New to PHP here. I'm working on a basic PHP project for university. I have a page with a list of patients. When you click on the patient it will take you to the patient page with more details of that patient.
I have a main patientDetails.php page which will display the details.
However I am a little puzzled. How do I get the "?Name.." part of the link to work. So how do I get the patientDetails page to load the specific patient details?
I have an index.php page which has the list of patients as below.
<td>Stuart</td></tr><tr> <td>2</td>
<td>Fred</td></tr><tr>
In the PatientDetails page I have a select statement to gather details from the Database but not sure where else to go from here.
$query = sprintf("select * from PHPEnrolment WHERE PatID = '$PatID' AND NAME = '$Name' AND Email = '$Email'");
$result = mysql_query($query, $link);
if ($result) {
while($row = mysql_fetch_array($result)) {
}
As you can see above the database table with patients has the PatID, Name and Email field.
Thanks
PLEASE NOTE: This is a basic project that I am working on, so I am aware some of the features are outdated but I need to get it working with these features if possible.
You can get the url variables using $_GET. I see that you are passing only Name parameter in the url. Use the below code.
$name = $_GET['Name'];
$query = "select * from PHPEnrolment WHERE NAME = '$name'";
$result = mysql_query($query, $link);
if ($result) {
while($row = mysql_fetch_array($result)) {
// display details here
}
}
As per your code there are several errors. and for getting value you can use $_GET['Name']. I am assuming that you have all other variables($Email and $PatID)
$Name = $_GET['Name'];
Your query is not right. Change it from
select * from PHPEnrolment WHERE PatID = '$PatID' AND $NAME = 'Name' AND Email $Email
To
select * from PHPEnrolment WHERE PatID = '$PatID' AND NAME = '$Name' AND Email = '$Email'
So your whole code should look like.
$query = "select * from PHPEnrolment WHERE PatID = '$PatID' AND NAME = '$Name' AND Email = '$Email'";
$result = mysql_query($query, $link);
NOTE : STOP USING MYSQL. It is deprecated now.
Related
Users can add their workouts into a database. There will have more than one workout per user. How can i display the workout names for the logged in user, underneath each other? I have no problem adding the workouts. When I echo the names of the added workouts, it display right next to each other like "workout1workout2". I want to be able to display the workouts underneath each other. I have no idea how to do it. It would be best if I could display each workout as a button.
$query = "SELECT * FROM info WHERE username='$_SESSION[username]' ";
$results = mysqli_query($db, $query);
while($data = mysqli_fetch_array($results)) {
echo $data['workout'];
}
Firstly, you shouldn't put that username in query like that. Look into SQL Injection and how to avoid that in PHP.
As for displaying buttons/workouts. You can embed HTML code inside the echo statement like below. You can add any html like that and it will render HTML.
$query = "SELECT * FROM info WHERE username='$_SESSION[username]' ";
$results = mysqli_query($db, $query);
while($data = mysqli_fetch_array($results)) {
echo "<button>$data['workout']</button></br>";
}
$query = "SELECT * FROM info WHERE username='$_SESSION[username]' Group By username";
I want to select email address from DB to send a email. Following is my query that I have made.
$userID=$_SESSION['userID'];
$select_query = mysql_query("SELECT * FROM employee WHERE emp_id = '$userID'");
$select_sql = mysql_fetch_array($select_query);
$name=$select_sql['manager_name'];
$select_query1 = mysql_query("SELECT email FROM employee WHERE employee.name='$name'");
$select_sql1 = mysql_fetch_array($select_query1);
$email=$select_sql1['email'];
But $select_query1 return "NULL Invalid address:" instead of the correct value. I could not found the problem with this. Please help !
You are using $_SESSION['userID'] to get all data from table employee so instead of doing two queries simply try this
$empID = $_SESSION['userID'];
$query = mysql_query("SELECT * FROM employee WHERE emp_id=$empID");
$result = mysql_fetch_array($query);
$email = $result['email'];
<?php
session_start();
$username = "root";
$password = "password";
$database = "meipolytechnic";
mysql_connect('localhost', $username,$password);
#mysql_select_db($database) or die(mysql_error());
$username=$_SESSION['MM_Username'];
$query = "SELECT rollno FROM users where username = '".$username."'";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_num_rows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
echo ($rows['rollno']);
?>
i want to retrieve only the logged in users roll no from users table in database
when i run this code
and log in as foo
i get the following stuff
Unknown column 'foo' in 'where clause'
There should be session_start() at the top of the page
query need to change as
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."' ";
EDIT
Please try something before posting a question here. Please google or go through www.w3school.com for clearing this kind of issues. Make a good knowledge about arrays and mysql connection. And mysql_query function won't work latest PHP version.
Please try following code.
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
To use an array inside a string you need to put a curly bracket before it and after it
so
$query = "SELECT rollno FROM users where username = {$_SESSION['MM_Username']}";
or
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];
First of all start session using start_session()
then change your query:
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];
then change:
$num = mysql_num_rows($result); instead of $num = mysql_numrows($result);
Try this query
$query = "SELECT rollno FROM users where username = ".$_SESSION[MM_Username]." ";
And start session on same page.
You can use
$username=$_SESSION[MM_Username];
$query = "SELECT rollno FROM users where username = '".$username."'";
and start the session by using session_start()
and you have used two closing tags omit one make it like below
echo ($rows['rollno']);
?>
This type of error occur when query goes false
May be becouse You have not start session, becouse if you dont have session_start(), then nothing will come in session variable... just try as
$query = "SELECT rollno FROM users where username = '".$_SESSION[MM_Username]."'";
may this help you
You must need to use session_start() before using $_SESSION variable in code.
So put below code at start,
session_start();
Then do some modification in query like,
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
first start your session
session_start();
and Change your query like this...
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
I am trying to get four different values from my database. The session variable username and usernameto are working, but I want to get 4 different values -- two each from username and usernameto:
<?php
session_start(); // startsession
$Username=$_SESSION['UserID'];
$Usernameto= $_SESSION['UserTO'];
$db = mysql_connect("at-web2.xxxxxx", "yyyyy", "xxxxxxx");
mysql_select_db("db_xxxxxx",$db);
$result1 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
while($myrow1)
{
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
while($myrow2)
{
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
?>
Edit - just realized that you didn't tell us what wasn't working about the code you provided. Are you getting an error message or are you not getting the correct data back? You still should fix your query, but we'll need some more information to know what's wrong.
Your query statements shouldn't have "and" between the select parameters, so it should be:
Edit 2 - I just noticed that you had a while loop that you don't need, try this:
$result1 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
if (isset($myrow1)) {
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
if (isset($myrow2)) {
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
An example from the php manual echoing an html table
I don't know if you can derive what you need from this?
More specific: You can use:
$line = mysql_fetch_array($result, MYSQL_ASSOC);
I need to read a text file, query the database table with that name, and store that table's data in another table. So far I have written this code but I don't know why it's not working.
foreach ($lindb as $namedb) {
$query = "SELECT * FROM ntable WHERE name =" .$namedb. "";
$result = mysql_query($query);
while ($r = mysql_fetch_array($result)) {
$query = "INSERT INTO ndtable (name,details,address,login,country) VALUES (\"".$r["name"]."\", \"".$r["details"]."\", \"".$r["address"]."\", \"".$r["login"]."\", \"".$r["country"]."\")";
mysql_query($query);
}
}
You don't have quotes around $namedb
ie. SELECT * FROM ntable WHERE name =" .$namedb. ""; should be SELECT * FROM ntable WHERE name ='" .$namedb. "'";
I suggest a SELECT INTO would be the better choice... and please post the error so we are able to help...