PHP - How Can I Keep Values Of This Type Drop Down List Has It Has 3 Selects For Date
Hi i have this drop down list for date which contain 3 selects DAY MONTH YEAR hwo can i make so that when update form select keeps the same value has before help please
<?php $months = array('','January','February','March','April','May','June','July','August','September','October','November','December'); echo '<select name="month_of_birth">'; for ($i=1;$i<13;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $months[$i] . '</option>'; } echo '</select>'; echo '<select name="day_of_birth">'; for ($i=1;$i<32;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $i . '</option>'; } echo '</select>'; echo '<select name="year_of_birth">'; $year = date("Y"); for ($i = $year;$i > $year-50;$i--) { $s = ($i == $year)?' selected':''; echo '<option value="' . $i . '" ' . $s . '>' . $i . '</option>'; } echo '</select>'; ?> Similar TutorialsHi i have a drop down menu for date which is meant to insert all 3 values into a date column bt is only sending the year how can i fix it <select name="date_of_birth"> <option value="1">January <option value="2">February <option value="3">March <option value="4">April <option value="5">May <option value="6">June <option value="7">July <option value="8">August <option value="9">September <option value="10">October <option value="11">November <option value="12">December </select> <select name="date_of_birth"> <option value="1">1 <option value="2">2 <option value="3">3 <option value="4">4 <option value="5">5 <option value="6">6 <option value="7">7 <option value="8">8 <option value="9">9 <option value="10">10 <option value="11">11 <option value="12">12 <option value="13">13 <option value="14">14 <option value="15">15 <option value="16">16 <option value="17">17 <option value="18">18 <option value="19">19 <option value="20">20 <option value="21">21 <option value="22">22 <option value="23">23 <option value="24">24 <option value="25">25 <option value="26">26 <option value="27">27 <option value="28">28 <option value="29">29 <option value="30">30 <option value="31">31 </select> <select name="date_of_birth" id="year"> <?php $year = date("Y"); for($i=$year;$i>$year-50;$i--) { if($year == $i) echo "<option value='$i' selected>Current Year</option>"; else echo "<option value='$i'>$i</option>"; } ?> Hi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Ok basic setup is Table name = unit_data I have a field unit_paid_date colum type = DATE unit_paid_date = 2011-03-02 thats yyyy-mm-dd lets say for example the DueDate is the Second of every month I need to select all records from unit_data that are PAST DUE so if the last paid date is 2011-01-02 that record will pop up as PAST DUE BUT if that paid date is say 2011-03-10 it will not be shown because the invoice was paid Ahead of due date I had this almost working properly -- but can use any help you guys offer. I was attempting to use mysql WHERE queries and do checks against paid_date but failed Thanks I have a dropdown list which have some elements. I want to get all the elements of the dropdown list into a variable of type array. Any Idea??? Hi there, i'm trying to sum up values from mysql based on their date, specifically only values from entries with the same date should be added together, my db looks like this: id, codes, value, date now i'm building a statistics page, the value field is always 50, for showing the total for all the entries i'm just doing Code: [Select] $totalincome = $codes * 50; and print that on the statistic, i want to have statistics for each day, the date is saved in this format: YYYY-MM-DD how could i do that ? Thanks ! I've got two tables (classOfferings, instructors). The fields I'm dealing with are 'co.instructorId', 'i.instructorId', 'i.fName', 'i.lName'. I have a form with dynamically generated drop downs. What I would like to do is check classOfferings table to see which instructors are teaching classes then display them in drop down with the 'instructorId' as the value and 'fName' and 'lName' as the user selectable part. So I have found the distinct 'instructorId', but I can't make 'fName' and 'lName' appear as the user selectable part. The code below produces a drop down which has invisible values, but still posts a value. <select size="1" name="instructor"> <option value="" selected>Search By Teacher...</option> <? $instrList=mysql_query("select distinct instructorId from classOfferings order by instructorId asc"); $instrNameList=mysql_query("select fName, lName from instructors where classOfferings.instructorId = instructors.instructorId order by lName asc"); // Show records by while loop. while($instructor_list=mysql_fetch_assoc($instrList)){ $instrNames = ($instr_Name['fName']) . ($instr_Name['lName']); ?> <option value="<? echo $instructor_list['instructorId']; ?>" <? if($instructor_list['instructorId']==$select){ echo "selected"; } ?>> <? echo $instrNames; ?></option> <? // End while loop. } ?> </select> Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. <?php // Daniel URL duplicator. // Input links list. $linksList = "links.txt"; // How many times to duplicate the url? $manyTimes = 1000; // Read in the list. for ($x = 0; $x <= $manyTimes; $x++) { $handle = fopen($linksList, "r"); $line = fgets($handle); echo $line; fclose($handle); } ?