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PHP - Strange Behavior With Strtotime()
Current time when testing was: 1291064453
I run the following: echo date('m/d/y', strtotime('first day', 1291064453)); Expecting: 11/1/10 What I actually get: 11/30/10 Can anyone explain this? Similar TutorialsSo i am currently coding database connection class and i have encountered very strange behavior from my script. base.class.php: Code: [Select] <?php class base{ private $settings; function get_settings(){ $settings["dbhost"] = 'localhost'; $settings["dbuser"] = '*****'; $settings["dbpass"] = '*****'; $settings["dbname"] = 'core'; return $settings; } } ?> database.class.php Code: [Select] <?php require_once 'base.class.php'; class database extends base{ private $query_now; private $link; public function __construct(){ $settings = base::get_settings(); $dbhost = $settings["dbhost"]; $dbuser = $settings["dbuser"]; $dbpass = $settings["dbpass"]; $dbname = $settings["dbname"]; $this->link = mysql_connect($dbhost, $dbname, $dbpass) or die ("Could not connect to the mysql database"); mysql_select_db($dbname, $this->link) or die ("Could not select the database"); } function query($query){ $this->query_now = $query; return mysql_query($query, $this->link); } function getArray($result){ return mysql_fetch_array($result); } } ?> When i try to create an instance of database class, i get mysql_connect error. I have tried to echo my array and it seems that correct information is being passed over. Now the strange thing is if i remove my password from the base class i don't get a mysql_connect error but this time instead i get "Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'core'@'localhost' (using password: YES) " In case you are wondering, does my mysql database user has a password, the answer is: yes for sure... (Also i have tried to setup a simple script for connecting to my database and everything worked fine) So any ideas? I'm returning a table row that contains information about a file, but it seems in IE versions older than 10, it is cutting off some of the returned json when being used.
The data is being returned properly as seen in the following json:
{"file_name":"<i class='video'><\/i> <a href=\"\/Development\/test(4).mp4\" class=\"is_file\" target=\"_blank\">test(4).mp4<\/a>"}
But when you use it, it cuts off the html. A simple alert will return
</i> test(4).mp4</a>and same when appending it and the sort. It is also happening for another part of HTML that is being returned properly in the json. It is working for everything else that is returned. I have been searching around for a very long time trying to find why this is happening. Has anyone other than me encountered this? Hi guys, In my connection.php I have:
$db->query("DROP TABLE IF EXISTS mydata") ;
$db->query("CREATE TABLE mydata
(
ID INT AUTO_INCREMENT NOT NULL PRIMARY KEY,
guid INT,
title VARCHAR(100),
body LONGTEXT,
term VARCHAR(100)
)");
and my query code :
$myarray = array (
guid => 100,
title => "title test",
body => "just a test",
term => "term test",
);
$myplaceholders[] = '(' . implode (", ", array_fill(0, count($myarray), '?')) . ')'; //also tried '(?,?,?,?)'
