PHP - Help Output Some Code During A Date Range For An Hour?

Hi i have the following code

Quote

$sql= "SELECT * FROM income WHERE month(date) between '04' and '12' and year(date) between 2020 and 2020";
 

This obviously selects all the information from April 2020 to December 2020

 

How do i change this to add a date as well for example if i wanted to display all the information from April 6th 2020 to December 5th 2020 ?

 

Setup:
Mysql table 1: acct#, client name, status(active/not active), Starting balance, ending balance
Mysql table 2: acct#, invoice amount, invoice date, total


how do i select all customers that are Active, and the starting balance != ending balance and then get the total for all invoices for that customer where the invoice date is between 2010-01-01 and 2010-03-01

Basically my out put needs to look like this.

Customer name, Total of all invoices for customer
john doe,  $10,564.21   
Susy Scott, $158.02
Bang Head, $837,294.85

Customer can have multiple invoices within the given date range. But I only need the total for all the invoices for each customer added up and echo'ed.

Quickness is also a factor on the query and echo.

Thank you for your valuable time.

 

I am working on a report generation. Here I need to count the number of months involved in the selected date range. I need to apply the monthly charges accordingly.

Example: If the user selects 25-08-2011 to 03-10-2011, it should return 3 as the number of months.

Yes, the number of days are less than 60 but still the date range is spread across 3 months.

Any solution.. Please.

Girish

I'm trying to wrap my head around how i should go about doing this. I have two dates...2011-05-03 and 2011-05-08. I want to write a function that creates an array of the days that consist of that date range. So say something like

Code: [Select]
<?php
$start = "2011-05-03";
$end = "2011-05-08";

function get_days($start, $end)
{
//code to get the days
}

echo get_days($start, $end);

//hopefully produce something like...
$day['1'] = "2011-05-03";
$day['2'] = "2011-05-04";
$day['3'] = "2011-05-05";
$day['4'] = "2011-05-06";
$day['5'] = "2011-05-07";
$day['6'] = "2011-05-08";

anybody know any idea how to do something like this? Pretty sure i'm going to need the mktime() function to account for ranges going across months and years. 

Hi,   I can I include a date range criteria to query with in the following code? The date field in the table (t_persons) is IncidentDate.

$criteria = array('FamilyName', 'FirstName', 'OtherNames', 'NRCNo', 'PassportNo', 'Gender', 'IncidenceCountryID', 'Status', 'OffenceKeyword', 'AgencyID', 'CountryID', 'IncidenceCountryID' );
    $likes = "";
    $url_criteria = '';
    foreach ( $criteria AS $criterion ) {
            if ( ! empty($_POST[$criterion]) ) {
                    $value = ($_POST[$criterion]);
                    $likes .= " AND `$criterion` LIKE '%$value%'";
                    $url_criteria .= '&'.$criterion.'='.htmlentities($_POST[$criterion]);
            } elseif ( ! empty($_GET[$criterion]) ) {
                    $value = mysql_real_escape_string($_GET[$criterion]);
                    $likes .= " AND `$criterion` LIKE '%$value%'";
                    $url_criteria .= '&'.$criterion.'='.htmlentities($_GET[$criterion]);
            } //var_dump($likes);
    }
    $sql = "SELECT * FROM t_persons WHERE PersonID>0" . $likes . " ORDER BY PersonID DESC";

Kind regards.  

I'm trying to do the following PHP. I have written it in English

if (today's date) => date1 AND <= date2 then {display image1}
elseif (today's date) => date3 AND <= date4 then {display image2}
else {display image 13}

There are 24 fixed dates and 13 fixed images.

I have tried using combinations of strtotime(), replacing the date value with variables, hard coding the dates within the program. I don't seem to be able to get any combination to work properly. I'm sure it's just a syntax error but I can't see it. When I've searched the web all the answers I've found relate to dates within databases but as I only have 24 dates it seems a bit of overkill. I would appreciate any pointers to the correct method I might be able to use. The dates need only a day and month as I would like this to repeat year after year.

<?php
$today = strtotime(date('d-m'));

if (strtotime($today) >= strtotime('28-10') && strtotime($today) <= strtotime('24-11')) {echo "<div>image1</div>";}
elseif (strtotime($today) >= strtotime('25-11') && strtotime($today) <= strtotime('22-12')) {echo "<div>image2</div>" ;}
elseif (strtotime($today) >= strtotime('23-12') && strtotime($today) <= strtotime('24-02')) {echo "<div>image3</div>" ;}
else {echo "<div>image13</div>";}
?>

The result I get from this is image1 appears on the web page but I would expect image3 as today is 22-01

Thank you in advance.

Andrew

Hi all This is a one I've spent DAYS on and can't figure out :-\ This is for a wedding venue and their pricing schedule is divided into low, medium, high and very high seasons. I.e. low season runs from 8 Jan 16 to 28 Feb 16, medium runs from 4 Mar 16 to 20 Mar 16, etc.

