Editing/Updating Existing Data in MySQL Database

Question

I am working on a appointment management website. I have designed a page for arranging an appointment (inserting in database) and its working fine. Now i want to view a existing appointment (view/fetch from database) and even modify if needed (updating). And i am facing an error in existing appointment.

here is the code for existing appointment:

<?php
mysql_connect("localhost", "root", "") or die("Could not connect.");

mysql_select_db("appointo") or die("could not find db.");

if(isset($_POST['uname'])){
$searchq = $_POST['uname'];

$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

$query = mysql_query("SELECT * FROM appointments WHERE username LIKE '$searchq'") or die("could not search.");
$count = mysql_num_rows($query);
while($row = mysql_fetch_array($query)){
        $name = $row['name'];
        $doctorname = $row['doctorname'];
        $purpose = $row['purpose'];

        //$oname = '<div>'.$name.'</div>';
        //$odname = '<div>'.$doctorname.'</div>';
        //$opurpose = '<div>'.$purpose.'</div>';
    }

}
?>

<form method="POST">

<input type="text" name="uname" placeholder="Seach using Name.." />
<input type="search" name="search" value="Search" /><br><br>

Output: 
<?php

//$name=$_SESSION['name'];
//$doctorname=$_SESSION['doctorname'];
//$purpose=$_SESSION['purpose'];

echo "'Name:''.$name'";
echo "Doctor Name:".$doctorname;
echo "Purpose:".$purpose;
?>
</form>

Still not able to fetch data. I want that if someone enters a username (say, nichani) which i already have stored in my database. And clicks on search button, he should see below output as:

Name: Nishant Nichani
Doctor Name: Dr. A
Purpose: General Checkup

help me. .fetch it from database.

There is **no more support** for `mysql_*` functions, they are [**officially deprecated**](https://wiki.php.net/rfc/mysql_deprecation), **no longer maintained** and were [**removed in PHP 7**](http://php.net/manual/en/function.mysql-connect.php#warning). You should update your code with [PDO](http://php.net/pdo) or [MySQLi](http://php.net/mysqli) to ensure the functionality of your project in the future. — j08691, Oct 12 '16 at 14:51
The issue is because your those variables were not defined in the scope where you are trying to use them, try to run your query against your DB and ensure you can fetch data out, then move the echo to inside the first tag — Babajide Apata, Oct 12 '16 at 14:54
What if i store it in another variable and use it outside tags, for the output. — Nishant Nichani, Oct 12 '16 at 15:00

MSS · Answer 1 · 2016-10-12T15:15:22.013

You should add uname input feild to your form.
You Should check if form is submitted or it's just first time page load.
You should also check for empty result.

And to do a LIKE search you should use % to specify any string with any length.

mysql_select_db("appointo") or die("could not find db.");

if(isset($_POST['uname'])){
$searchq = $_POST['uname'];

$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

$query = mysql_query("SELECT * FROM appointments WHERE username LIKE '%$searchq%'") or die("could not search.");
$count = mysql_num_rows($query);
while($row = mysql_fetch_array($query)){
        $name = $row['name'];
        $doctorname = $row['doctorname'];
        $purpose = $row['purpose'];

        //$oname = '<div>'.$name.'</div>';
        //$odname = '<div>'.$doctorname.'</div>';
        //$opurpose = '<div>'.$purpose.'</div>';
    }

}
?>

<form method="POST">
Uname:
<input name="uname" value="<?php echo @$_POST['uname']; ?>" placeholder="search name"/>

Output: 
<?php
//$name=$_SESSION['name'];
//$doctorname=$_SESSION['doctorname'];
//$purpose=$_SESSION['purpose'];
if(isset($_POST['uname'])){
 if(!empty(isset($query))){
   echo "'Name:''.$name'";
   echo "Doctor Name:".$doctorname;
   echo "Purpose:".$purpose;
}else{
  echo "NOT FOUND";
}

}
?>
</form>

Error are solved but still i am not able to fetch data from the database. — Nishant Nichani, Oct 12 '16 at 15:04
You should use % in your query to say the string can start with any string and ends with any string. I mention it in the answer — MSS, Oct 12 '16 at 15:16

score 0 · Answer 2 · answered Oct 12 '16 at 15:07

0

Define a blank varible before entering if statement, as $doctorname = ''; $purpose = ''; It should get your work done.

answered Oct 12 '16 at 15:07

Kangkon Saikia

57
3

nothings happens. still the same. – Nishant Nichani Oct 12 '16 at 15:10
Check whether your query is correct.. echo $query, as what it returns, – Kangkon Saikia Oct 12 '16 at 15:26
If soo.. debug what it is fetching... you may also use print_r($query)... and fetch data if row count is grater than 0. – Kangkon Saikia Oct 12 '16 at 15:28

Editing/Updating Existing Data in MySQL Database

2 Answers2