We are having troubles with our php session.
Trying to add a patient to a database, linked to the doctor that is curing the patient.
We have a select from the existing doctors in the database. They show up in the dropdown list.
But when we are trying to send the selected doctor to the database (the php session), there is no addition to the database. All the other inputs (patient name, patient birth date, etc) are put in the database, except the data from the dropdown list.
Add_patient.php
Doctor:<br>
<select name="doctor">
<option value="">--Select--</option>
<?php
$config = parse_ini_file("divkey.ini.php", true);
include("connect/connect_mysql.php");
$opdracht = "SELECT * FROM gebruiker ORDER BY id";
$resultaat = mysql_query($opdracht);
while ($rij = mysql_fetch_array($resultaat)) {
$id = $rij['id'];
$name = $rij['name'];
$fname = $rij['fname'];
?>
<option value ="<?php $id;?>"><?php echo"$name $fname" ?></option>
<?php
} ?>
</select>
Session_add.php
$doctor = $_POST['doctor'];
# query
Our query from session_add.php works. Just the not for the $_POST['doctor'].
# query
$opdracht = "INSERT INTO patient ( `name`, `fname`, `geslacht`, `doctor`, `straatnaam`, `huisnummer`, `postcode`, `gemeente`, `telefoonnummer`, `patientnummer`, `land`, `bloedgroep`, `gsmnummer`, `geboortedatum`, `geboorteplaats`, `taal`, `nationaliteit`, `rijksregisternummer`, `huisarts` )
VALUES ('".$name."', '".$fname."', '".$geslacht."', '".$doctor."','".$straatnaam."' ,'".$huisnummer."' , '".$postcode."' , '".$gemeente."', '".$telefoonnummer."',
'".$patientnummer."','".$land."', '".$bloedgroep."', '".$gsmnummer."', '".$geboortedatum."', '".$geboorteplaats."', '".$taal."', '".$nationaliteit."','".$rijksregisternummer."', '".$huisarts."')";
# other values are not important, it's in Dutch and these values are sent to the database
# doing query
$result = mysql_query($opdracht) or die(mysql_error());
# we use or die(mysql_error())
The query is executed, a 0 (zero) is added to the database instead of the selected doctor.
<option value ="<?php echo $id;?>"><?php echo $name.$fname; ?></option>
Try this.
We found the problem, it was in the sql query. With var_dump($_POST)
I checked the values that were sent to the session, these were the right ones.
In the SQL query, there was '".$doctor."'
This was wrong, it should be '.$doctor.' (recognised as id)
Thanks for the help!
Related
I have a list code but it didn't get further)
<?
include("connect.phtml");
$r= mysql_query("SELECT name_goods FROM goods")
or die ("!1");
echo "<select name='product'>";
while($row=mysql_fetch_array($r))
{
echo "<option value='".$row['name_goods']."'>".$row['name_goods']."</option>";
}
mysql_close();
?>
You should put your name_goods id into your option value instead of the name.
Then you will need to insert a $_POST['product'] (which will be the selected value) to your database with the appropriated query.
Something like :
$good_id = $_POST['product']
$query = $pdo->prepare("INSERT INTO table(good_id) VALUES (".$good_id.")");
$query->execute();
I created 3 tables in my database "Colleges" in PhpMyAdmin. The names of the tables are "cool", "data" and "tab". The first table "cool" consists of the the names of the states of India. It has two columns : ID and Statename. From the data in this table, I created a drop down list in HTML form. Further the HTML form consists of the name, email id, contact and the address. The user has to fill in the details and select his/her state from the drop down list. Now, the user input consisting of name, email id, contact and the address goes into the second table "data" and the selected state from the drop down list goes into the 3rd table "tab", "tab" consists of 2 columns ID and stat where the state name gets stored here. I joined the above two tables "data" and "tab" using inner join of SQL. When I fetched the data in another web page, the name, email id, contact and address are getting printed but not the the statename. Instead of the state name, ID of the statename (as given in table cool) is getting printed.
I want state name to get printed.
Here is the drop down list created using the data from my database :
<td>State :</td>
<td>
<?php
$mysqli = new mysqli('localhost', 'root', '', 'colleges');
$resultset = $mysqli->query("SELECT ID, Statename from cool");
?>
<select name="state">
<?php
while($rows = $resultset->fetch_assoc())
{
$ID = $rows['ID'];
$Statename = $rows['Statename'];
echo "<option value='$ID'>$Statename</option>";
}
?>
</select>
</td>
</tr>
And what are the changes to be done here to insert the selected dropdown state name into the table in my database ?
