Use dropdown list selection as mySQL query parameter - php

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.

Related

MySQLi_query Only Running Once/Consuming Connection Variable

I'm writing an application in PHP and am currently designing a form which includes two drop-down style boxes. These boxes need to be populated with data from a database I've got access to. The database contains two stored procedures, each returning the data to populate one of the drop-down boxes.
In my PHP, I prepare the queries I need to run like:
$viewLevelQuery = "CALL pr_getViewLevels();";
$publishLevelQuery = "CALL pr_getPublishLevels();";
And, in the lines immediately following, execute them like:
$con=mysqli_connect($host,$user,$pass,$dbnm);
$viewLevelResult = mysqli_query($con,$viewLevelQuery);
$publishLevelResult = mysqli_query($con,$publishLevelQuery);
mysqli_close($con);
Later in the PHP, I render and populate the drop-down boxes like:
echo "<div class='col-sm-10'>
<select class='form-control' id='viewLvl' name='viewLvl'>";
while($row=mysqli_fetch_array($viewLevelResult))
{
echo "<option>".$row['viewName']."</option>";
}
echo "</select> </div>";
And:
echo "<div class='col-sm-10'>
<select class='form-control' id='pubLvl' name='pubLvl'>";
while($row=mysqli_fetch_array($publishLevelResult))
{
echo "<option>".$row['publishName']."</option>";
}
echo "</select> </div>";
My issue is that only one of the boxes is being populated and it seems as though only the first of the MySQLi_query calls is being performed. No error is returned and the page continues to render as it should.
If the MySQLi_query lines at the beginning are switched, the other drop-down box is populated as expected which leads me to believe the queries to the database are correct.
Does the call to MySQLi_query "consume" the $con variable? If I execute the queries from PHP like:
$con=mysqli_connect($host,$user,$pass,$dbnm);
$viewLevelResult = mysqli_query($con,$viewLevelQuery);
$con=mysqli_connect($host,$user,$pass,$dbnm);//NOTICE Extra call to connect
$publishLevelResult = mysqli_query($con,$publishLevelQuery);
mysqli_close($con);
Both boxes get populated correctly. Doing this, however, doesn't feel right. Am I missing something? Or is this the way to go?.
Many thanks
EDIT: In case anyone with similar issues stumbles across this, as discovered below the error I was getting was because of mysqli's use of unbuffered queries. This means that it can't hold the results of two queries at once and so this memory needs to be freed once you have stored/used the results in your PHP. I simply copied the resultset to a PHP array and then included a:
while(mysqli_next_result($con))
{
mysqli_free_result($con);//Clears results
}
after each query. My understanding of what is going on is a little flakey, but this seems to make sense. It works, anyhow.

retrieving data from database with php and send to another page

I am now trying to create a page which shows restaurants in a list for people to comment and rate, the restaurant info is in the database. I've already get the data I want using while() function, but I am struggling to pick one of them and pass to another page. Below is my code. I tried to use sessions to store the data, like the "$_SESSION['rid']" for storing the restaurant ids. I have 8 rows in the restaurant table, and when I click on the first restaurant, the number shows on the other page is 8.
<?php
$sql_restaurant = "select * from tbl_restaurants";
$results = mysql_query($sql_restaurant);
while($restaurant_row=mysql_fetch_assoc($results)){
$_SESSION['rid'] = $restaurant_row['restaurant_id'];
echo "<a><span style = 'color:black'>".$restaurant_row['restaurant_name']."</span></a>";
echo "<dd>";
echo "<a><span style = 'color:black'>".$restaurant_row['restaurant_address']."</span></a>";echo '</dd>';
echo '<i class="mod_subcate_line"></i>';
echo 'Rate it!';
echo 'Comment!';
echo '<br/>';
echo '<br/>';
}
?>
I want it to show the right restaurant id when I click on different restaurants. How can I solve this?
Right now your code is changing $_SESSION['rid'] every time you cycle through the loop. So it will have the last cycle of the loop.
You shouldn't do this with sessions, always keep in mind that HTTP is a stateless system and sessions are stateful. It's always better to not rely on state.
As mentioned in the comments, you should write the restaurant ID as part of the URL of the link in a query parameter that you can then access through the $_GET array.
If you want to use $_SESSION, choose a different name for every loop. Here, $_SESSION['rid'] get the value of the last loop.

How to put MySQL table into session variable and using the table on next page?

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.