$mykeys = implode(', ', array_keys($myarray));
array_push($values, ...array_values($myarray)); //also tried $values = array_values($myarray) ;
$res = $db->prepare("INSERT INTO mydata ($mykeys) VALUES " . join(', ', $myplaceholders)) ;
if ($res->execute($values)) {
echo 'data inserted';
} else {
echo 'error in query';
}
After executing the code, the table is created but no data is inserted. The strange thing is when I leave the create table statement out of the connection.php and run the code the data is inserted. Any ideas where I'm going wrong? I am working with the Amazon API, and I am trying to display a default image if a product does not have an image associated with it. The query is coming back as an array. Typically, each product image array looks like this: $d = SimpleXMLElement Object ( http:// => [url]http://ecx.images-amazon.com/images/I/51aUIul6XjL._SL160_.jpg [Height] => 160 [Width] => 112 ) The code goes like this: Code: [Select] if ($d=$E->MediumImage) { $iu=$d->URL; $ih=$d->Height; $iw=$d->Width; echo count($d); if (strlen($iu) > 0) {echo "<center><a href='$url' target='_blank'><img src='images/amazon_noimage.jpg' width='175' height='175' border='0'></a></center>";} else {echo "<center><a href='$url' target='_blank'><img src='$iu' width='$iw' height='$ih' border='0'></a></center>";} } However, images/amazon_noimage.jpg never shows up (even though it is linked correctly, as I've tested this link). I have tried the following: if (strlen($iu) > 0) if (count($iu) > 0) if (strlen($d->URL) > 0) if (count($d->URL) > 0) if (isset($d)) if (!isset($d)) etc ... If I display the following, where there is no image, I get nothing displayed: echo $iu; print_r($iu); echo $d->URL; etc ... However, if there is an image, I get a link, such as the following: http://ecx.images-amazon.com/images/I/51c2BFpDN0L._SL160_.jpg There seems to be NOTHING that I can do to trigger the 'ELSE' part of the if statement. This is a total enigma to me ... any ideas?? Hey all, when job_title property is equal to null, I want this to happen: Welcome to the blog of John Merlino. If it is not null then: Welcome to the blog of John Merlino, a web designer. //where web designer refers to the value stored in job_title So I come up with this: Code: [Select] echo "Welcome to the blog of " . $blogger->first_name . ' ' . $blogger->last_name . (!is_null($blogger->job_title)) ? ', ' . $blogger->job_title . '.' : '.'; But when job_title is null, all the page renders is this: Code: [Select] , . That's right. Just a comma, then a space, and then a period. What am I missing here? Thanks for response. This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=331683.0 Ok so I have a simple database table set up, and I am trying to get the results into an HTML table, however I have a question. One loop I write in a for loop will dump the data I want to the table, the other returns no results. For example I have the code... Code: [Select] //SQL Lookup //SQL Lookup $sqlAppt = mysql_query("SELECT * FROM Appointment WHERE TTUNumber = \"$TTUNumber\" && ShippingNumber = \"$ShippingNumber\" LIMIT 1") or die('Could not execute SQL statement ' . mysql_error()); $numbRows = mysql_num_rows($sqlAppt); echo "<p>Number of rows found: $numbRows </p>\n"; So When I create the first whlile loop, i do not get any results from my database. Code: [Select] while($r=mysql_fetch_array($sqlAppt)){ $ApptDate=$r['Date']; $ApptTime=$r['Time']; $ApptNumber=$r['ApptNumber']; $ApptDock=$r['Dock']; echo '<h1>I got here</h1>'."\n"; echo "<table>\n"; echo "<tr><td>Date</td><td>$ApptDate</td></tr>\n"; echo "<tr><td>Time</td><td>$ApptTime</td></tr>\n"; echo "<tr><td>Appt Number</td><td>$ApptNumber</td></tr>\n"; echo "<tr><td>Dock</td><td>$ApptDock</td></tr>\n"; echo "</table>\n"; } The Results is this: Quote Number of rows found: 1 I got here Date Time Appt Number Dock However, when I create a different loop structure... Code: [Select] for($i = 0; $i < $numbRows; $i++) { $r=mysql_fetch_array($sqlAppt); $ApptDate=$r['Date']; $ApptTime=$r['Time']; $ApptNumber=$r['ApptNumber']; $ApptDock=$r['Dock']; echo '<h1>I got here</h1>'."