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PHP - Need Help Creating A Select Time Drop Down
Hey Guys. I am trying to create a time dropdown and increment eat by an an hour until it is lless than or equal to the end time. I am using the DateTime class to accomplish this.
Now when I am trying to use a for loop to accompish this it does not work. Can anyone please help me out with this issue?
Below is the code that I have
$start_hour = new DateTime("now",new DateTimeZone("America/New_York"));
$start_hour->setTime(6,00);
$formatted_start_time = $start_hour->format("H:i:s");
$end_hour = mktime(11,45);
for($start_hour; $start_hour <= $end_hour; $start_hour->modify("+60 minutes"));{
echo "<select name='cat_display_timeslot'>";
echo "<option>{$formatted_start_time}</option>";
echo "</select>";
}
Thanks!
Edited by eldan88, 24 August 2014 - 07:21 PM. Similar TutorialsHey guys, I want to create stats of visitors and some other values in the database depending on time. The stats are supposed to be created one time per hour. What can I use for it? Cronjob? What will be if it doesn't work?? Thanks for helping! Hello, I am completely new to php so please forgive me ahead of time. I am trying to create a link button on a website for a client that only directs to a certain external website on certain days and hours. Specifically, this is a web radio button that the client only wants to link to the web radio site when their live broadcast is on the air during certain hours on the weekend (i.e. Sundays 8-10 PM). Any other time, the client wants the web radio button to link to a default page since the web radio station plays unrelated music all other hours of the week. Does anyone know the best way to create such a time sensitive link for a button on a webpage? Thanks. Hi there, I am just wondering which way is the correct one if you want to create object properties/attribute dynamically at run time using magic method __set(), __get(), for example: class Foo{ public $data = array(); public function __set($name,$value){ $this->data[$name] = $value; } public function __get($name){ if ( isset($this->data[$name]) ){ return $this->data[$name]; } } } Or using the below way class Foo{ public function assign(array $data){ foreach($data as $key => $value){ $this->$key = $value; } } } I just encountered the second way in many places and I am totally confused . Is the __set() method called implicitly by the PHP engine in the second way or what? how come this is considered as a valid code ? your usual help is appreciated Hello.. I'm a somewhat beginner in php, but i decided to create my own basic forums system with a content managing system. Right now I'm trying to figure out how to make a thread link be at the top of the list for the forum display page, when someone posts a response or creates a new thread. HELP MUCH APPRECIATED I have database "raj" with table "pagination"
In table pagination have "id", "actualtime" and "created" field
id - auto_increment actualtime- varchar created- datetime
that looks like this CREATE TABLE pagination( id int auto_increment, actualtime varchar(55), created datetime )
I want to display all rows which are created on todays date and will display from today upto yesterday at 06:00 pm. after that the content will be refreshed based on based on a DATETIME field called 'created' that holds the date and time of each record's creation.
this is my query to fetch rows but it display value after 06:00 pm on yesterday but i want to display all data before 06:00 pm from currentdate. After 06:00 pm data will be refreshed and clear. plz help me......
SELECT actualtime FROM pagination WHERE created BETWEEN date_add(date_sub(curdate(), INTERVAL 1 day), INTERVAL 18 hour) AND curdate()";
This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=310181.0 guys, please help i have table with datas like speed of vehicle, position e.t.c from morning 8 a.m. to 8 p.m. I need to get these details, but after every 15 minutes i.e. after selecting datas at 8 a.m. it shd select datas @ 8.15 a.m. then 8.30 a.m. hw can i write a mysql query for this ? or a PHP approach is appreciated Hi everyone - this code is doing my head in! You might want to take a look at the page as it will help me explain it better (www.bradleystokejudoclub.co.uk/inttest.php Basically what I have is a database with the results from international competitions, and I am trying to build a kind of search for it. I have 3 drop down boxes, one with player names, one with competiton and one with year. The only one that works is the competiton. I can do competition + year, but not year by itself, and player doesnt work at all .... this is the relevant code: (before head) <?php include("includes/dbconnect120-gem.php"); include("includes/db_auth_bits.php"); include("includes/db_stp.php"); if($_POST) { if($name == 'select') { $sql1 = ""; } else { if(($award == 'select') && ($year == 'select')) { $sql1 = "r.pname_id = '$pname'"; } else { $sql1 = "r.pname_id = '$pname' AND "; } } if($award == 'select') { $sql2 = ""; } else { if($year == 'select') { $sql2 = "r.comp_id = '$comp'"; } else { $sql2 = "r.comp_id = '$comp' AND "; } } if($year == 'select') { $sql3 = ""; } else { $sql3 = "r.year_id = '$year'"; } if(($sql1 == "") && ($sql2 == "") && ($sql3 == "")) { $where = ""; } else { $where = " WHERE "; } $sql = "select p.pname, i.comp, m.place_name, yr.year_full from intcomp_result r left join playername p on r.pname_id=p.name_id left join intcomp i on i.comp_id = r.comp_id left join place m on m.place_id = r.place_id left join yearname yr on r.year_id = yr.year_id $where $sql1 $sql2 $sql3 order by r.year_id desc, r.comp_id, r.place_id"; $search_result = mysql_query($sql); } ?