So I have a chatroom type of database where the text that a user inserts gets stored into a databse as their username in one field and their message in the other. I want to have my page output the database info, so that people can see each others messages.
How do I do this?
Also, is it possible to make a for loop that checks to see if the database has been updated with a new message, therefore it reloads the page? (Then the page outputs the database info again to update everyones messages)
Please help.. i'm so confused.
Take a look at MySQL functions in PHP manual. You need to connect to the server/database and run a select query to get the data from tables.
As for the loop: you could use JavaScript setInterval function and combine that with AJAX call to periodically poll for new records.
Like the others have said, you will want to connect to your database and then query the table that you have the data in.
while($row = mysql_fetch_assoc($results))
{
echo $row['username'] . " said: " . $row['message'] . "<br />";
}
I use mysql_fetch_assoc() instead of mysql_fetch_array() since the arrays are associative arrays (not indexed by integers, but rather by names (associations))
As for displaying the update on the page dynamically, that involves AJAX. Basically what that means is that your page will call out to a background script to get the new records from the database. This would require a new field in your 'messages' table, something like 'msg_delivered' that you could set to '1' when it has been fetched.
You should check out this if you are interested in making an AJAX chat client: http://htmltimes.com/javascript-chat-client-in-jquery.php
To read anything from a mysql database you would use the mysql_connect() and the mysql_query() functions
eg:
$link = mysql_connect('localhost', 'root', '');
$results = mysql_query('select * from messages');
while($row = mysql_fetch_array($results))
{
echo $row['username'] . ': ' . $row['message'].'<br />';
}
To display new messages the best way would be to use AJAX and poll the database from there, either loading a separate page into a DIV or getting XML back and placing into HTML tags. I would recommend using JQuery for these kinds of tasks. Check http://www.sitepoint.com/article/ajax-jquery/ for an example.
Related
I have two PHP pages. On page1 a temporary table is created and filled with data from a mysql database. I am trying to store this table into a $_SESSION variable so that I can put the table onto page2.
Right now this has been my approach:
This is (part) of the code on page1:
ob_start();
session_start();
//Select data from temporary table
$result = mysqli_query($mysqli,"SELECT * FROM table");
//store table into session variable
$_SESSION['fase1result'] = $result;
This is the code on page2:
ob_start();
session_start();
$table = $_SESSION['fase1result'];
echo "<table border='1'>
<tr>
<th>ProductID</th>
<th>ProductName</th>
<th>Fase1</th>
</tr>";
while($row = mysqli_fetch_array($table))
{
echo "<tr>";
echo "<td>" . $row['ProductID'] . "</td>";
echo "<td>" . $row['ProductName'] . "</td>";
echo "<td>" . $row['Fase1'] . "</td>";
echo "</tr>";
}
echo "</table>";
Unfortunately, up until now these scripts return me an error on page2. At this moment, the echoing of the table on page2 is just to test and verify that the table is actually passed on. At a later moment I want to be able to use MySQL queries to further add data to the table. Hope you could help me.
UPDATE:
Error that I'm getting is:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in domain/page2.php on line 32
With line 32 in page2 being:
while($row = mysqli_fetch_array($table))
To better explain my question, I have posted another question which can be found here:
Modifying MySQL table on different pages with scores from a HTML form
On page1 a temporary table is created and filled with data from a mysql database. I am trying to store this table into a $_SESSION variable so that I can put the table onto page2.
That's impossible.
And shouldn't be used anyway.
something wrong with your design. Most likely such a table is superfluous and you don't actually need it at all.
As of the real problem behind this one - better ask another question, explaining the real life task for which you decided to use a temporary table passed between pages.
Responding to your question one by one:
Error you are Getting
The error that you are getting normally is the result of incorrect spelling or reference of table name, field name or any other variable in the MySQL query. In your case, it may be due to incorrect calling/storing your Session Variable. For example,
//Instead of "table", you typed "tabel". This is just an example.
$result = mysqli_query($mysqli,"SELECT * FROM table");
Share your code so that I can try picking up this error. Above is just an example.
Storing values in Session Variable is not Recommended
Suppose your user fills in the form and moves on to the next phase. The data from the first phase is transferred to the second phase via Session Variable. What if the user simply closes the tab and restarts the process? Session Variable will still be set and the previous data may interfere with the new one and can produce unexpected results.
Ideal Solution
It is better to store the values in JavaScript Array and then transfer to the next page by Hidden Input field. Some of the benefits of using this logic are:
Fast Performance
More Secure
Easily Manageable
Reference Code
If you are taking the values from HTML Forms, then it is very simple to have the value in POST. Using the JQuery UI selection, you can add the selected values in a JavaScript Array.
