Inserting holiday fines Using php

[mugiwara05]

View attachment 324779 View attachment 324780View attachment 324781View attachment 324782

Patulong naman po about sa programming sa php, Yung sa holiday meron pa po ba kayong alternative way sa pagkuwa ng Fines/penalties sa library system php? Kase nahihirapan po ako sa holiday pero pag due date at return lang ok naman yung mismong pag insert lang ng holiday mismo. Sana matulungan niyo po ako eto po yung code ko

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
    $returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila')); 
    $currentdate = new Datetime('Asia/Manila');

    $returndate->setTime(0,0);
    $currentdate->setTime(0,0);
    $borrowdate->setTime(0,0);

    $interval = DateInterval::createFromDateString('1 day');
    $period = new DatePeriod($returndate, $interval, $borrowdate);

    $borrowdate->format("D");
    $weekendDays = 0;
    $totalDays = 0;
    $holiday = 0;

    
     $query_holiday =mysqli_query($dbcon,"SELECT * FROM holiday_tbl"); 
     $fi = $row['borrow_details_id'];
    $row=mysqli_fetch_array($query_holiday);
    $decholi = $row ['holiday']; 
   
    
    foreach ($period as $p )
    {
        $totalDays++;
        if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
      
        foreach ($row as $decholi) {
            if ($p->format('Y-m-d')  == $decholi) $holiday++ ;
        } 
        
    } 
     
    echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;
   

    $minusdays = ($totalDays - $holiday);
    $fines = ($minusdays - $weekendDays) * 5;

    if($type == 'Teacher') {
        $fines = 0;
      
      
    }
    echo "₱ ". $fines;
    mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

 

[jskhulitz]

Please do the necessary changes, nag-assume lang ako ng mga field names at variables.

#1

	// query for holiday dates - "SELECT * FROM holiday_tbl where date>='$dateBorrowed' and date<='$dateReturned'";
	$tempArray = array();
	while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC))
		$tempArray[$rows['holidayDate']] = ''
	} 

    foreach ($period as $p )
    {
        $totalDays++;
        if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
      
		if(isset($tempArray[$p])){ $holiday++; }
    }

#2

$holiday = count($tempArray);
echo $holiday;

Code Explanation:

ilagay mo sa temporary array yung mga dates na makukuha mo from your holiday query, sa query, isama mo as parameter yung date borrowed, at dat returned, kasi within those 2 dates lang dapat ang makuha na holiday. Once na nasa array na yung mga holiday dates, pwede mo na magamit sya to count kung ilang holidays ang natamaan mo.

Check #2, di mo na kailangan mag loop, display mo na lang yung array size ng holiday date result mo

 

[mugiwara05]

Maraming salamat po sir pasensya na po di ko talaga alam yung array pero sa ibang paraan ko lang ginagawa kaso siguro etong holiday na to kailangan talaga array gamitin kaya medyo nahihirapan ako. Bali may error po

CATCHABLE FATAL ERROR: OBJECT OF CLASS DATETIME COULD NOT BE CONVERTED TO STRING IN C:\WAMP64\WWW\STI_LMS\LIBRARIAN\VIEW_BORROW.PHP ON LINE 87

 Tama po ba ginawa ko?

 $borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila')); 
$currentdate = new Datetime('Asia/Manila');

$returndate->setTime(0,0);
$currentdate->setTime(0,0);
$borrowdate->setTime(0,0);

$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("D");
$weekendDays = 0;
$totalDays = 0;
$holiday = 0;


 $result=mysqli_query("SELECT * FROM holiday_tbl where date>='$borrowdate' and date<='$returndate'"); <---- Dito po may error
 $tempArray = array();
 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
     $tempArray[$rows['holidayDate']] = '';
 }
 foreach ($period as $p )
 {
     $totalDays++;
     if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
   
     if(isset($tempArray[$p])){ $holiday++; }
 }
 
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;


$minusdays = ($totalDays - $holiday);
$fines = ($minusdays - $weekendDays) * 5;

if($type == 'Teacher') {
    $fines = 0;
  
  
}
echo "₱ ". $fines;
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

Maraming salamat po sir eto na lang kulang ko sa thesis ko inserting holiday sana matulungan niyo po ako susubukan ko aralin yung array once na natapos ko na thesis na to salamat po

 

[jskhulitz]

Kung yung mismong query ang error, subukan mo iquery direkta sa phpmyadmin, tignan mo kung anong error ang ilalabas. Kasi kung valid SQL naman yung nasa statement, walang eror para lumabas dun...