> (body) <form method="post" action="<?php echo $PHP_SELF;?>"> Name:<select name="name"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select distinct r.pname_id, p.pname from intcomp_result r left join playername p on r.pname_id=p.name_id order by p.pname"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $nameid=$row['pname_id']; $pname=$row['pname']; ?> <option value="<?php echo $nameid; ?>"><?php echo $pname;?></option> <?php } ?> </select><br /> Competition:<select name="comp"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select * from intcomp order by comp_id asc"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $compid=$row['comp_id']; $comp=$row['comp']; ?> <option value="<?php echo $compid; ?>"><?php echo $comp;?></option> <?php } ?> </select><br /> Year:<select name="year"> <option value="select" selected="selected" >-SHOW ALL-</option> <?php $sql="select distinct r.year_id, y.year_full from intcomp_result r left join yearname y on r.year_id=y.year_id order by y.year_id"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $yearid=$row['year_id']; $year=$row['year_full']; ?> <option value="<?php echo $yearid; ?>"><?php echo $year;?></option> <?php } ?> </select><br /> <input name="Submit" type="submit" class="button" tabindex="14" value="Submit" /> </form> <?php if(isset($search_result)) { while($row = mysql_fetch_array($search_result)) { echo $row['pname'].' - '.$row['comp'].' - '.$row['year_full'].' - '.$row['place_name'].'<br />'; } } ?> Hope you can help! Thanks Gem Hey there, i have a cookie which echos ok but would like to know how to automatically select the corresponding option from a drop down list when the page loads. Normal List <form id="switchform"> <select name="switchcontrol" size="1" class="topusernav" onChange="chooseStyle(this.options[this.selectedIndex].value, 60)"> <option value="none">-----</option> <option value="noir">Noir</option> <option value="crimson">Crimson</option> <option value="forrest">Forrest</option> <option value="ocean">Ocean</option> <option value="petal">Petal</option> </select> </form> works fine but <form id="switchform"> <select name="switchcontrol" size="1" class="topusernav" onChange="chooseStyle(this.options[this.selectedIndex].value, 60)"> <option value="none" <?php if (!(strcmp("none", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>-----</option> <option value="noir" <?php if (!(strcmp("noir", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Noir</option> <option value="crimson" <?php if (!(strcmp("crimson", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Crimson</option> <option value="forrest" <?php if (!(strcmp("forrest", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Forrest</option> <option value="ocean" <?php if (!(strcmp("ocean", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Ocean</option> <option value="petal" <?php if (!(strcmp("petal", "$_COOKIE[\"mysheet\"];")) {echo "selected=\"selected\"";} ?>>Petal</option> </select> </form> gives me a whitespace error any thoughts?? thanks Hey guys i am new around here and i need some help with php. So basically i want to make a submit form and i found a problem for myself here's some images to understand what i am talking about. so here's what i've made at first. This is meant for new data. <p><label>Status :</label> <select name="status" > <option value=""></option> <option value="Ongoing">Ongoing</option> <option value="Completed">Completed</option> </select> basically what I want is when I edit the data, that selection returns the value that was stored in the database rather than just null value. See image below to understand what i am trying to do. i dont really get how to do it with select. Please help me and thanks before. Sorry if my question was already asked before. Tried to search but didnt really found what i am looking for. Hi, I'm no pro at PHP but I am trying to get a drop down menu to a authenticate before moving to the next part of the form. What I want is once a selection has been made, ONLY THEN can the user move on, OTHERWISE a message echo appears. This is the html menu box <select size="1" name="title"> <option>Please Select</option> <option value="Mr">Mr</option> <option value="Mrs">Mrs</option> <option value="Miss">Miss</option> <option value="Ms">Ms</option> <option value="Dr">Dr</option> </select> Then this is what I have in the form PHP: $visitortitle = $_POST['visitortitle']; if ( HOW DO I GET THIS PART TO AUTHENTICATE AN OPTION HAS BEEN SELECTED? ) { echo "<p>Please enter a title correctly<br />before you try submitting the form again.