I want to send some information from one php file to another.
I've read about the use of $_SESSION and $_POST, but they're giving me some problems.
My code looks something like this:
<form action="booking.php" method="post">
<select name="bookingflight">
<?php
$query = "SELECT Flight, Name
FROM Airport";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)) {
echo "<option value=\"". $row['Flight'] . ',' . $row['name'] . "\">" . $row['Flight'] . ' ' . $row['name'] . "</option>";
}
?>
</select></p>
<input type="submit" value="book flight"/>
</form>
This gives me a dropbox list of all the flights, along with the name of the flights.
If I select an element it's stored in $_POST["bookingflight"] which I can access in booking.php, which is fine.
However, it's given as a string, while I should be able to handle flight and name separately.
Ideally, I'd have two variables, one for flight and one for name, which I can access in booking.php.
How should I do this? With $_SESSION I don't even know how to assign a selected item from the list to a variable.
Alternate Solution: If Flight is unique data
while($row = mysql_fetch_array($result)) {
$flight=$row['Flight'];
$name=$row['Name'];
echo "<option value='$flight'>$flight $name</option>";
}
PHP : booking.php
<?php
$flight=$_POST['bookingflight'];
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT Flight,Name FROM Airport WHERE Flight='$flight'");
$row = mysqli_fetch_array($result);
$name= $row['Name'] ; // Here you get Name of selected Flight
mysqli_close($con);
?>
Or
If your Airport table contains any unique data/id , you can pass that data as option value too
You started this the wrong way...
Give your Airport table a unique primary key (named id, INT, auto-increment) and pass that as a value in generated options:
echo "<option value=\"". $row['id'] ."\">" . $row['Flight'] . ' ' . $row['name'] . "</option>";
Now when you POST your form you get that ID value in booking.php.
Because every flight has a unique ID you can just issue another query and you get your result as an array there:
$query = "SELECT Flight, Name FROM Airport WHERE id = $id";
Put this in your booking.php...
if($_POST)
{
$values = $_POST['bookingflight'];
$val= explode(',',$values);
$_SESSION['flightnumber'] = $val[0];
$_SESSION['flightname'] = $val[1];
}
Related
I have a page that I have been working on. It runs several queries to get existing data from several tables in my DB. There is a table that shows the result of three queries. The first query gets the extension and the secret of phones, the 2nd query gets MAC addresses of phones, and finally the third query gets the names of templates for the phones. The results of the last two queries (with the help of others) are setup as dropdowns in the 3rd and 4th columns of the table created to show the extensions. This way I can select the MAC of the phone I want to assign to the extension and then the template to make the phone work the way I want. The whole page is set as a form and I am using $post to the insert page. My goal here is to take the information (array) that is created by the user making their selections and insert ALL the 4 columns of information into a new table, from there I want to create files using that information to setup the phones. Here is the code I have for now.
<?php
error_reporting(E_ALL);
ini_set('display_errors','On');
$link = mysql_connect("localhost", "root", "cacti") or die ('Error connecting to mysql' . mysql_error());
mysql_select_db("cqadmin");
$sql2 = "SELECT extension, secret from extensions;";
$result2 = mysql_query($sql2) or die(mysql_error());
echo "<table border='3'>
<tr>
<th>Extension #</th>
<th>Secret</th>
<th>MAC Address</th>
<th>Template</th>
</tr>";
while($row = mysql_fetch_array($result2))
{
$sql = "SELECT id , mac FROM phones order by mac;";
$result = mysql_query($sql) or die(mysql_error());
$sql1 = "SELECT id , templatename FROM templates order by templatename;";
$result1 = mysql_query($sql1) or die(mysql_error());
echo "<tr>";
echo "<td>" . $row['extension'] . "</td>";
echo "<td>" . $row['secret'] . "</td>";
echo "<td> <select name='phone'>";
while($rowA = mysql_fetch_array($result)) {
echo '<option value="' . $rowA['id'] . '">' . $rowA['mac'] . '</option>';
}
echo "</select></td>";
echo "<td><select name='template'>";
while($rowB = mysql_fetch_array($result1)) {
echo '<option value="' . $rowB['id'] . '">' . $rowB['templatename'] . '</option>';
}
echo "</select></td>";
echo "</tr>";
}
echo "</table>";
?>
<input type="submit" value="Submit your selections">
</body>
</html>
And my insert page
<?php
echo "You got here";
//***********Get the Assignment information *************
$values = array_values($_POST);
print_r($values);
?>
The resulting print shows this
Array ( [0] => 324 [1] => 24 )
Looking at my db table 324 is the index id of the last phone scanned and in the template table 24 is the last template created, No info on the extension or the secret.