>Hey Guys, I'm stuck on this simple bit of code lol what I'm trying to do is load in a list of urls: site1.com site2.com site3.com etc For each site that is read, I'm trying to duplicate it X times, above would print to screen the same url 1000 times, then move onto the next print it 1000 times etc until the list is done (or how ever many times I select) I can't think of the best way to do it! any help would be appreciated guys! Graham Hello, I have a datepicker that when a date is selected, the post to a php file will get the sum values for that selected day, month, and year. I have the "by day" working with this... <?php $choice = (isset($_POST['choice'])) ? date("Y-m-d",strtotime($_POST['choice'])) : date("Y-m-d"); $con = mysql_connect("localhost","root","xxxxxxxxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("inverters", $con); $sql = "SELECT sum(power/1000) AS choice FROM feed WHERE date = '".$choice."' group by date"; $res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error()); $row = mysql_fetch_assoc($res); echo $row['choice'].'<br />'; ?> But by month and year are not clicking... Help Please Alan i'm trying to print out values from a table into a drop down box, but my code doesnt seem to work. when i test it, there are no values in the drop down box. if someone could find the problem, that would be great. <form name="add_sub_section" method="post" action="<?php echo $_SERVER['../PHP_SELF']; ?>"> Sub Section Name <input type="text" name="sub_section_title" maxlength="200" /><br/><br/> Sub Section desc <textarea name="sub_section_desc" rows="4" columns="30" /></textarea><br/><br/> SECTION <?php $list = "SELECT section_title FROM section_main"; echo "<select name='section_title'>"; while ($a = mysql_fetch_row($list)) { for ($j = 0; $j < mysql_num_fields($list); $j++) { echo "<option value=". $a[$j] . ">". $a[$j] . "</option>"; } } echo "</select>"; ?> <br/><br/> <input type="submit" name="submit" value="Add Section"/> </form> the connection to the table works, so i havnt put that code in. there's no formatting for the html yet, i just want to get the php working first Thanks 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. I have a "select"-drop down bar and I want to have a numbered list in it, i've tried but it doesn't seem to be possible. Is there any way that i'm able to do this? Hi ,
I am pretty new to php. I would like to create a php website based on folders.
I have a code so far that shows the names of folder in a selection box. What I would like to achieve now is how i get the images inside the folder after I click on submit. I know how to get the images with a glob function. What I don't know is how to get the value of each specific folder name. How would I do that?
I have this code so far to get the names of the folders in the selection box:
<form action="test.php" method="post"> <select name="myDirs"> <option value="" selected="selected">Select a genre</option> <?php $dirs = glob("*", GLOB_ONLYDIR); foreach($dirs as $val){ echo '<option value="'.$val.'">'.$val."</option>\n"; } ?> </select> <input type="submit" name="submit2"> <?phpif (isset($_POST['submit2'])) Helllo Every1 well im kinda new to php and i needed some help i was working on a page all my text boxes and check boxes are at the bottom of my file and like this //====================================================================== </script> </head> <body> <div style="margin-left: 170px"> <input type="checkbox" id="getitems" checked value="1">Run Plugin? <input type="checkbox" id="sellitm" checked value="1">Sell Items?<br><br> Item Name:<br><input type="text" id="itemd" value=""><br><br>How Many Cycles?:<br><input type="text" id="runtm" value="10"></div> <div style="margin-left: 170px"><br> <button id="btn_save" style="color:white;background-color:#00660F;border-width:1px;border-style:solid; "> Save settings </button> </div> '; echo $this->ObjectTable(); echo ' </body> </html> '; } } ?> and well i decided to do away wit one of the text boxes and use a drop down list instead that is populate from a text file and the only code that i could find that i was able to get working was this 1 <?php $text = file_get_contents("itemlist.txt"); $array = explode("\n",$text); echo "<select>"; foreach ($array as $value) { echo "<option value='$value'>$value</option>"; } echo "</select>"; and the problem im having is with that one it just stays at the top of the page when loaded like i have no way to position it... and im really stuck i tried saving it in another php and i tried using <?php include(); ?> function but it did not work if any1 could help me out that would be awsome. T.I.A Hi all, i want my list/menu field values to come from my database. how can i accomplish that? thanks i did Code: [Select] <select name="select"> <option value="0">--select below--</option> <option value="1">Me</option> <?php require_once '../konnect/konex.php'; $result = mysql_query("SELECT * FROM is_clients"); while($row = mysql_fetch_array($result)) { echo "<option value ='2'>".$row"['name']</option>"; echo "<br />"; } ?> </select> Hi all.