</p>\n"; die ( '<a href="pef.html">click here go back and try again</a>' ); echo $id;} If anyone can help me sort out this part of the form I can move on as the rest is working fine? Thanks Gary Hello. I have a mysql database with an 'entrytime' field which contains the unix timestamp the record was added. How do I create a mysql query to select all records for a given month/year? For example, if I wanted to create a query to display all records for October 2010 how do I go about it? I am at a loss and can't find my answer via google. Thanks in advance for any help! I am using jquery .change function to perform an operation when a month is selected from a drop down menu. The change works but I am unable to update the value of the drop down menu with the updated month. My drop down shows the starting value as default even on change. Can anyone help. Following is the code snippet that does change and then the drop down menu form. Code: [Select] $("#monthName").change(function() { alert($("#monthName").val()); if ($("#post").val() == 1) { $("#monthselect").submit(); } }); Code: [Select] <form id="monthselect" action="<?=$_SERVER['PHP_SELF']?>" method="get"> <input id="post" type="hidden" name="post" value="1"> <label>SELECT MONTH</label> <select id="monthName" name="monthName"> <option value="January">January</option> <option value="February">February</option> <option value="March">March</option> <option value="April">April</option> <option value="May">May</option> <option value="June">June</option> <option value="July">July</option> <option value="August">August</option> <option value="September">September</option> <option value="October">October</option> <option value="November">November</option> <option value="December">December</option> </select> </form> Even after the change, January shows up by default even if I select say June or July. I tried something like following but did not work. Code: [Select] $("#monthName option[value=" + $("#monthName").val() +"]").attr("selected","selected") ; the same page? I am trying to update a mysql table called AvItems with the value 'Torso' in the Equip "section?" I have been through the forums and cannot see anything to match. I dont mind if the page looses the onsubmit() and has a button instead. Though I would like to update the database and link back to the same page: There is a display that shows the item that is currently equiped, I have put this in to show it works, or doesn't as the case may be. Hope I got the code /code right this time. many thanks in advance Andy Curtis Code: [Select] create table Items( ItemID integer unsigned auto_increment primary key, ItemName varchar(20) not null, Type varchar(10), UsedOn varchar(10), ); create table AvItems( AvItemID integer unsigned auto_increment primary key, AvID integer unsigned, ItemID integer unsigned, Equip varchar(8)); <?php $username="root"; $password="MyPassword"; $database="MyDataBase"; $AvName = "AndyJCurtis"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $AvAccR = mysql_query( " SELECT AvID FROM AvAcc WHERE AvName = '$AvName' " ); $AvID = mysql_result($AvAccR, 0, 'AvID'); /////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// $Torso = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' AND UsedON = 'Body' "); $TorsoE = mysql_query(" select ItemName from AvItems, Items where AvItems.itemID = Items.itemID AND AvItems.AvID = '$AvID' And UsedON = 'Body' AND Equip = 'Body' "); if(mysql_num_rows($TorsoE) != 0) { $TorsoItem = mysql_result($TorsoE ,0,"ItemName"); //mysql_close(); ?> <title></title> <head></head> <body> <form action="http://localhost/CI/Equip2.php" method="post"> <table border=1> <tbody> <tr> <td>Torso<BR> <?PHP echo "$TorsoItem <BR>"; ?> <select name="Torso" onchange="submit();" value =" Update"> <?PHP while($TorsoRow = mysql_fetch_array($Torso)) { echo "<option value=\"".$TorsoRow['ItemName']."\">".$TorsoRow['ItemName']."\n </option>"; } ?> </select> </td> </tr> //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// <?php if($_POST['Torso'] == 'Update') { mysql_query("update AvItems set Equip = '' where Equip='Torso'") or die("cant update unequip"); mysql_query("update AvItems set Equip = 'Torso' where ItemID='{$_POST['ItemName']}'") or die("cant update equip"); } ?> /////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// </tbody> </table> </.form> </body> </html> Hi, I am trying to create a drop down menu that will select a value that is stored in the database - right now the code creats a dropdown (with nothing selected) - hope someone can help. in the database, the values are stored as --null- Option1 Option2 Option3 my code is Code: [Select] $instruction = $_GET['instruction']; <?php <select id="instruction" name="instruction"> <option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option> <option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1') echo 'selected = "selected"'; ?>>Option1</option> <option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2') echo 'selected = "selected"'; ?>>Option2</option> <option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3') echo 'selected = "selected"'; ?