functional example (and typed straight in here so I apologise if there are any typos!).

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