Subukan mo nga i-echo itong statement na ito. Kung ano ang idisplay sa screen, copy mo then paste mo sa phpmyadmin. Tignan mo kung ano result.

echo "SELECT * FROM holiday_tbl where date>='$borrowdate' and date<='$returndate'";

-----------------------

Most probable cause: ano yung content ng $borrowdate and $returndate variables mo.

 

[mugiwara05]

Kung yung mismong query ang error, subukan mo iquery direkta sa phpmyadmin, tignan mo kung anong error ang ilalabas. Kasi kung valid SQL naman yung nasa statement, walang eror para lumabas dun...

Subukan mo nga i-echo itong statement na ito. Kung ano ang idisplay sa screen, copy mo then paste mo sa phpmyadmin. Tignan mo kung ano result.

echo "SELECT * FROM holiday_tbl where date>='$borrowdate' and date<='$returndate'";

-----------------------

Most probable cause: ano yung content ng $borrowdate and $returndate variables mo.

kapag ineecho ko po ayaw pa din po ganyan pa din po error na lumalabas kahit tinanggal ko na yung query pinalit ko yan ayaw talaga po e. eto po yung contents ng mga dates https://imgur.com/a/fwOIK

 

[jskhulitz]

Nasaan pala yung holiday_tbl mo?

Tsaka ano yung value na naka-assign sa variables na $borrowdate and $returndate.

 

[mugiwara05]

Nasaan pala yung holiday_tbl mo?

Tsaka ano yung value na naka-assign sa variables na $borrowdate and $returndate.

eto po yung sa holiday_tbl https://imgur.com/a/iiYrx , hindi po ba eto yung tinutukoy niyo https://imgur.com/a/fwOIK ? naka new Datetime po lahat nung dates e hindi po strtotime

 

[jskhulitz]

subukan mo nga muna maglagay ng temporary value for those two variables

$borrowdate = '2017-09-02';
$returndate = '2017-09-21'

then proceed with the query, check mo kung ano result

 

[mugiwara05]

subukan mo nga muna maglagay ng temporary value for those two variables

$borrowdate = '2017-09-02';
$returndate = '2017-09-21'

then proceed with the query, check mo kung ano result

eto yung nilagay kong code

$borrowdate = new DateTime('2017-09-02');
$returndate = new DateTime('2017-09-21');

CATCHABLE FATAL ERROR: OBJECT OF CLASS DATETIME COULD NOT BE CONVERTED TO STRING IN

eto yung line nung error

$result=mysqli_query("SELECT * FROM holiday_tbl where date>='$borrowdate' and date<='$returndate'");

siguro sir naka datetime siya e kase pang string type ata yung code

 

[jhefreyzz]

try mo ito idol:
instead of $borrowdate->format("D"); // this results to Mon-Sun
gawin mong $borrowdaate->format("Y-m-d"); // while this results to 2017/09/18

if hindi gumana post mo lang yung error na lumabas

 

[mugiwara05]

try mo ito idol:
instead of $borrowdate->format("D"); // this results to Mon-Sun
gawin mong $borrowdaate->format("Y-m-d"); // while this results to 2017/09/18

if hindi gumana post mo lang yung error na lumabas

same error pa din po siya pero may pinalitan po ko

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$borrowdates=date_format($borrowdate, 'Y-m-d');
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila'));
$returndates= date_format($returndate, 'Y-m-d');
$currentdate = new Datetime('Asia/Manila');
  


$borrowdate->setTime(0,0);
$returndate->setTime(0,0);
$currentdate->setTime(0,0);



$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("Y-m-d");
$weekendDays = 0;
$totalDays = 0;
$holiday = 0;

$result=mysqli_query("SELECT * FROM holiday_tbl where date>='$borrowdates' and date<='$returndates'");  --- -line 92 error
$fi = $row['borrow_details_id'];
 $tempArray = array();
 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
     $tempArray[$rows['holiday']] = '';
 }
 foreach ($period as $p )
 {
     $totalDays++;
     if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
   
     if(isset($tempArray[$p])){ $holiday++; } -- line 103 error
 }
 
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;


