|
PHP - Round () Syntax Help
This works
echo $data3['value']; But this doesn't echo round($data3['value'], 2); Can someone please advise me as to what I am doing wrong? Similar Tutorialshello guys, im creating a search result page that uses pagination, and im trying to figure out how many total pages their are going to be in my links with the page numbers. this is probably something really simple, ive been writing scripts for the last few hours and have a bit of a headache, any help would be appreciated here's the coding and the results i want: <?php $per_page = 10; $rows = mysql_query("SELECT COUNT(*) FROM articles WHERE text LIKE $string"); // this will return 31 $pages = $rows / $per_page; // returns 3.1 $total_pages = round($pages); // returns 3, want to round 3.1 up to 4 here ?> hi know who to round a number to the nearest whole number, if i use echo round(9.5); it echo's 10 and if i use echo round(9.2); it echo's 9 i want it so it always rounds up? can anyone help? cheers matt HOw to get the AVERAGE of my entire database? $query = "SELECT ROUND(AVG(scores)) FROM table"; Thanks! Bickey I need to take a time and convert so it is rounded up or down to the nearest 0 or 5. Code: [Select] function roundMe ($time, $upDown) { if($upDown==1) { //round up } else { //round down } return $x; } echo roundMe("16:22", 1); // returns 16:25 echo roundMe("16:22", 0); // returns 16:20 echo roundMe("16:25", 1); // returns 16:25 echo roundMe("16:59", 1); // returns 17:00 echo roundMe("16:59", 0); // returns 16:55 i have a price i want to add 10%-15% to the base price and that the result will be rounded to 9 in the ones digit place. for example 85 +15%=97.7 $result=99 117+15%=134.55 $result = 139 played with round() and ceil() I am trying to output only 40 of the latest files uploaded on the database server. But when I am trying to see the files uploaded on the server I am getting this error: Warning: round() expects parameter 2 to be long, string given in C:\wamp\www\mshare\lista.php on line 8 Call Stack The code: Code: [Select] <?php function bytestostring($size, $precision = 0) { $sizes = array('YB', 'ZB', 'EB', 'PB', 'TB', 'GB', 'MB', 'kB', 'B'); $total = count($sizes); while($total-- && $size > 1024) $size /= 1024; return round($size, $precision)." ".$sizes[$total]; } echo '<div class="header">40 last uploaded files:</div>'; $sql = "SELECT * FROM filer order by id desc limit 40"; $q = mysql_query($sql); echo '<table style="width: 100%;">'; while ($r = mysql_fetch_array($q)) { $i++; $id = $r['id']; $filnamn = $r['name']; $filtyp = explode("/", $r['type']); $filtyp = ucwords($filtyp[0]); if($filtyp == "") { $filtyp = "Unkown"; } $filstorlek = bytestostring($r['size'], ""); $a = $i%2; echo'<tr><td class="lista">'; echo ''.$i.'. <td class="lista">'.$filnamn.'<td class="lista">'.$filtyp.' <td class="lista">'.$filstorlek.' <td class="lista"><a href="download.php?id='.$id.'"'.$a.'"><img src="download-icon.jpg" class="imagedow"></a>'; } echo "</table>"; ?> I am new to PHP coding so you guys just know. Code: [Select] $newrating = round($imageinfo['rating'],2); this outputs: 4.49 in my db rating is 4.4884 all i need is the 4.... would i use some type of preg_replace or something just to grab the first 1 before the decimal? (all i need is the first number before the decimal) I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you
<title>Inputing Travel Detials</title>
<header>
<h1 align="center"> Adding Travel Detials </h1>
<body>
<p> <center><img src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center>
<table border="1">
<tr>
<td><a href="index.php"> Home Page </a></td>
<td><a href="administratorhomepage.html">Administrator Home Page </a></td>
<td><a href="viewhomepage.html">View Home Page </a></td>
<td><a href="Inputhomepage.html">Input Home Page </a></td>
<td><a href="traveldetials.html">Enter More Travel Detials </a></td>
</table>
</p>
<?php
include "connection.php";
$Applicant_ID = $_POST["Applicant_ID"];
$Method_Of_Travel = $_POST["Method_Of_Travel"];
$Cost = $_POST["Cost"];
$ETA = $_POST["ETA"];
$Main_Gate_Advised = $_POST["Main_Gate_Advised"];
$query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'");
$result = mysqli_query($dbhandle, $query)
or die(mysqli_error($dbhandle));
if($result){ echo "Success!"; }
else{ echo "Error."; }
// successfully insert data into database, displays message "Successful".