I think I am close but I do not know where to go from here.
PS. I know I need to use mysqli or pdo, not sure how to change over yet.
Hey i need to get information from a database and display it on my index.php.
<html>
<body>
<?php
$connection = mysql_connect('localhost', 'root', ''); //The Blank string is the password
mysql_select_db('users');
$query = "SELECT * FROM employee"; //You don't need a ; like you do in SQL
$result = mysql_query($query);
echo "<table>"; // start a table tag in the HTML
while($row = mysql_fetch_array($result)){ //Creates a loop to loop through results
echo "<tr><td>" . $row['name'] . "</td><td>" . $row['age'] . "</td></tr>"; //$row['index'] the index here is a field name
}
echo "</table>"; //Close the table in HTML
mysql_close(); //Make sure to close out the database connection
?>
</body>
</html>
however this will show the information for all of the users in the database.
I have a primary key in the database, which is the account number, could we use this to only display information about that account on the page only?
I will also need to save the details into variables on the php page so some guidance doing that would be greatfull.
Thanks Oliver
In order to retrieve data according to a specific user you must explicitly tell mysql. So, add WHERE id={the I'd u want the info for}.
For your second question. "How do u get the id when the user logs in".
The answer to this is you can (a) how a session initialized at login or (b) how another file or even a loop in the script you need the Id for to run a query for the users name and then you get the Id.
The draw backs of option b is that you are slowing this done and are doing alot of call which can crash your overall site.
The choice is your this is how you can get a particular I'd for a user.
<?php
include('session.php');
?>
<html>
<body>
<?php
$connection = mysql_connect('localhost', 'root', 'Oliver'); //The Blank string is the password
mysql_select_db('users');
$query = "SELECT * FROM username WHERE username='$login_session'"; //You don't need a ; like you do in SQL
$result = mysql_query($query);
echo "<table>"; // start a table tag in the HTML
while($row = mysql_fetch_array($result)){ //Creates a loop to loop through results
echo "<tr><td>" . $row['name'] . "</td><td>" . $row['username'] . "</td></tr>"; //$row['index'] the index here is a field name
echo "<tr><td>" . $row['sex'] . "</td><td>" . $row['phone'] . "</td></tr>"; //$row['index'] the index here is a field name
echo "<tr><td>" . $row['email'] . "</td><td>" . $row['password'] . "</td></tr>"; //$row['index'] the index here is a field name
}
echo "</table>"; //Close the table in HTML
mysql_close(); //Make sure to close out the database connection
?>
</body>
</html>
I got the WHERE username="$login_session"
and the login_session value is a unique value for each of the users in the database so it can never be the same, so i can use this to pull up information for that user only
You can use session variables to store the information in the case you are working with sessions, if not, you can use a regular variable to store the information or maybe use post or get variables
First off, I want to store the names of these checkboxes which are submitted, and not their values.
This is my code:
<?php
$con=mysqli_connect("localhost","root","","notifier");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM student");
echo "Enter the attendance. Please untick for 'ABSENT' students and submit";
echo "<br>";
echo "<form action=\"d.php\" method=\"post\">";
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
}
echo "<input type=\"submit\" name=\"submit\" value=\"submit\">";
echo "</form>";
?>
This code simply fetches a column of student rollnumberss from student table, prints them, and as well as prints a checkbox infront of them which is checked by default.
Names of checkboxes will be the student id (varchar, another column).
Now since All Checked checkboxes, that is the checboxes which will be submitted to next page will have same default value "P", I m not concerned about their values.
How do I store the names of these checkboxes in an array, and later on use it to perform updation in table for all these student id's?