The database for this is laid out as attached.   I would like people to be able to enter their arrival and departure date and for it to be able to choose which seasons the date range falls under. I.e. If they stay 8 to 12 Jan, that's no problem to calculate and falls under LOW season. However, if they stay 27 Feb to 10 Mar, this falls under both LOW and MID seasons, with the majority of the stay in the MID season (which we'd price with but can use PHP to choose the 'highest' season from the array). I'm a bit stumped on this one so any help is hugely appreciated! I'm sure it's something quite simple that I'm missing. Attached Files  Screen Shot 2015-01-10 at 16.58.17.png   146.3KB   0 downloads

I have the week date range displaying the week range from sunday - saturday   $current_dayname = date("0"); // return sunday monday tuesday etc.
echo $date = date("Y-m-d",strtotime('last sunday')).'to'.date("Y-m-d",strtotime("next saturday"));   Which outputs in the following format 2015-01-25to2015-01-31   However, when I press next or previous, the dates don't change.   <?php
$year = (isset($_GET['year'])) ? $_GET['year'] : date("Y");
$week = (isset($_GET['week'])) ? $_GET['week'] : date('W');
if($week > 52) {
    $year++;
    $week = 1;
} elseif($week < 1) {
    $year--;
    $week = 52;
}
?>

<a href="<?php echo $_SERVER['PHP_SELF'].'?week='.($week == 52 ? 1 : 1 + $week).'&year='.($week == 52 ? 1 + $year : $year); ?>">Next  Week</a> <!--Next week-->
<a href="<?php echo $_SERVER['PHP_SELF'].'?week='.($week == 1 ? 52 : $week -1).'&year='.($week == 1 ? $year - 1 : $year); ?>">Pre Week</a> <!--Previous week-->

<table border="1px">
    <tr>
        <td>user</td>
<?php
if($week < 10) {
    $week = '0'. $week;
}
for($day= 1; $day <= 7; $day++) {
    $d = strtotime($year ."W". $week . $day);

    echo "<td>". date('l', $d) ."<br>". date('d M', $d) ."</td>";
}
?>
    </tr>
</table>

Hi. Im searching for a way to enter multiple records at once to mysql databased on a date range. Lets say i want to insert from date 2010-11-23 to date 2010-11-25 a textbox with some info and the outcome could look in database like below.

----------------------------------------------------------
| ID   |   date                |   name   |    amount   |
-----------------------------------------------------------
| 1     |   2010-11-23    |   Jhon     |        1         |
| 2     |   2010-11-24    |   Jhon     |        1         |
| 2     |   2010-11-25    |   Jhon     |        1         |
-----------------------------------------------------------

The reason i need the insertion to be like this is because if quering by month and by user to see vacation days then vacations that start in one month and end in another month can be summed by one month. If its in one record then it will pop up in both months.

Help is really welcome.

Hello,   I've been going about reading different posts on here and other forums but so far I haven't been able to come across something that works for my purpose so after all the reading I thought I'd finally just ask. I'm rather new at php/sql queries so please bare with me.   First what I'm trying to accomplish.   I need a form with several fields (options) to fetch information from a table that will then display the results based on the options selected.   - from date - to date - employee name - department - branch - product line   In other words, the purpose is to be able to choose a date range (from one day to up to a year or more) and then to be able to choose either ONE employee name to view statistics invididually from a given department and branch or to select a department to view all the statistics for everyone under that given department and/or brach and/or product line.   Now the tricky part is that besides fetching the records, calculations need to be made before the records are displayed and that part right here is what is giving me a huge headache.   This is the query that I have.

SELECT
report_daily_id,
report_date,
emp_id,
emp_fullname,
emp_dept,
emp_branch
prod_line
calls,
tk_time,
hld_time,
ac_time,
tran_calls,
work_time,
tran_rate,
ah_time
FROM daily_report
GROUP BY emp_id, emp_fullname, emp_branch, emp_dept, prod_line

The calculations based on the date range and one or more of the other options need to give me the following results.

SUM(calls) AS 'Total Calls'
SUM(tk_time) / SUM(calls) AS 'Talk Time'
SUM(hld_time) / SUM(calls) AS 'Held Time'
SUM(ac_time) / SUM(calls) AS 'AC Time'
SUM(tran_calls) AS 'Total Trans'
SUM(tran_calls) / SUM(calls) AS 'Tran Rate'
SUM(ah_time) / SUM(calls) AS 'AH Time'

I don't know how else to explain myself past this point but if you have any questions, perhaps I can answer it and give more details.   Thank you,

I spent the last hour or so typing this code up, and for some reason I am getting a query error. I have reviewed & revised the code up and down for the past half hour and can't seem to figure out the problem.