>>Option3</option> </select> <? Hello guys I've hit a problem whicle trying to validate my form. I have 3 drop down boxes where the user chooses from three options. But I can't seem to figure out how to set the validation so the user does not select the same option in each drop down. Can anyone help me solve this please. Thank you. <p><b>Course Choice 1</b> <select name="course1"> <option value="0"></option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <p><b>Course Choice 2</b> <select name="course2"> <option value="Leave Blank">Leave Blank</option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <p><b>Course Choice 3</b> <select name="course3"> <option value="Leave Blank">Leave Blank</option> <option value="Business computer Systems">Business computer Systems</option> <option value="Business computer Science">Business computer Science</option> <option value="Business computer Science (Games)">Business computer Science (Games)</option> <option value="Business Information Systems">Business Information Systems</option> <option value="Digital Media Development">Digital Media Development</option> <option value="Digital Media">Digital Media</option> </select></p> <div align="centre"><input type="submit" name="submit" value="send request" /></div> //Validate course choice 1 if (!empty($_REQUEST['course1'])) { $course1 = $_REQUEST['course1']; } else { $course1 = NULL; echo '<p><font color="red">Please enter your first choice</font></p>'; } //Validate course choice 2 if (!empty($_REQUEST['course2'])) { $course2 = $_REQUEST['course2']; } else { $course2 = NULL; echo '<p><font color="red">Please enter your second choice</font></p>'; } //Validate course choice 3 if (!empty($_REQUEST['course3'])) { $course3 = $_REQUEST['course3']; } else { $course3 = NULL; echo '<p><font color="red">Please enter your third choice</font></p>'; } //If everything is ok, print the message if ($name && $email && $course1 && $course2 && $course3) { echo "<p>Thank you, <b>$name</b>, You have chosen the following courses for information:<br /><br /> <b>$course1</b><br /> <b>$course2</b><br /> <b>$course3</b></p> <p>We will reply to you at <i>$email</i>.</p>\n"; } else { // One form element was not filled out properly echo '<p><font color="red">Please go back and fill out the form again.</font></p>'; } Hi there, i am relatively new to php, mysql, css etc but learning fast. My problem is such; i have a php file which is doing a SELECT mysql_query, WHILE results to strings, then ECHO the resulting rows to produce a list formatted using <table> and finally this <table> is inside a <form> which will POST the changes back to the specific database.tble.row. I wish to have a drop down menu within the <form><table> which will be populated from a separate database.table. I have accomplished the drop down menu outside the <?php ?> tags inside <form><table> which POSTS to a php file but my problem is to add the populating drop down menu inside <?php ?> an already ECHOing resulting rows from the sql query. i.e <?php blurb and stuff ?> <form><table><tr><td> <select name etc> <?php $result = mysql_query("SELECT * FROM tbl WHERE string = tble.rw ORDER BY column"); while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['column'].">".$row['column']."</OPTION>"; } ?> </select> WORKS!!!! but placing this inside <?php $x =mysql_query[select] while {strings = conditions; echo ("<form><table><tr><td> insert populated drop menu here </td> etc "); echo"";}?> doesnt work and just leaves the select drop menu blank Hoep you understand my problem. I do not think i can attached the population WHILE loop to a string and just insert the string to the form but maybe i am wrong. thanks in advance and if you go tthis far reading you must be on lots and lots of coffee zark I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue... Here is my select box code which is working perfect: <select name="city"> <?php $sql = "SELECT id, city_name FROM cities ". "ORDER BY city_name"; $results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!"); while($row = mysqli_fetch_array($results_set)) { echo "<option value=$row[id]>$row[city_name]</option>"; } ?> </select> Are any variables defined in a while loop global to the while loop only? Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form. Any ideas for a workaround to get this value passing as normal? if (isset($_POST['addPosting'])) { $query = "INSERT INTO Postings (id, city_id, title, description) VALUES ('','$row[id]','$_POST[title]','$_POST[description]')"; Hello, I have the following function function make_agent_drop($dropname,$parent=''){ $agents = mysql_query("SELECT * FROM ad_category WHERE cat_status='1' AND parent_id='".(is_numeric($parent)?$parent:"0")."'") or die(mysql_error()); $anum = mysql_num_rows($agents); if($anum>0){ $agentdrop='<select style="width:150px; height:20px; margin-left:100px; font-size:11px;" name="'.$dropname.'" id="'.$dropname.'" class="text" '.(is_numeric($parent)?'':'onchange="update_subcatdrop($(this).val());').'"> <option value="0">Select a Category</option>'; while($row= mysql_fetch_array($agents)){ $agentdrop.='<option value="'.$row['cat_id'].'">'.$row['cat_name'].'</option>'; } $agentdrop.='</select>'; }else{ $agentdrop= 'No '.(is_numeric($parent)?'Sub':'').'Categories Found.'; } return $agentdrop; ; } I creates a drop down from database cats and sub cats.. I am trying to figure out how to add a submit button to dynamically appear when it displays the sub category... Thanks! Dan |
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