\n"; echo "<table>\n"; echo "<tr><td>i:</td><td>$i</td></tr>\n"; echo "<tr><td>Date</td><td>$ApptDate</td></tr>\n"; echo "<tr><td>Time</td><td>$ApptTime</td></tr>\n"; echo "<tr><td>Appt Number</td><td>$ApptNumber</td></tr>\n"; echo "<tr><td>Dock</td><td>$ApptDock</td></tr>\n"; echo "</table>\n"; } Naturally I get a different result! Quote Number of rows found: 1 I got here i: 0 Date 1982-12-26 Time 08:00:00 Appt Number 123 Dock 34 I can't figure out why for the life of me that my while loop would not return anything on here>? Does anyone have any idea's? Thanks hello; my webhost made a change to my php.ini file yesterday. since then php is escaping single quotes that it receives from flash. Code: [Select] //-- AS3 var myVariables:URLVariables = new URLVariables(); myVariables.pvs_params = "'h'e'l'l'o" ; var myURLRequest:URLRequest = new URLRequest("mysql_task_mgr.php"); myURLRequest.data = myVariables ; myURLRequest.method = "get" ; navigateToURL( myURLRequest, '_blank' ) ; Code: [Select] //-- php print( $_GET[ "pvs_params" ] ) ; // --> \'h\'e\'l\'l\'o any thoughts? my webhost is stumped; I'm trying to have a running total of the number of views an image gets. My test file works perfectly every time. The actual file, image_win.php, does not and after much testing can't see what's wrong. This files STRANGENESS: the first two views count fine, with 1 added the total in the field image_visit. But every time after that 2 is added to the total in image_visit. Very bizarre. Thanks for taking a look -Allen Code: [Select] <?php ////////TEST IMAGE_WIN //////////THIS WORKS include_once "scripts/connect_to_mysql.php"; include_once "scripts/paths.php"; $art_id = 372; $QUERY="SELECT user_id FROM artWork WHERE art_id = '$art_id'"; $res = mysql_query($QUERY); $num = mysql_num_rows($res); if($num >0){ while($row = mysql_fetch_array($res)){ $owner_id = $row['user_id']; } } mysql_query("UPDATE userInfo SET image_visit = image_visit +1 WHERE user_id = '$owner_id'"); ?> <?php //////// image_win.php /////////THIS DOES NOT WORK $user_id=$_SESSION['user_id']; include_once "scripts/connect_to_mysql.php"; include_once "scripts/paths.php"; $image_link = $_GET['image_link']; $art_id = $_GET['art_id']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <meta name="viewport" content="width=device-width, height=device-height, target-densityDpi=device-dpi"> <title>Image Window</title> </head> <body bgcolor="#000000" leftmargin="0"> <div align="center" > <img src="slir/w640-h535/<?php echo $path.$image_link.$postCat; ?>" /> <?php $QUERY="SELECT user_id FROM artWork WHERE art_id = '$art_id'"; $res = mysql_query($QUERY); $num = mysql_num_rows($res); if($num >0){ while($row = mysql_fetch_array($res)){ $owner_id = $row['user_id']; } } mysql_query("UPDATE userInfo SET image_visit = (image_visit +1) WHERE user_id = '$owner_id' "); ?> MOD EDIT: code tags added. So, my problem is that I need to edit my Xmas calculator to understand when this year's Xmas is over, it will automatically jump to the next one (2011-12-25). I have no idea how to do this (noob to php...) Thanks in advance. Here's my code: Code: [Select] <?php $time=time(); $xmas=strtotime("2010-12-25 00:00:00"); $diff = $xmas - $time; $days=intval($diff/86400); $left=$diff%86400; $hs=intval($left/3600); $left=$left%3600; $mins=intval($left/60); $secs=$left%60; echo "<font size=12 face=corbel> Xmas is after:<br> <strong>$days days</strong><br> <strong>$hs hours</strong><br> <strong>$mins mins $secs secs</strong>! </font>"; ?> I'm trying to add 5 days to $today date: Code: [Select] <?php $exp_date = "2011-09-11"; $todays_date = date("Y-m-d"); $today = strtotime("+5 days", $todays_date); $expiration_date = strtotime($exp_date); if ($expiration_date > $today) { echo "Valid: Yes"; } else { echo "Valid: No"; } ?> Is +5 days used wrong? First time posting code and very new to php aka learning as I go... The below code works as shown, but in the 'else' clause, I'd like to replace the +4 with a variable. How to do this? I've tried, Code: [Select] $renewaldate = strtotime($subscriptiondate); $sft = "' +" . $duesmultiplier . " year'"; $next_year = date('Y-m-d',strtotime($sft,$renewaldate)); but this doesn't return what I expect. Perhaps this can't be done using this technique? Code: [Select] if($n == 0) { $renewaldate = strtotime($subscriptiondate); $next_year = date('Y-m-d',strtotime('next year',$renewaldate)); echo $next_year; } else { $renewaldate = strtotime($subscriptiondate); $next_year = date('Y-m-d',strtotime('+4 year',$renewaldate)); echo $next_year; } Thank you when i use strtotime('+3 HOURS') everything is fine. so why cant i use strtotime('+3.5 HOURS')? what would be the proper way to do this? Hello,