$minusdays = ($totalDays - $holiday);
$fines = ($minusdays - $weekendDays) * 5;

if($type == 'Teacher') {
    $fines = 0;
  
  
}
echo "₱ ". $fines;
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

May alam ka po ba alternative na pagcalculate ng fines o penalties? gamit strtotime? kase nahihirapan na po ko dito sa code na to e pahelp po

View attachment 324867 View attachment 324868

 

[jskhulitz]

Yung error mo sa line 92 - mysqli_query() - expects 2 parameter, 1 lang nilagay mo. Ang correct syntax nyan ay:

$result = mysqli_query($connectionString, $queryString);

Sa current code mo, $queryString lang ang pinapasan mo sa mysqli_query

 

[mugiwara05]

Yung error mo sa line 92 - mysqli_query() - expects 2 parameter, 1 lang nilagay mo. Ang correct syntax nyan ay:

$result = mysqli_query($connectionString, $queryString);

Sa current code mo, $queryString lang ang pinapasan mo sa mysqli_query

Maraming salamat po sa reply niyo po sir eto po yung panibagong error line 95

WARNING: MYSQLI_FETCH_ARRAY() EXPECTS AT MOST 2 PARAMETERS, 3 GIVEN IN

while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) {

Maraming salamat po sir!

 

[jskhulitz]

papost nga ng current code mo

 

[mugiwara05]

papost nga ng current code mo

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$borrowdates=date_format($borrowdate, 'Y-m-d');
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila'));
$returndates= date_format($returndate, 'Y-m-d');
$currentdate = new Datetime('Asia/Manila');
  


$borrowdate->setTime(0,0);
$returndate->setTime(0,0);
$currentdate->setTime(0,0);



$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("D");
$weekendDays = 0;
$totalDays = 0;
$holiday = 0;

$result=mysqli_query($dbcon,"SELECT * FROM holiday_tbl where date>='$borrowdates' and date<='$returndates'");
 $fi = $row['borrow_details_id'];
 $tempArray = array();
 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) { ----- line 95 error
     $tempArray[$rows['holiday']] = '';
 }
 foreach ($period as $p )
 {
     $totalDays++;
     if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
   
     if(isset($tempArray[$p])){ $holiday++; } --- line 103 error
 }
 
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;


$minusdays = ($totalDays - $holiday);
$fines = ($minusdays - $weekendDays) * 5;

if($type == 'Teacher') {
    $fines = 0;
  
  
}
echo "₱ ". $fines;
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

eto po yung error View attachment 324884

Pero nagddisplay na po yung fines yung holiday na lang po talaga
View attachment 324885

salamat po sir!!

- - - Updated - - -

papost nga ng current code mo

EDIT:

Pag hindi pa siya lumagpas sa Due Date eto lang error niya Hindi lumalabas yung error na

: ILLEGAL OFFSET TYPE IN ISSET OR EMPTY IN

View attachment 1221229

Pero pag lumagpas na sa due date eto error na lumalabas View attachment 1221231 Kada Araw na lumalagpas Dumadagdag yung error pababa ng pababa tulad ng ganito View attachment 1221233 Naka September 30 ako kaya ang lumabas na error is 7 kase due date September 22 , sa 23 counting

- - - Updated - - -

papost nga ng current code mo

Sir nagsearch ako sa google pano makuwa error in specific way eto nakita ko sir


View attachment 324896
tinanggal ko yung date eto nilagay ko

$result=mysqli_query($dbcon,"SELECT * FROM holiday_tbl where '$borrowdates' >= '$returndates'");

bali eto na lang error niya lahat

WARNING: ILLEGAL OFFSET TYPE IN ISSET OR EMPTY IN line 107

 

[jskhulitz]

Bakit ganito ang query mo? bakit variable to variable ang comparison mo? Ang kino-compare sa database ay column to variable.
Tapos ang logic pa ng comparison mo ay "Date nung hiniram" >= "Date nung isinauli". Kahit kailan hindi ka magkakaroon ng $returndates na mas maaga compared sa $borrowdates. Ano un, nag-timetravel ka sa past?