PHP Script for Datepicker to select and display database from Mysql

After spending 3 days on internet and struggling with so many different forums , i have found a match and similar case of my problem here.
Friends, I am zero in PHP, but still i have managed to do something to fulfill my requirement.
I am stuck with one thing now..So i need help on....
I am using one html+php form to submit database into mysql.
I created a display of that table through php script on a webpage.
Now i want a datepicker option on that displayed page by which i should able to select the date range and display the data of that date range from my mysql table.
And then take a export of data displayed of selected date range in excel.
This displayed page is login protected, so i want after login the next thing comes in should show a sate selection option which should be fromdate to to date , and then records should displayed from the database and i can take export of those displayed results in excel file.
The code i am using on this page is below which do not have any thing included for excel export and date picker script, I am pasting the code here and request you to please include the required code in it as required.
Thanks In advance
<?php
//database connections
$db_host = 'localhost';
$db_user = '***********';
$db_pwd = '*************';
$database = 'qserves1_uksurvey';
$table = 'forms';
$file = 'export';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
// sending query
$result = mysql_query("SELECT * FROM {$table} ORDER BY date desc");
if (!$result) {
die("Query to show fields from table failed");
}
$num_rows = mysql_num_rows($result);
$fields_num = mysql_num_fields($result);
echo "$num_rows";
echo "<h1></h1>";
echo "<table border='1'><tr>";
// printing table headers
for($i=0; $i<$fields_num; $i++)
{
$field = mysql_fetch_field($result);
echo "<td>{$field->name}</td>";
}
echo "</tr>\n";
// printing table rows
while($row = mysql_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>\n";
}
mysql_free_result($result);
?>
</body></html>
This isn't a "write my code for me, please" site, so you're going to need to be a little more engaging and pro-acive. But we can certainly provide some guidance. Let's see...
Currently you have a page which displays all records from a given table, is that correct? And you need to do two things:
Before displaying any records, have the user select a date range. And keep the date range selection on the page so the user can re-select.
Provide a button which lets the user export the selected records to Excel.
For either of these, you're going to need to add an actual form to the page. Currently there isn't one. For the date picker, I recommend (naturally) using the jQuery UI datepicker. So the form for that would look something like this:
<form method="POST" action="myPHPFile.php">
<input type="text" id="fromDate" name="fromDate" />
<input type="text" id="toDate" name="toDate" />
<input type="submit" name="filterDate" value="Submit" />
</form>
<script>
$(function() {
$("#fromDate").datepicker();
$("#toDate").datepicker();
});
</script>
You may have to wrap the JavaScript in a $(document).ready(){} in order to make it work correctly, you'll want to test that. Anyway, this will give you a form to submit the dates to your script. Wrap the parts of your script which output data in a conditional which determines if the form values are present or not. If they're not, don't fetch any records. If they are, do some basic input checking (make sure the values are valid values, make sure fromDate is before toDate, etc.) and construct your SQL query to filter by date range. (Do take care to avoid SQL injection vulnerabilities here.)
For the Excel output, you may be able to find a ready-made solution for you that just needs a little tinkering. If I were to create one from scratch, I'd probably just output to a .csv file rather than a full Excel file. Most users don't know/care the difference. In that case, you'd just want to either create a second script which is nearly identical to the existing one or add a flag to the existing one which switches between HTML and CSV output, such as via a hidden form field.
For the output of the CSV, first make sure you set your response headers. You'll want to write a header to tell the browser that you're outputting a CSV file rather than text/html, and possibly suggest a file name for the browser to save. Then, the form inputs the SQL query will all be pretty much the same as before. The only difference is in the "HTML" that's being output. Rather than HTML tags, you'd wrap the records in commas, double-quotes (where appropriate), and carriage returns.
There's really nothing special to outputting a "file" vs. "HTML" because the HTTP protocol has no distinction between the two. It's always just text with headers.
Now, I'm sure you have more questions regarding this. And that's fine. In fact, we like to encourage asking (and, of course, answering) questions here. So please feel free to ask for clarification either in comments on this answer (or other answers), or by editing and refining your original question, or by asking an entirely new question if you have a specific topic on which you need help. Ideally, a good question on Stack Overflow consists of sample code which you are trying to write, an explanation of what the code is supposed to be doing, a description of the actual resulting output of the code, and any helpful information relevant to the code. As it stands right now, your question provides code somewhat unrelated to what you're asking, and you're just requesting that we add some features to it outright for you.

Display database contents? PHP / MySQL

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.

Categories