I'm trying to use strtotime to add time to a mysql type timestamp. Any help would be appreciated. I think I have tunnel vision from looking at it when I know it has to be something obvious...lol
$start = "2014-05-22 09:16:24";
$type = 30;
$endtime = date($start,strtotime("+ '.$type.' minutes"));
echo "End: ".$endtime."</br>Start: ".$start;
Right now it returns:
End: 2014-05-22 09:16:24
Start: 2014-05-22 09:16:24
Hi, I have a bit of an issue with my application, I didn't notice until today, Sunday... Apparently PHP sees Sunday as the first day of the week, which I cannot fathom where it got hat notion from... but anyway, so I have the following samples of my code: Code: [Select] $mondayoflastweek = date('Y-m-d H:i:s', strtotime('monday last week')); $sundayoflastweek = date('Y-m-d 23:59:59', strtotime('sunday last week')); The crazy and very troublesome thing about the results of that code is that if you run that today, Sunday, it shows today as last week... So therefore, the next bit of code is just as problematic: Code: [Select] $mondayofthisweek = date('Y-m-d H:i:s', strtotime('monday this week')); $sundayofthisweek = date('Y-m-d 23:59:59', strtotime('sunday this week')); As you can imagine, for payroll, this is causing a big problem... My week starts at precisely 12:00 am on Monday and ends at precisely 11:59 pm on Sunday... Could someone please help me out with this... Hello! A user is going to input a date in the form "mm/dd/yyyy". I'd like to convert this to a time stamp so that I can store it via MySql. I've heard about the "strtotime" function in PHP but I'm not sure how it can tell that the user 03/04/2011 represents March,4th 2011 and not, say April,3rd 2011 (European version). Help with the correct syntax for conversion would be greatly appreciated. Thank you, Eric Hi I am new to PHP and trying to convert a date from a registration form (in the form 01/01/2011) to Y-m-d so that it can be stored in the database. This is the code I have, pretty sure it worked before but not it has stopped working! Any ideas? $dateformat = $_POST['dob']; $correctformat = date('Y-m-d',strtotime($dateformat)); I have checked what $_POST['dob'] is printing and that is correct, however the $correctformat is printing 1970-01-01 everytime. Any help would be greatly apreciated. Thank you Good Day, I'm struggling to get strtotime() to return anything. I have the following code.. $fields['birth']['day'] = 9; $fields['birth']['month'] = 7; $fields['birth']['year'] = 1982; $dob = $fields['birth']['day'].' '.$fields['birth']['month'].' '.$fields['birth']['year']; // Returns: 9 7 1982 $dob = strtotime($dob); die($dob); No matter what I do $dob isn't returning anything once strtotime() is used. I've tried matching my day/month/year 's with examples in the manual on the strtotime() page but using date() on the fields, but it still doesn't want to return anything. I feel I'm missing something simple here. Any help is appreciated. Regards, Ace how do I use strtotime() instead of mktime()? $matchTimea = mktime($mHour, $mMinute, 0, $mMonth, $mDay, $mYear); if($matchTimea-time() < $cutoffTimea*60) { Little confuzzled. I have got a bunch of listings that will be printed at the end of the page but I need to subtract a time specified in a textbox (send it to itself) on the page. Code: [Select] $row['active'] = (strtotime('$row[etd]') + $rwy1 = $_POST['tmatttextfield']);Well this just converts the time in the database which is always going to be a four character number such as: 1300 for 1PM into a time and adds the textbox value to the time. Code: [Select] <td width="50" bgcolor="<?php echo $bc?>"><strong><?php echo date('hi', $row['active']);?></strong></td>That just prints it all out formatting the time to only Hours and Minutes. So just to clarify the user would put in say '10' in the textbox which would constitute 10 minutes. The code above should then define that '10' as 10 minutes. Define the database result, ie. 1500 as 3PM and add the 10 minutes to it so it would become 1510. I have been getting some wierd and wonderful results whilst trying to get it to work. Mainly just 1200 though. Harry. |
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