$result = mysqli_query($dbcon,"SELECT * FROM holiday_tbl where '$borrowdates' >= '$returndates'");

Above query should be:

$result = mysqli_query($dbcon,"SELECT * FROM holiday_tbl where holiday between $borrowdates and $returndates");
//OR
$result = mysqli_query($dbcon,"SELECT * FROM holiday_tbl where ($borrowdates<=holiday and $holiday<=$returndates"));

Dito naman sa current code mo

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$borrowdates=date_format($borrowdate, 'Y-m-d');
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila'));
$returndates= date_format($returndate, 'Y-m-d');
$currentdate = new Datetime('Asia/Manila');

$borrowdate->setTime(0,0);
$returndate->setTime(0,0);
$currentdate->setTime(0,0);

$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("D");
$weekendDays = 0;
$totalDays = 0;
$holiday = 0;

$result=mysqli_query($dbcon,"SELECT * FROM holiday_tbl where date>='$borrowdates' and date<='$returndates'");
 $fi = $row['borrow_details_id'];
 $tempArray = array();
 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) { ----- line 95 error
     $tempArray[$rows['holiday']] = ''; //$rows['holiday'] - date format = yyyy-mm-dd
 }
 foreach ($period as $p )
 {
     $totalDays++;
     if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
	
     //if(isset($tempArray[$p])){ $holiday++; } --- line 103 error - Illegal offset type
	 // ang error mo dito ay pinapasa mo as INDEX yung $p na object. remember, si $p ay object, hindi mo sya ginamit as variable. To compensate:
	 $tempDate = $p->format("Y-m-d");
	 if(isset($tempArray[$p])){ $holiday++; }
 }
 
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;


$minusdays = ($totalDays - $holiday);
$fines = ($minusdays - $weekendDays) * 5;

if($type == 'Teacher') {
    $fines = 0;
}
echo "₱ ". $fines;
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

 

[mugiwara05]

Dito naman sa current code mo

$borrowdate = new Datetime($row['date_return'],new DateTimeZone('Asia/Manila'));
$borrowdates=date_format($borrowdate, 'Y-m-d');
$returndate = new Datetime($row['due_date'],new DateTimeZone('Asia/Manila'));
$returndates= date_format($returndate, 'Y-m-d');
$currentdate = new Datetime('Asia/Manila');

$borrowdate->setTime(0,0);
$returndate->setTime(0,0);
$currentdate->setTime(0,0);

$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($returndate, $interval, $borrowdate);

$borrowdate->format("D");
$weekendDays = 0;
$totalDays = 0;
$holiday = 0;

$result=mysqli_query($dbcon,"SELECT * FROM holiday_tbl where date>='$borrowdates' and date<='$returndates'");
 $fi = $row['borrow_details_id'];
 $tempArray = array();
 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) { ----- line 95 error
     $tempArray[$rows['holiday']] = ''; //$rows['holiday'] - date format = yyyy-mm-dd
 }
 foreach ($period as $p )
 {
     $totalDays++;
     if($p->format( "w" )== 0 or $p->format( "w" )== 6 ) $weekendDays++;
	
     //if(isset($tempArray[$p])){ $holiday++; } --- line 103 error - Illegal offset type
	 // ang error mo dito ay pinapasa mo as INDEX yung $p na object. remember, si $p ay object, hindi mo sya ginamit as variable. To compensate:
	 $tempDate = $p->format("Y-m-d");
	 if(isset($tempArray[$p])){ $holiday++; }
 }
 
echo "<p>Total  days: $totalDays</p><p>Weekend days: $weekendDays </p> <p> Holidays: $holiday </p>" ;


$minusdays = ($totalDays - $holiday);
$fines = ($minusdays - $weekendDays) * 5;

if($type == 'Teacher') {
    $fines = 0;
}
echo "₱ ". $fines;
mysqli_query($dbcon,"update borrowdetails set fines='$fines' where borrow_details_id = '$fi'");

Sir pinalitan ko yung code mo tempDate into tempArray kaya wala na siyang error, yung holiday 0 pa din po pero nung echo ko yung tempArray nagecho yung holiday ano po kaya problema, sir maraming salamat po malapit na ata siya magawa!!!

Inedit ko po yung date ko sa september 26 kase ginawa kong holiday yung sept 25

View attachment 325075 View attachment 325076

 

[jskhulitz]

Wut? Bakit mo pinalitan yung $tempDate at ginawa mong $tempArray? Kaya walang lumabas na problema at walang naidisplay na holiday ay dahil na si $tempArray na dati ay isang array ay naging single value variable.