Use the following code:
while($row = mysqli_fetch_array($result))
{
echo '<br>' .$row['classrollno'] . ' <input type="checkbox" name="studentId[]" value="' . $row['studentid'] . '" checked />';
}
Then, when you process the form, the $_POST['studentId'] variable will contain an array with all the id's.
Since the value that will probably be inserted in the db is 'P' for every student, you wouldn't need to include it in your form, but just hardcode it in your query.
Keep adding the names to an array. Its straight forward.
Declare $allStudentIds = array(); outside while loop. Then, to store in that array,
$allStudentIds[] = $row['studentid'];
Since you wanted to use these values later, you can directly store them inside a session variable:
$_SESSION['allStudentIds'][] = $row['studentid'];
In above case, $_SESSION['allStudentIds'] will be an array of all student ids selected.
Note: You need to start session using session_start() as the first line in the script after opening <?php tag.
Simply, in the fetching while loop, define an array and set each checkbox value to one of its elements then assign it as a session variable:
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
$names[] = $row['studentid'];
}
Then,
$_SESSION['names'] = $names;
Your confusion seems to stem from the fact that you are mixing the View (the name of the checkbox in HTML) and the Model/Data (which the student_id you are getting from your DB query ie. the $row = mysqli_fetch_array($result) in the while loop).
All you need to do is create an empty array (eg. $studentid_arr) before the loop and after the echo statement which is just contributing to the view (the HTML) you do some work with your data. What you want to do currently is to store the student_ids (and not the name of the checkbox) in your $studentid_arr.
That can be done with a simple array_push ($studentid_arr,$row['studentid']);
So your while loop would look like
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
array_push ($studentid_arr,$row['studentid']);
}
Now you can just POST this PHP array to your next script which is expecting these values. (which is what I assume you mean by submitting to the next page)
First of all, I am a newbie when it comes to coding, so please be kind and patient :)
What I am trying to do is to select two rows ('ID', 'name') from a MySQL table (categories), populate a drop down list with one row ('name'), and on submission of a form, pass the other ('ID') to another table.
Now, I can populate the drop down list, no problem. I have populated this with both 'ID' and 'name' to test that both of the variables I am using to hold this information, contain the correct data. But I cannot seem to $_POST the information.
I guess I am either looking at the wrong part of the array, or I am simply using the wrong code.
This is the code to create a new product, under a category from the database.
<?php
include 'db_config.php';
?>
<form enctype="multipart/form-data" action="insert.php" method="post">
<h3>Add New Product</h3>
Category:
<!-- START OF categories (table) names (row) SQL QUERY -->
<? $sql = "SELECT ID, name FROM categories";
$result = $mysqli->query($sql);
echo "<select name='category_name'>";
while ($row = $result->fetch_assoc()) {
$cat_ID=$row['ID'];
$cat_name=$row['name'];
extract($row);
echo "<option value='" . $cat_ID . $cat_name . "'>" . $cat_ID . " " . $cat_name ."</option>";
}
echo "</select>";
?>
<!--END OF SQL QUERY -->
<br>
Code: <input type="text" name="code"><br>
Name: <input type="text" name="prod_name"><br>
Description: <input type="textarea" name="description"><br>
Image: <input type="file" name="image"><br>
<input type="Submit">
</form>
For now, I am just echoing this out in the insert.php script, to test the code above. This is a snippet of the insert.php script.
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
echo "Code: ". $_POST['code'] . "<br>";
echo "Name: " . $_POST['prod_name'] . "<br>";
echo "Description: ". $_POST['description'] . "<br>";
echo "Image: " . $_POST['image'] . "<br>";
Don't worry about the last line above. I know this needs to be $_FILES, and I have all this covered. I have stopped writing the data to the table until I get my issue fixed. In the full script, image are being upload to "/images" and the location stored in the table. This all works fine.
The problem is with the first two lines, as they are blank when returned. I thought I was storing the information correctly, as I am calling the same variables to populate the drop down list, but I cannot seem to $_POST it.
Does that makes sense?
Thanks to all who help me. Once day I will be as good as you....I hope.