if($query){
echo "Successful";
}
else {
echo "Data not Submitted";
}
//closing the connection
mysqli_close($dbhandle)
?>
Ok this is puzzleing. I am geting "Could not delete data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.
<?php $con = mysqli_connect("localhost","user","password","part_inventory"); // Check connection if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } else { $result = mysqli_query($con, "SELECT * FROM amp20 "); $amp20ptid = $_POST['amp20ptid']; // escape variables for security $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']); mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'"); if (!mysqli_query($con, $amp20ptid)); { die('Could not delete data: ' . mysqli_error($con)); } echo "Part has been deleted to the database!!!\n"; mysqli_close($con); } ?> Hi.. I encountered problem in rounding of numbers into two decimal places. here is my sample code: if($W4_STATUS == 1 AND $DEPENDENTS == 0 AND $TotEarn >= 7917 AND $TotEarn <= 12500) { $TAX = ($TotEarn - 7917); $TAX = (937.50 + ($TAX * .25)); $TAX = number_format($TAX, 2, '.', ''); } for example from this: $TAX = ($TotEarn - 7917); $TAX = (937.50 + ($TAX * .25)); the output is: 1417.615 using this: $TAX = number_format($TAX, 2, '.', ''); the output was : 1417.61 but it should be : 1417.62 Thank you Hi guys
I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?
<?php
$con = mysql_connect("localhost:3306","root","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('sfapp', $con);
$sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')
VALUES
('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
?>
<?php
//if "email" variable is filled out, send email
if (isset($_REQUEST['pe_add'])) {
//Email information
$admin_email = $_REQUEST['pe_add'];
$forename = $_REQUEST['forename_add'];
$email = "autoreply@testing.com";
$subject = "Application";
$desc =
"Dear $forename
Thank you for submitting your online application, we will be in touch shortly.
"
;
//send email
mail($admin_email, "$subject", "$desc", "From:" . $email);
//Email response
echo "Thank you for contacting us!";
}
//if "email" variable is not filled out, display the form
else {
?>
If you are seeing this, you need to go back and fill out the Personal Email section!
<?php
}
header("location:complete.php");
mysql_close($con)
?>
Thanks in advance.
hello, if i have the following, how would i display it? Code: [Select] function minutes_round ($hour = "$signintime", $minutes = '5', $format = "H:i") { $seconds = strtotime($hour); $rounded = round($seconds / ($minutes * 60)) * ($minutes * 60); return date($format, $rounded); } Hey guys, I use the code below to add a corners to an image and it works great but when i add a border to the image it add a square.... any ideas how i can add a border to the rounded image? <?php $image_file = $_GET['src']; $corner_radius = isset($_GET['radius']) ? $_GET['radius'] : 20; // The default corner radius is set to 20px $topleft = (isset($_GET['topleft']) and $_GET['topleft'] == "no") ? false : true; // Top-left rounded corner is shown by default $bottomleft = (isset($_GET['bottomleft']) and $_GET['bottomleft'] == "no") ? false : true; // Bottom-left rounded corner is shown by default $bottomright = (isset($_GET['bottomright']) and $_GET['bottomright'] == "no") ? false : true; // Bottom-right rounded corner is shown by default $topright = (isset($_GET['topright']) and $_GET['topright'] == "no") ? false : true; // Top-right rounded corner is shown by default $imagetype=strtolower($_GET['imagetype']); $backcolor=$_GET['backcolor']; $endsize=$corner_radius; $startsize=$endsize*3-1; $arcsize=$startsize*2+1; if (($imagetype=='jpeg') or ($imagetype=='jpg')) { $image = imagecreatefromjpeg($image_file); } else { if (($imagetype=='GIF') or ($imagetype=='gif')) { $image = imagecreatefromgif($image_file); } else { $image = imagecreatefrompng($image_file); } } $size = getimagesize($image_file); // Top-left corner $background = imagecreatetruecolor($size[0],$size[1]); imagecopymerge($background, $image, 0, 0, 0, 0, $size[0], $size[1], 100); $startx=$size[0]*2-1; $starty=$size[1]*2-1; $im_temp = imagecreatetruecolor($startx,$starty); imagecopyresampled($im_temp, $background, 0, 0, 0, 0, $startx, $starty, $size[0], $size[1]); $bg = imagecolorallocate($im_temp, hexdec(substr($backcolor,0,2)),hexdec(substr($backcolor,2,2)),hexdec(substr($backcolor,4,2))); $fg = imagecolorallocate($im_temp, hexdec(substr($forecolor,0,2)),hexdec(substr($forecolor,2,2)),hexdec(substr($forecolor,4,2))); if ($topleft == true) { imagearc($im_temp, $startsize, $startsize, $arcsize, $arcsize, 180,270,$bg); imagefilltoborder($im_temp,0,0,$bg,$bg); } // Bottom-left corner if ($bottomleft == true) { imagearc($im_temp, $startsize, $starty-$startsize,$arcsize, $arcsize, 90,180,$bg); imagefilltoborder($im_temp,0,$starty,$bg,$bg); } // Bottom-right corner if ($bottomright == true) { imagearc($im_temp, $startx-$startsize, $starty-$startsize,$arcsize, $arcsize, 0,90,$bg); imagefilltoborder($im_temp,$startx,$starty,$bg,$bg); } // Top-right corner if ($topright == true) { imagearc($im_temp, $startx-$startsize, $startsize,$arcsize, $arcsize, 270,360,$bg); imagefilltoborder($im_temp,$startx,0,$bg,$bg); } $newimage = imagecreatetruecolor($size[0],$size[1]); imagecopyresampled($image, $im_temp, 0, 0, 0, 0, $size[0],$size[1],$startx, $starty); // Output final image header("Content-type: image/png"); imagepng($image); imagedestroy($image); imagedestroy($background); imagedestroy($im_temp); ?> Hi there. I am trying to display numbers if they have the numbers like: number_format($price,2) This gives us two decimal places i.e 7.23 Now thats fine, I would like to round the last two decimal places. I would like to make it 7.25, (rounding it from 7.23 to 7.25) Similarly making: 12.44 to 12.45 33.18 to 33.20 52.13 to 52.15 So actually rounding the last two places. Coz obviously customers won't be able to pay 52 dollars 13 cents. So therefore rounding the last cent's (decimal part) making it to 52.15 Thank you. All comments and feedback are welcomed. (d Hi.. I create mysql syntax for query testing before i input to my php code here is my mysql code: Code: [Select] set @t = 0; set @rqty=31968; SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN @t+OUTPUT_QTY > @rqty THEN @rqty -@t ELSE OUTPUT_QTY END as qty, @t := @t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (@t < @rqty); and i attach the sample output of the above query. Now that query test is work i will input that code to my php codes. $sql = "SELECT SKUCode, Materials, Comp, Qty FROM bom WHERE SKUCode = '$SKUCode'"; $res = mysql_query($sql, $con); ($row = mysql_fetch_assoc($res)); $Materials = $row['Materials']; $Qty = $row['Qty']; $Comp = $row['Comp']; //P28 //-----Compute Req Qty and Save to table---// $ReqQty = $Qty * $POReq; // 31968 $sql = "UPDATE bom SET ReqQty = '$ReqQty' WHERE SKUCode = '$SKUCode' AND Materials = '$Materials'"; $resReqQty = mysql_query($sql, $con); $t = 0; $sql = "SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN $t+OUTPUT_QTY > $ReqQty THEN $ReqQty -$t ELSE OUTPUT_QTY END as qty, $t := $t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '$Comp' AND ($t < $ReqQty)"; when I echo the query: I got this: SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN 0+OUTPUT_QTY > 