	 [COLOR="#FF0000"]$tempDate = $p->format("Y-m-d");[/COLOR] // ito ba kamo ang pinalitan mo? ginawa mong $tempArray = $p->format("Y-m-d");
	 // kung ito ang pinalitan mo, ano na lang ang iccheck ng if condition na ito sa baba? siyempre since nagng variable na lang si $tempArray at hindi na 
	 // sya array, puro false na ang ire-return na result nung condition, hindi na magi-increment si $holiday
	 if(isset($tempArray[$p])){ $holiday++; }

Actually, ako ang may fault sa error, that chunk of code should be like this. Nakita mo yung difference? imbes na si $p ang gawin natin index ni $tempArray (at mag-issue sya ng error na LLEGAL OFFSET TYPE IN ISSET OR EMPTY), iko-convert muna natin yung date format ng object na si $p at in-assign sa isang temporary variable na $tempDate. Once na mai-assign kay $tempDate, sya ang gagawin natin index ni $tempArray at iccheck ang condition na isset

	 [COLOR="#FF0000"]$tempDate[/COLOR] = $p->format("Y-m-d");
	 if(isset($tempArray[[COLOR="#FF0000"]$tempDate[/COLOR]])){ $holiday++; }

to check kung may laman ang $tempArray array, do this. Then post mo dito yung result

echo '<pre>';
	print_r($tempArray);
echo '</pre>';

 

[mugiwara05]

Wut? Bakit mo pinalitan yung $tempDate at ginawa mong $tempArray? Kaya walang lumabas na problema at walang naidisplay na holiday ay dahil na si $tempArray na dati ay isang array ay naging single value variable.

	 [COLOR="#FF0000"]$tempDate = $p->format("Y-m-d");[/COLOR] // ito ba kamo ang pinalitan mo? ginawa mong $tempArray = $p->format("Y-m-d");
	 // kung ito ang pinalitan mo, ano na lang ang iccheck ng if condition na ito sa baba? siyempre since nagng variable na lang si $tempArray at hindi na 
	 // sya array, puro false na ang ire-return na result nung condition, hindi na magi-increment si $holiday
	 if(isset($tempArray[$p])){ $holiday++; }

Actually, ako ang may fault sa error, that chunk of code should be like this. Nakita mo yung difference? imbes na si $p ang gawin natin index ni $tempArray (at mag-issue sya ng error na LLEGAL OFFSET TYPE IN ISSET OR EMPTY), iko-convert muna natin yung date format ng object na si $p at in-assign sa isang temporary variable na $tempDate. Once na mai-assign kay $tempDate, sya ang gagawin natin index ni $tempArray at iccheck ang condition na isset

	 [COLOR="#FF0000"]$tempDate[/COLOR] = $p->format("Y-m-d");
	 if(isset($tempArray[[COLOR="#FF0000"]$tempDate[/COLOR]])){ $holiday++; }

to check kung may laman ang $tempArray array, do this. Then post mo dito yung result

echo '<pre>';
	print_r($tempArray);
echo '</pre>';

Ohh kaya pala di nagiincrement si holiday kase false pala yung binabato kaya hindi tinatanggap.
Opo sir pinalitan ko na po gamit yung code niyo

eto po yung result sa array walang display po

Sinubukan ko po mag echo sa loob ng if pero wala po lumalabas tingin ko hindi siya pumapasok sa if ano po kaya problema
sinubukan ko mag echo sa labas pero nageecho siya depende sa dami ng holiday ng nadaanan nya at sinubukan ko echo yung tempDate meron naman po laman
View attachment 325105
View attachment 325103

View attachment 325102

 

[jskhulitz]

yung $tempDate, meron talaga laman yan dahil ang ipinapasa nating value sa kanya ay yunng inclusive dates between date borrowed at date returned. Dahil yan dito:

$tempDate = $p->format("Y-m-d");

Ang kailangan natin malaman ay kung may laman ba yung $tempArray natin. Di ba may ganito ka:

 while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) { ----- line 95 error
     $tempArray[$rows['holiday']] = ''; //$rows['holiday'] - date format = yyyy-mm-dd
 }

//dugtungan mo nito
echo '<pre>';
print_r($tempArray);
echo '</pre>';

Also, ano yung nasa first 2 columns, yung may nakalagay na 2017-09-12 at 2017-09-15