TIA
Smurf.
this bellow:
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
is wrong, select element has its name category_name, so, instead of this, you should
do:
echo "Category: " . $_POST['category_name'] . "<br>";
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
There aren't any form elements with those names in your form ($row["ID"] and $row["name"]). Those would be really strange names for a form element anyway. The form element you're creating is:
<select name='category_name'>
So the selected value would be posted as:
$_POST['category_name']
The option elements for that select appear to have values which are a combination of ID and Name:
"<option value='" . $cat_ID . $cat_name . "'>"
Thus, if the user selects an option with a value of 1SomeName then $_POST['category_name'] will evaluate to '1SomeName'.
It's certainly unconventional to use the combination of ID and Name for the option values, but it should work. The problem is presents is that you now have a composite string which needs to be parsed in order to be useful. Generally what one would do is just use the ID as the value and the Name as the display. All you should need to use it throughout the code is the ID.
The $_POST variable you want, is inside category_name
Cuz your select is...
<select name='category_name'>
So you need to get it by...
$_POST['category_name'];
Which will return whatever you've assigned to the select options...ie
2 Name
2 being the ID, and Name being the name
But if you then want to use that ID to retrieve from DB or anything, you're gonna have to explode that apart...like so....to get each piece.
$array = explode(" ", $_POST['category_name']);
That will leave you with...
$array[0] = ID
$array[1] = Name
But I would avoid all that part, by just assigning the ID to the value only...like so..
echo "<option value = '".$cat_ID."'> ".$cat_name." </option>";
That way you just pass the ID and have access to it on the other side.
I have a database that I want to get data out onto a website. It contains states listed by name and id. Counties listed by id, namne , and state that contains thems ID and then clubs that exist , with a reference to the county id's that they exist in and columns for their actual data.
What I've got :
A drop down menu that populates itself with state id and name.
What I'd like to accomplish:
On selection of state , let's say ny , take it's id and use this in gathering another mysql array for the county drop down. I'd like it to dynamically occur on selection of state , maybe even giving a count of results next to the drop down.
$resstate = mysql_query("SELECT * FROM state ORDER by longstate;") or die("Note: " . mysql_error());
State:
<select name="State" size=1>
<?
while( $rs = mysql_fetch_array( $resstate ) ) {
echo "<option value=" .$rs['id'] . ">" . $rs['longstate'] . "</option>";
}
echo "</select>";
?>
I know I could use a JavaScript onChange="this.form.submit()" on the first drop down, but it's my understanding that I'd then be making a new page at that point and don't know if I could keep the functionality of the state drop down, say if you accidentally chose new Hampshire when you wanted New York.
here's an example of the current array filling the drop down :
http://snowmobileamerica.com/countytest.php
----EDIT---
Using Dagons Advice , I looked into Ajax.
I made a php file that's supposed to query the database based on a reference to getcounty.php?q=
The file is created as follows :
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
mysql_select_db("snowusa_clubs", $cn);
$sql="SELECT * FROM county WHERE state_id = '".$q."' ORDER by name";
$result = mysql_query($sql);
echo "<select name="County" size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value=" .$rc['id'] . ">" . $rc['name'] . "</option>";
}
echo "</select>";
mysql_close($cn);
?>
If i try to run it manually http://www.snowmobileamerica.com/getcounty.php?q=33 I get a 500 internal server error...
Any ideas where I went wrong?
try adding an id to the element, then make an ajax call to a handler with jquery:
$("#State").change(function() {
$.post("path/to/request handler/" , { "State" : $(this).val() },
function(data){
if (data == "OK"){
//add some elements here
} else {
//handle an error here
}
});
});
not able to comment yet.
but for the second question try:
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
echo "Conn ok<br>";
mysql_select_db("snowusa_clubs", $cn);
echo " Database opened<br>";
$sql="SELECT * FROM county WHERE state_id = '$q' ORDER by name";
$result = mysql_query($sql);
echo " Database queried <br>";
echo "<select name='County' size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value='" .$rc['id'] . "'>" . $rc['name'] . "</option>";//added single quotes in the value
}
echo "</select> ";
mysql_close($cn);
?>