how can i make the values show like a list. I tried html line break "<br>" and php \n but all to no avail. It just show all the values in one straigth line.
example of what i want is for the values to appear like this:
1234567890
0987654345
4567890675
instead of :
1234567890 0987654345 4567890675
Thanks
<form data-abide method="post" action=""> <div> <select name=""> <option value="name"> <?php $stmt = $pdo->query("SELECT acct_num FROM table order by id desc"); while ( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) { echo $row['acct_num']; } ?> </option> </select> </div> <div> <label>New Password <small>required</small></label> <input type="password" name="password" id="password" required> <small class="error">New password is required and must be a string.</small> </div> <div> <label>Confirm New Password <small>required</small></label> <input type="password" name="password2" id="password2" required> <small class="error">Password must match.</small> </div> <input name="submit" type="submit" class="button small" value="Change Password"> </form> Edited by Mr-Chidi, 13 November 2014 - 01:34 AM. hello there.. i have a problem with my php coding where i want to keep date choose by user in the database. this is the drop down date Code: [Select] <select name="Date_Day"> <option> - Day - </option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> <option value="11">11</option> <option value="12">12</option> <option value="13">13</option> <option value="14">14</option> <option value="15">15</option> <option value="16">16</option> <option value="17">17</option> <option value="18">18</option> <option value="19">19</option> <option value="20">20</option> <option value="21">21</option> <option value="22">22</option> <option value="23">23</option> <option value="24">24</option> <option value="25">25</option> <option value="26">26</option> <option value="27">27</option> <option value="28">28</option> <option value="29">29</option> <option value="30">30</option> <option value="31">31</option> </select> <select name="Date_Month"> <option> - Month - </option> <option value="01">January</option> <option value="02">Febuary</option> <option value="03">March</option> <option value="04">April</option> <option value="05">May</option> <option value="06">June</option> <option value="07">July</option> <option value="08">August</option> <option value="09">September</option> <option value="10">October</option> <option value="11">November</option> <option value="12">December</option> </select> <select name="Date_Year"> <option> - Year - </option> <option value="2010">2010</option> <option value="2011">2011</option> <option value="2012">2012</option> <option value="2013">2013</option> <option value="2014">2014</option> <option value="2015">2015</option> <option value="2016">2016</option> <option value="2017">2017</option> <option value="2018">2018</option> <option value="2019">2019</option> <option value="2020">2020</option> <option value="2021">2021</option> <option value="2022">2022</option> <option value="2023">2023</option> <option value="2024">2024</option> <option value="2025">2025</option> <option value="2026">2026</option> </select> the code to connect to the database Code: [Select] $date_year= ($_POST['Date_Year']); $date_month=($_POST['Date_Month']); $date_day=($_POST['Date_Day']); $date=$date_year."-".$date_month."-".$date_day; $query="INSERT INTO aduan (date) VALUES ('date($date)')"; $result=mysql_query($query); if($result){ echo 'Registration success.'; ?><script>window.location ='thanks.php'</script> <?php } else echo 'Registration failed';} when enter a value of date, the database will just show '0000-00-00'.. really hope for your help.. hello fellas, need some help please if possible. i have created a date of birth section in my form where the user selects his/her date of birth from the dropdown menu. they would first select the day then month then year of their birthday. how would i setup the database to get this to work? i currently have: Code: [Select] day VARCHAR( 2 ) NOT NULL , month VARCHAR( 4 ) NOT NULL , year VARCHAR( 4 ) NOT NULL , is this correct? many thanks 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_lineThe 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, |