//Declare Global JavaScript Variable on Page Load. This will be at the end of <head>
$(document).ready(function() {
window.fase1result = [];
} );
After this, on each click event where you want to add the data to be taken to the next page, use the following code to add the value to this array.
fase1result.splice(indexOf_to_add, 1, "SelectedValue");
To understand .splice better, click here.
One selection, e.g. clicking on a Div or link, add the value to a fase1result and on submit add the array value to Input Hidden by using the following:
Add a Javascript Function on form's onsubmit.
<form id="myForm" method="POST" action="fase2.php" onsubmit="return fase1Values()">
Add <input type="hideen" name="fase1values_input" id="fase1values_id"> in the form.
Below is the JavaScript onsubmit function just before </body>.
function fase1Values() {
$( '#fase1values_id' ).val( JSON.stringify(fase1result) );
}
Note that JSON.stringify is required in order to set the Array as an input value.
$decode_fase1result = json_decode( $_POST['fase1values_input'] );
Now you have transferred the fase 1 selection data using an Array from Page 1 to Page 2 without storing data in any temporary table.
Hope this answers your question and solves your problem as well.
I am trying to send a specific cell to a PHP page and deleting that row which contains this specific cell. But it appears I am doing something wrong.
This is how I sending the information:
echo "<td><a href='delete.php?id= '$row['pid']''>Delete</a>< /td>";
and using it like this:
$del = "DELETE FROM sca WHERE pid = $_GET['id']";
My database connections are working well but I couldn't manage to send the 'pid' integer to a PHP page.
Thanks
Probably you misstiped some character. Try with this :
echo "<td><a href='delete.php?id=".$row['pid']."'>Delete</a></td>";
echo '<td>Delete</td>';
echo "<td><a href='delete.php?id={$row['pid']}'>Delete</a></td>";
Try to keep this code mixture (html and php) readable or improve it for future changes and revisions.
Make sure to receive correct value in $row['pid'].
Also as advice you should protect your data and clean it before any db interaction.
I had asked a similar question a few days ago but think I was trying to do to much at one time. I am hoping someone can help get me started on this.
I have two drop down lists, one will be populated with years (2012, 2011 etc) and I have some mySQL databases called "db_2012", "db_2011" etc. In these databases are tables representing months.
I would like the user to select a year and then use that selection to query the correct db and return a list of table names which will be used to populate the second drop down list. Then click a button "See Results" to query the selected table and show the results.
I am putting this on a wordpress website and am using a php template file that I created. This is still new to me and what I have so far doesnt work like I want it too, it is basically setup now that you select a year and select a month (not populated from db) and click a button. It makes the query and my table is displayed, but I need this solution to be more dynamic and work as described above. Thanks for the help.
echo '<form action="" method="post">';
echo'<select name="years" id="years">';
foreach($yearList as $year){
echo'<option value="'.$year.'">'.$year.'</option>';
}
echo'</select><br />';
echo '<select name="monthList" id="months">';
foreach($monthList as $month) {
echo'<option value="'.$month.'">'.$month.'</option>';
}
echo '</select>';
echo '<input type=\'submit\' value=\'See Results\'>';
echo '</form'>
$yearList and $monthList are just pre populated arrays. So now from here I want to click the See Results button and query my sql database using the parameters from the drop down selections.
$database = $_POST['yearList'];
$month = $_POST['monthList'];
$wpdbtest_otherdb = new wpdb('Username', 'Password', $database, 'localhost');
$qStr = "SELECT * FROM $month";
$myResults = $wpdbtest_otherdb->get_results($qStr, OBJECT);
It sounds like you want to send an AJAX call to a separate php page for security and processing, then have the PHP return XML that you parse back into the second selection box via the AJAX callback. It can be a little messy, but it allows you to check for weird form values that users might inject.
Edit: The PHP will receive your AJAX parameters as parts of the $_GET or the $_POST array. From there, you can do your checks and db call (or not), then add header("Content-Type:text/xml"); so the server sends it back with the correct header. After that you'll need to echo the XML-formatted data you want the JavaScript to receive. Just remember not to echo anything other than the XML if the request is supposed to go through.
I am writing a simple user/login system in Php with postgresql.
I have a function that confirms whether username/passwords exists, which gets activated when a user presses the Login button.
public function confirmUserPass($username, $password){
$username=pg_escape_string($username);
/* Verify that user is in database */
$q = "SELECT password FROM users WHERE email = '$username'";
$result = pg_query($this->link,$q);
/* Do more operations */
}
I want to print the query stored in $results such that I can see it on the browser. When I do it in phppgAdmin using SQL it shows me the output but I cannot see it on the browser. I tried echo and printf but I could not see anything on the browser. I also tried to see view source from the browser but it shows nothing.