Can someone look after this for me and tell me what I could be doing wrong?

Yes, I know my code is a bit sloppy and may use bad practice techniques, but it works for me.

Its a survey that I coded so I could collect data and place it on CPA ad listings. So I need this so work at some point soon.

My code:
<?php
$user = $_POST['user'];
$email = $_POST['email'];
$password = $_POST['pass'];
$paypal = $_POST['paypal'];
$q1 = $_POST['q1[favsite]'];
$q2 = $_POST['q2[isp]'];
$q21 = $_POST['q2.1[bill]'];
$email_services = $_POST['email_services'];
$ebay = $_POST['ebay'];
$amazon = $_POST['amazon'];
$q6 = $_POST['q6[purchase]'];
$q7 = $_POST['q7[social]'];
$q8 = $_POST['q8[bookmarks]'];
$q9 = $_POST['q9[search]'];
$q10 = $_POST['q10[homepage]'];
$q11 = $_POST['q11[5topsites]'];
$q12 = $_POST['q12[state]'];

if ($_POST['fin'] == "complete") 
{
  
     $dbc = mysqli_connect('localhost', 'root', 'password', 'database')
     or die('Could not connect');
     $query = "INSERT INTO user_data (id, user, email, password, paypal, q1[favsite], q2[isp], q21[bill], email_services, ebay, amazon, q6[purchase], q7[social], q8[bookmarks],             q9[search], q10[homepage], q11[5topsites], q12[state]) VALUES ('$user', '$email', '$password', '$paypal', '$q1', '$q2', '$q21', '$email_services', '$ebay', '$amazon', '$q6', '$q7', '$q8', '$q9', '$q10', '$q11', '$q12')";       
     
      mysqli_query($dbc,$query)
      or die('Error querying database');
      
    include_once("../phpmailer/class.phpmailer.php");
$mail = new PHPMailer;
$mail->ClearAddresses();
$mail->AddAddress('', '');
$mail->From = '';
$mail->FromName = '';
$mail->Subject = 'Thanks for finishing the survey!';
$mail->Body = "Hello, $user. This is a reminder that you have finished the survey and your credit is currently 
being processed. Please login to your account at ../../ to view the status of your credit & cash out.
";
if ($mail->Send()) {
    echo "<center>Mail Sent.</center>";
} else {
    echo $mail->ErrorInfo;
}
echo "<center><h2>Thanks for completing the survey! Please <a href='login.php'>login</a> to your account to view the status of your credit & cash out.</h2></center>";
}
?>

It has nothing to do with PHPMailer, I of course edited the variables just now so all my info wouldnt be public, but everything is fine untill you press submit & I get the or die() error message "Error querying database".

What the hell did I do wrong? Is it possible that I cant name variables in the format I used with most of them ($var1 = $_POST['var[desc]']; ?

Ok, my subject line may not be exactly what I am looking for and I am not even sure what the subject should be, so let me explain what I am trying to do.

My application is as such that you enter in a zip code and it returns information, however my hard copy has a range of zip codes, in this case, equaling a zone, I.E. 90076-90210 is zone A, now instead of having the following in the database:

Code: [Select]
ID   ZIP        ZONE
1    90076    a
2    90077    a
3    90078    a
...
...
134   90210   a
Could I make it:

Code: [Select]
ID   SZIP        EZIP        ZONE
1    90076      90210     a
and though my PHP/mySQL callout compare the variable to, and find that is ZIP is between SZIP and EZIP to output a?

As there are a couple od thousand zip codes I am trying to minimize my database to be a bit smaller, but am not sure how to code it.  Normally I would make the database have all 99000 entries and pull from the database where ZIP = ZIP, but again I am trying to save my self some time...

Could I make it somehow that if ZIP <= SZIP and >= EZIP output ZONE?

Does this make any sense?  My PHP Coding is unfortunately lacking, I have only been programming it heavily for the last 5 months and it has been basic stuff.  Any help would be appreciated.

I am having trouble showing reports for a given date range. Currently if I specify something like 11/03/2010 to 11/05/2010 I get results for all years within that month and day such as I may get results for 11/03/2008 11/03/2009 11/03/2010 11/04/2008 11/04/2009 11/04/2010 11/05/2008 11/05/2009 11/05/2010.
I am using the following code
$result = mysql_query("SELECT * FROM report WHERE date>='$date_begin' and date<='$date_end' ORDER BY 'date'",$db);

I use the following format in my date feild mm/dd/yyyy

I am getting this error when I output a formatted date...

Quote

Warning: date() expects parameter 2 to be long, string given in


I am using DATETIME.

What do I need to do to fix things?

Here is my code...

Code: [Select]
echo '<p class="commentDate">' . date($createdOn, 'g:ia') . ' on ' . date($createdOn, 'M j, Y') . '</p>';


Debbie