<?php
$connection = mysqli_connect("localhost", "root", "", "colleges");
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$email = $_POST['email'];
$contact = $_POST['contact'];
$address = $_POST['address'];
$state = $_POST['state'];
if($name !=''||$email !=''){
//Insert Query of SQL
$insert = "INSERT Into data(student_name, student_email, student_contact, student_address) values ('$name', '$email', '$contact', '$address')";
$query = mysqli_query($connection, $insert);
$insert2 = "INSERT Into tab(stat) values ('$state')";
$query2 = mysqli_query($connection, $insert2);
echo "<br/><br/><span>Data has been inserted successfully</span>";
}
else{
echo "<p>Insertion Failed <br/> Some Fields are Blank</p>";
}
}
mysqli_close($connection); // Closing Connection with Server
?>
3). The printing part:
<?php
$hostname = "localhost";
$dbname = "colleges";
$username = "root";
$password = "";
$conn = mysqli_connect("$hostname","$username","","$dbname");
if(mysqli_connect_errno())
{
echo "Failed to Connect MySQL (phpmyadmin) Database: ".mysqli_connect_error();
}
$query = ("select student_name, student_email, student_contact, student_address, stat from data t2 inner join tab t3 on t2.ID=t3.ID");
$result = mysqli_query($conn, $query);
echo "<center>";
echo "<h1>Student list</h1>";
echo "<hr/>";
echo"<table border = '1'>
<tr>
<th>Name</th>
<th>Email</th>
<th>Contact</th>
<th>Address</th>
<th>State</th>
</tr>";
while ($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>".$row['student_name']."</td>";
echo "<td>".$row['student_email']."</td>";
echo "<td>".$row['student_contact']."</td>";
echo "<td>".$row['student_address']."</td>";
echo "<td>".$row['stat']."</td>";
echo "</tr>";
}
echo "</table>";
echo "</center>";
mysqli_close($conn);
?>
echo "<option value='$ID'>$Statename</option>";
and
$state = $_POST['state'];
and
$insert2 = "INSERT Into tab(stat) values ('$state')";
You are actually not inserting the name of the state, but it's id, as the select return the selected option's value, not it's content. You can either change the first line to
echo "<option value='$Statename'>$Statename</option>";
which is a weird solution, or just use a join in your mysql query to get the statename
JOIN ON cool.ID = tab.stat
Extention:
When you use select, the innerHTML of option is displayed, but the value is sent in POST. So if you write
<select name="state">
<option value="1">name</option>
</select>
You will see name name, but the 1 will be sent.
On the next page, $State = $_POST['state']; will have $State the value of '1'. In the SQL you put this value to your stat field in your database, with a generated ID. It means, your ID won't hold any data, but the state field will!
So when printing on the third page, you have to join the list of states by it's ID with the stored stateid in the 'state' field. Then you will be able to print the statenames.
"SELECT * FROM t2 LEFT JOIN t3 ON t2.?? = t3.?? LEFT JOIN t1 ON t1.ID = t3.state"
Or something similar. Do you have a field you can use to join the t2 and t3 tables? It seems you are losing data when inserting to your database.
I want to insert the value of a selected 'select form' into my mysql database
this 'select from' is connected to database too
<label>Kategori</label>
<select class="form-control">
<?php include "config.php";
$kat = mysql_query("select * from kategori order by kode_kategori");
while ($hasil = mysql_fetch_array($kat)){
echo "<option value = '$hasil[kode_kategori]'>$hasil[nama_kategori]</option>";
}
?>
</select>
Image
Firstly, learn SQL. Learn the basics
To insert something into something, here is the syntax
$sql = "INSERT INTO mytable ('username', 'password') VALUES ('$_POST[username]', '$_POST[password]');";
You can insert value in database like below code:
<?php
if(isset($_POST['submit_button_name'])){
$insert_value = mysql_query("INSERT INTO mytable ('Column name1', 'Colum name2' , 'Column name3') VALUES ('$_POST['file_name1']', '$_POST['filed_name2']','$_POST['filed_name3']'");
if($insert_value){
echo "value insert successfully..."
}
else{
echo "error while insert data.."
}
}
?>
In our system they asked us to add research interest but when we add new user, we should assign an interest to him/her. It's adding successfully to the database table, we have problem which is we get zero record in the interest_id database table, why?
Here all research interest added successfully to database:
As you see in picture below when admin add a new user he chooses multiple interests and assign them to a user:
Here is screenshot of user profile as you can see interests that admin assigned to new user:
The problem is that it don't write interest id to database?