Can somebody help me with that?
Regards
From your code: $result = pg_query($this->link,$q);
As you've found already, trying to display the contents of $result from the line above will not give you anything useful. This is because it doesn't contain the data returned by the query; it simply contains a "resource handle".
In order to get the actual data, you have to call a second function after pg_query(). The function you need is pg_fetch_array().
pg_fetch_array() takes the resource handle that you're given in $result, and asks it for its the next set of data.
A SQL query can return multiple results, and so it is typical to put pg_fetch_array() into a loop and keep calling it until it returns false instead of a data array. However, in a case like yours where you are certain that it will return only one result, it is okay to simply call it once immediately after pg_query() without using a loop.
Your code could look like this:
$result = pg_query($this->link,$q);
$data = pg_fetch_array($result, NULL, PGSQL_ASSOC);
Once you have $data, then you've got the actual data from the DB.
In order to view the individual fields in $data, you need to look at its array elements. It should have an array element named for each field in the query. In your case, your query only contains one field, so it would be called $data['password']. If you have more fields in the query, you can access them in a similar way.
So your next line of code might be something like this:
echo "Password from DB was: ".$data['password'];
If you want to see the raw data, you can display it to the browser using the print_r() or var_dump() functions. These functions are really useful for testing and debugging. (hint: Wrap these calls in <pre> tags in order for them to show up nicely in the browser)
Hope that helps.
[EDIT: an after-thought]
By the way, slightly off-topic, but I would like to point out that your code indicates that your system may not be completely secure (even though you are correctly escaping the query arguments).
A truly secure system would never fetch the password from the database. Once a password has been stored, it should only be used in the WHERE clause when logging in, not fetched in the query.
A typical query would look like this:
SELECT count(*) n FROM users WHERE email = '$username' AND password = '$hashedpass'
In this case, the password would be stored in the DB as a hashed value rather than plain text, and the WHERE clause would compare that against a hashed version of the password that has been entered by the user.
The idea is that this allows us to avoid having passwords accessible as plain text anywhere in the system, which reduces the risk of hacking, even if someone does manage to get access to the database.
It's not foolproof of course, and it's certainly not the whole story when it comes to this kind of security, but it would definitely be better than the way you seem to have it now.
You must connect to database , execute query, and then fetch results.
try this example from php.net
<?php
public function confirmUserPass($username, $password){
$username=pg_escape_string($username);
// Connecting, selecting database
$dbconn = pg_connect("host=localhost dbname=publishing user=www password=foo")
or die('Could not connect: ' . pg_last_error());
// Performing SQL query
$query = "SELECT password FROM users WHERE email = '$username'";
$result = pg_query($query) or die('Query failed: ' . pg_last_error());
// Printing results in HTML
echo "<table>\n";
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
// Free resultset
pg_free_result($result);
// Closing connection
pg_close($dbconn);
?>
}
?>
http://php.net/manual/en/book.pgsql.php
After years of false starts, I'm finally diving head first into learning to code PHP. After about 10 failed previous attempts to learn, it's getting exciting and finally going fairly well.
The project I'm using to learn with is for work. I'm trying to import 100+ fixed width text files into a MySql database.
So far so good
I'm getting comfortable with sql, and I'm learning some php tricks, but I'm not sure how to tie all the pieces together. The basic structure for what I want to do goes something like the following:
Name the text file I want to import
Do a LOAD DATA INFILE to import the data into one field it to a temporary db
Use substring() to separate the fixed width file into real columns
Remove lines I don't want (file identifiers, subtotals, etc....)
Add the files in the temp db, to the main db
Drop the temp db and start again
As you can see in the attached code, thigns are working fine. It gets the new file, imports it to the temp table, removes unwanted lines and then moves the content to final main database. Perfect.
Questions three
My two questions are:
Am I doing this 'properly'? When I want to run a pile of queries one after anohter, do I keep assinging mysql_query to random variables?
How would I go about automating the script to loop through every file there and import them? Rather than have to change the file name and run the script every time.
And, last, what PHP function would I use to 'select' the file(s) I want to import? You know, like attaching a file to an email -> Browse for file, upload it, and then run the script on it?
Sorry for this being an ultra-beginner question, but I'm having trouble seeing how all the pieces fit together. Specifcally I'm wondering how multiple sql queries get strung together to form a script? The way I've done it below? Some other way?
Thanks x 100 for any insights!