31968 THEN 31968 -0 ELSE OUTPUT_QTY END as qty, 0 := 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (0 < 31968) then I run it to the sql and I got an error: Error Code : 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ':= 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '' at line 1 (0 ms taken) Any help is highly appreciated Thank you so much Which of the below syntax is the better to be using? $bal = 100; if($bal > 100) { echo "you have over 100 pounds; } else { echo "you have less than 100 pounds"; } if ($bal > 100); echo "you have over than 100"; else ; echo "you have less than 100"; endif; The issue is there seems to be NO syntax error. 1. There is no relevant code before or after this line. 2. Yes, one would think a ! should be there as did I (I didn't write the code), however, even with the ! it still gives the same error. if (function_exists('gzcompress')) die(FUNCTION_NOT_FOUND); is giving me a syntax error, unexpected 'if', expecting 'function' or 'const' I am updating this code from PHP5.3 to PHP7.4 and I can't figure out what the syntax problem is since PHP allows this. I am using Eclipse PHP to do the conversion. Hi, i,m trying to echo a "a href" based on the language session of my page like this: Code: [Select] ?php if ($_SESSION['session_idioma'])=="eng" { echo '<a href="preordering.php"><img src="img/top_banner_'.$_SESSION['session_idioma'].'.jpg" width="744" height="182"></a>'; } { else echo '<a href="ordering.php"><img src="img/top_banner_'.$_SESSION['session_idioma'].'.jpg" width="744" height="182"></a>'; }?>but I'm getting a blank page wit no errors...must be a syntax error that I can't spot... hey all, here is the xml code Code: [Select] <title>Dogged determination</title> <link rel="alternate" href="http://video.news.com.au/1959670201/Dogged-determination" /> <author> <name>mr mittens</name> </author> now to parse this to get the title i would do this but what if i want the link? Code: [Select] <?php$feed = simplexml_load_file('http://video.news.com.au/feed.atom');$link = (string) $feed->entry->title;echo $link;?> but what do i do to get the 'link'? i tried link an that obviously didnt work. cheers nick Hi there, I was trying to modify the code from the tutorial "PHP Basic Database Handling" and the code uses something I haven't seen before, but looks intriguing. Can someone explain how to use this: echo <<<SOMELABEL stuff SOMELABEL; Or let me know why my code didn't work? The error is an unexpected end statement on the last line of the file. My Code: Code: [Select] <?php $mystring = $_GET['FirstLast']; if($mystring == "") { echo "You reached this page by mistake, click on the link in your email"; }else { //echo $mystring; // connect to db $conn = mysql_connect('localhost','user','pw') or trigger_error("SQL",E_USER_ERROR); $db = mysql_select_db('tsec',$conn) or trigger_error("SQL",E_USER_ERROR); $sql = "SELECT `First Name`, `Last Name`, `Title`, `Org Parameter 1`, `Org/Person Name`, `Attendee Parm Value 1`, `Attendee Parm Value 2`, `Attendee Parm Value 3`, `Attendee Parm Value 4`, `Attendee Parm Value 5`, `Attendee Parm Value 6`, `Attendee Parm Value 7`, `Attendee Parm Value 8` FROM `attendees` WHERE FirstLast = '$mystring'"; $result = mysql_query($sql,$conn) or trigger_error("SQL", E_USER_ERROR); $list = mysql_fetch_assoc($result); echo <<<LISTNAME <form action = '{$_SERVER['PHP_SELF']}' method = 'post'> <table border = '1'><tr><td colspan = '2'>{$list['First Name']} {$list['Last Name']}</td></tr> <tr><td>Title</td><td><input type = 'text' name = 'Title' value = {$list['Title']}</td></tr> </table></form> LISTNAME; mysql_close($conn); } ?> Thanks. |
|||||||