Below you can see the codes and SQL table:
<p>
<label>Research Areas</label>
<select name="interest_id[]" class="small-input" multiple>
<?php
$query = "SELECT * FROM research_interest ORDER BY name ASC";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_object($result))
{
?>
<option value="<?php echo $row->id; ?>"><?php echo $row->name; ?></option>
<?php
}
?>
</select>
</p>
Here is also codes for MySQL:
if(isset($_POST['action'])) {
switch($_POST['action']) {
case 'user_add':
$user_id = mysqli_real_escape_string($link,$_POST['user_id']);
$email = mysqli_real_escape_string($link,$_POST['email']);
$password = rand_pass(8);
$title_id = intval($_POST['title_id']);
$user_type_id = intval($_POST['usertype_id']);
$department_id = intval($_POST['department_id']);
$view_publication = mysqli_real_escape_string($link,$_POST['publication']);
$referee = mysqli_real_escape_string($link,$_POST['referee']);
$gender = mysqli_real_escape_string($link,$_POST['gender']);
$creation_date = date('Y.m.d');
//Interest
if(!empty($_POST['interest_id']) && is_array($_POST['interest_id'])) {
foreach ($_POST['interest_id'] as $interest)
{
$query = "INSERT INTO user_interest (user_id, interest_id)
VALUES ('$user_id', '$interest')";
$result = mysqli_query($link,$query);
}
}
//SQL Query
$query = "INSERT INTO user (user_id, email, title, user_type_id, department_id, referee, gender, password, view_publication, creation_date)
VALUES ('$user_id', '$email', '$title_id', '$user_type_id', '$department_id', '$referee', '$gender', '$password', '$view_publication', '$creation_date')";
$result = mysqli_query($link,$query);
echo '<script type="text/javascript">';
echo 'window.location = "mail.php?action=user_mail&user_id='.$user_id.'"';
echo '</script>';
break;
}
}
And this is a user_interest table shows that these three research IDs assign to user 1
but here it show that interest_id is zero , it don't shows ids of research interest that added to user 1
You are not including interest_id in your Insert query to the users table. So the table is assigning a default value of 0.
You're not inserting anything into the "interest_id" column in the second query so it's defaulting to zero. It looks like you're trying to set up a many to many relationship (one user can have many interests and one interest can belong to many users). In that case you'll need to set up a junction table (http://en.wikipedia.org/wiki/Junction_table) to handle the relationship.
Am facing troubles in this code, i just want to get all data from table row if the user selected "show all" from the select drop menu.
here is the select menu !
so, this menu grab data from this table, but if he selects All, what is the suitable code to echoing in between option value :)
<b>speciality:</b> <select id="main_mav" name="speciality">
<option value="none">Select speciality:</option>
<option value=""> All specialities </option>
<?php
$result = mysql_query('SELECT speciality FROM visits') or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row['speciality'].'">'.$row['speciality'].'</option>';
}
?>
</select><br />
That's the Submit form !
if ($region=="All regions" ){
$region=$_POST['""'];
}
else ( $region=$_POST['region']);
$date1 =$_POST['from_date'];
$date2 = $_POST['to_date'];
$product=$_POST['product'];
$speciality=$_POST['speciality'];
$type=$_POST['visit_type'];
sql="SELECT id, customer_name, seller_1_name, seller_2_name FROM visits Where (speciality ='$speciality') AND (visit_type ='$type') AND (product ='$product') AND (region ='$region') AND (visit_date BETWEEN '$date1' AND '$date2')";
$result=mysql_query($sql); ## This line is new.
$num=mysql_numrows($result);
$row = mysql_fetch_array($result);
What's the correct code to enter if user selected " show all in drop menu " ?!
You really need to sanitize your inputs, at least with mysql_real_escape_string!
On to your actual question: just check if $speciality is empty, and generate a different query without the (speciality ='$speciality') condition.
Since your HTML referenced 'specialties' and your PHP referenced 'regions' I'm gonna just stick with 'regions', but here's the idea.
if ($region=="All regions" ){
$sql = 'SELECT id, customer_name, seller_1_name, seller_2_name, FROM visits';
} else {
$region = mysql_real_escape_string($_POST['region']);
$date1 = mysql_real_escape_string($_POST['from_date']);
$date2 = mysql_real_escape_string($_POST['to_date']);
$product = mysql_real_escape_string($_POST['product']);
$speciality = mysql_real_escape_string($_POST['speciality']);
$type = mysql_real_escape_string($_POST['visit_type']);
$sql = "SELECT id, customer_name, seller_1_name, seller_2_name FROM visits Where (speciality ='$speciality') AND (visit_type ='$type') AND (product ='$product') AND (region ='$region') AND (visit_date BETWEEN '$date1' AND '$date2')";
}
$result = mysql_query($sql); ## This line is new.
$num = mysql_numrows($result);
$row = mysql_fetch_array($result);