Terry
<?php
// 1. Create db connection
$connection = mysql_connect("localhost","root","root") or die("DB connection failed:" . mysql_error());
// 2. Select the database
$db_select = mysql_select_db("pd",$connection) or die("Couldn't select the database:" . mysql_error());
?>
<?php
// 3. Perform db query
// Drop table import if it already exists
$q="DROP table IF EXISTS import";
//4. Make new import table with just one field
if ($newtable = mysql_query("CREATE TABLE import (main VARCHAR(700));", $connection)) {
echo "Table import made successfully" . "<br>";
} else{
echo "Table import was not made" . "<br>";
}
//5. LOAD DATA INFILE
$load_data = mysql_query("LOAD DATA INFILE '/users/terrysutton/Desktop/importmeMay2010.txt' INTO table import;", $connection) or die("Load data failed" . mysql_error());
//6. Cleanup unwanted lines
if ($cleanup = mysql_query("DELETE FROM import WHERE main LIKE '%GRAND%' OR main LIKE '%Subt%' OR main LIKE '%Subt%' OR main LIKE '%USER%' OR main LIKE '%DATE%' OR main LIKE '%FOR:%' OR main LIKE '%LOCATION%' OR main LIKE '%---%' OR `main` = '' OR `main` = '';")){
echo "Table import successfully cleaned up";
} else{
echo "Table import was not successfully cleaned up" . "<br>";
}
// 7. Next, make a table called "temp" to store the data before it gets imported to denominators
$temptable = mysql_query("CREATE TABLE temp
SELECT
SUBSTR(main,1,10) AS 'Unit',
SUBSTR(main,12,18) AS 'Description',
SUBSTR(main,31,5) AS 'BD Days',
SUBSTR(main,39,4) AS 'ADM',
SUBSTR(main,45,4) AS 'DIS',
SUBSTR(main,51,4) AS 'EXP',
SUBSTR(main,56,5) AS 'PD',
SUBSTR(main,100,5) AS 'YTDADM',
SUBSTR(main,106,5) AS 'YTDDIS',
SUBSTR(main,113,4) AS 'YTDEXP',
SUBSTR(main,118,5) AS 'YTDPD'
FROM import;");
// 8. Add a column for the date
$datecolumn = mysql_query("ALTER TABLE temp ADD Date VARCHAR(20) AFTER Unit;");
$date = mysql_query("UPDATE temp SET Date='APR 2010';");
// 8. Move data from the temp table to its final home in the main database
// Append data in temp table to denominator table
$append = mysql_query("INSERT INTO denominators SELECT * FROM temp;");
// 9. Drop import and temp tables to start from scratch.
$droptables = mysql_query("DROP TABLE import, temp;");
// 10. Next, rename the text file to be imported and do the whole thing over again.
?>
<?php
// 5. Close connection
mysql_close($connection);
?>
If you have access to the command like, you can do all your data loading right from the mysql command line. Further, you can automate the process by writing a shell script. Just because you can do something in PHP doesn't mean you should.
For instance, you can just install PHPMyAdmin, create your tables on the fly, then use mysqldump to dump your database definitions to a file. like so
mysqldump -u myusername -pmypassword mydatabase > mydatabase.backup.sql
later, you can then just reload the whole database
mysql -u myusername -pmypassword < mydatabase.backup.sql
It's cool that you are learning to do things in PHP, but focus on doing the stuff you will do in PHP regularly rather than doing RDBMS stuff in PHP which is not where you should do it most of the time anyway. Build forms, and process the data. Learn how to build objects, and why you might want to do that. Head over and check out Symphony and Doctrine. Learn about the Front Controller pattern.
Also, look into PDO. It is very "bad form" to use the direct mysql_query() functions anymore.
Finally, PHP is great for templating and including disparate parts to form a cohesive whole. Practice making a left and top navigation html file. Figure out how you can include that one file on all your pages so that your same navigation shows up everywhere.
Then figure out how to look at variables like the page name and highlight the navigation tab you are on. Those are the things PHP is well suited for.
Why don't you load the files and process them in PHP, and use it to insert values in the actual table?
Ie:
$data = file_get_contents('somefile');
// process data here, say you dump it into a 2d array like
// $insert[$rows][$cols]
// then you can insert these into the db, ie:
$query = '';
foreach ($insert as $row) {
$query .= "INSERT INTO table VALUES ({$row[1]}, {$row[2]}, {$row[3]});";
}
mysql_query($query);
The purpose behind setting mysql_query to a variable is so that you can get the data you were querying for. In the case of any other query than SELECT, it only returns true or false.
So in the case where you are using if ($var = mysql...) you do not need the variable assingment there at all as the function returns true or false.
Also, I feel like doing all your substring and data file processing would be MUCH better suited in PHP. you can look into the fopen function and the related functions on the left side of that page.