jdorma0 Posted March 11, 2010 Share Posted March 11, 2010 Basically, I need to prefill a select drop down box with the sunday of each week. I would like to it to basically be in the format of: <select name="row_date"> <option value="2010-01-01">2010-01-01</option> <option value="2010-01-08">2010-01-08</option> <option value="2010-01-15">2010-01-15</option> <option value="2010-01-22">2010-01-22</option> etc. etc. </select> But it would automatically select the current week (as the default value) and allow the user to change that.. but only from the displayed dates. I wouldn't need more than 6 weeks to be shown, but that's not too important. I've researched this forum, php.net and Google to no avail. Hopefully one of you bright people can help me out! Thanks! =] Joey Quote Link to comment https://forums.phpfreaks.com/topic/194859-display-a-certain-list-of-days-using-date/ Share on other sites More sharing options...
mikesta707 Posted March 11, 2010 Share Posted March 11, 2010 $time = strtotime("next sunday"); $week = 60*60*24*7;//60 secs a min * 60 mins an hour * 24 hours a day * 7 days a week for ($i = 0; $i < 10; $i++){ echo date("Y-m-d", $time) . "<br />"; $time += $week; } that should help get you started. its gets the next sunday of the week, and outputs the next 10. its pretty simple logic wise, but you may want to put in a check to see if its a sunday, if you want to list the current day if its a sunday. Quote Link to comment https://forums.phpfreaks.com/topic/194859-display-a-certain-list-of-days-using-date/#findComment-1024564 Share on other sites More sharing options...
jdorma0 Posted March 11, 2010 Author Share Posted March 11, 2010 $time = strtotime("next sunday"); $week = 60*60*24*7;//60 secs a min * 60 mins an hour * 24 hours a day * 7 days a week for ($i = 0; $i < 10; $i++){ echo date("Y-m-d", $time) . "<br />"; $time += $week; } that should help get you started. its gets the next sunday of the week, and outputs the next 10. its pretty simple logic wise, but you may want to put in a check to see if its a sunday, if you want to list the current day if its a sunday. Thanks so much! With a little modification: $week = 60*60*24*7;//60 secs a min * 60 mins an hour * 24 hours a day * 7 days a week $time = strtotime("last sunday"); for ($i = 0; $i < 10; $i++){ $etime = "<option value=\"".date("Y-m-d", $time)."\""; if($i==1) { $etime .="selected"; } $etime .= ">".date("Y-m-d", $time) . "</option>"; echo $etime; $time -= $week; } It works like a charm! I appreciate it. Quote Link to comment https://forums.phpfreaks.com/topic/194859-display-a-certain-list-of-days-using-date/#findComment-1024566 Share on other sites More sharing options...
mikesta707 Posted March 11, 2010 Share Posted March 11, 2010 no problem. if your topic is solved you can click the mark solved button at the bottom of the thread Quote Link to comment https://forums.phpfreaks.com/topic/194859-display-a-certain-list-of-days-using-date/#findComment-1024567 Share on other sites More sharing options...
jdorma0 Posted March 11, 2010 Author Share Posted March 11, 2010 no problem. if your topic is solved you can click the mark solved button at the bottom of the thread Yes sir. I know I should start another topic for this, but maybe you will know the answer. I need to prefill a select field for each row of my form. echo "<div align=\"center\"><font color=\"red\"><b>WARNING:</b> Only use the results for the week ending on <b>".$vars['emp_date']."</b></font></div><br />"; ?> <table align="center" class="style6"> <tr> <th>Employee</th> <th>Star</th> <th>CSAT</th> <th>CPH</th> <th>PA</th> <th>Credit</th> <th>Attachments</th> </tr> <?php echo '<form action="insert.php" method="post">'; $i=0; while($i<$vars['emp_num']) { $emp = "<tr><td><select name=\"row_associd_$i\">"; $res = mysql_query("SELECT * FROM employees WHERE department='".$_POST['emp_department']."'"); $emp .= "<option \"\">Select Employee</option>"; while($emps = mysql_fetch_array($res)) { $emp .= "<option value=".$emps['associd']; // $emp .=" selected"; // Need to decide how to get this to output, skipping the last person $emp .=">".$emps['fname']." ".$emps['lname']."</option>"; } $emp .= "</select></td>"; $emp .= "<td><input type=\"text\" size=\"5\" name=\"row_star_$i\">"; $emp .= "</td><td><input type=\"text\" size=\"5\" name=\"row_csat_$i\">"; $emp .= "</td><td><input type=\"text\" size=\"5\" name=\"row_cph_$i\">"; $emp .= "</td><td><input type=\"text\" size=\"5\" name=\"row_pa_$i\">"; $emp .= "</td><td><input type=\"text\" size=\"5\" name=\"row_credit_$i\">"; $emp .= "</td><td><input type=\"text\" size=\"5\" name=\"row_attachments_$i\">"; $emp .= "</td></tr>"; echo $emp; $i++; } ?> </table> <input type="hidden" name="emp_i" value="<?php echo $i;?>"> <input type="hidden" name="action" value="emp_processrows"> <input type="submit" name="emp_submit" value="rank employees"> </form> Basically, I would like to have selected each person for the row. Example: Bob Jones - selected Mary Beth - not selected Frank Sinatra - not selected Bob Jones - not selected Mary Beth - selected Frank Sinatra - not selected Bob Jones - not selected Mary Beth - not selected Frank Sinatra - selected I was playing around with it but I couldn't get it figured out. I'd appreciate it if you could help me with that one too. Quote Link to comment https://forums.phpfreaks.com/topic/194859-display-a-certain-list-of-days-using-date/#findComment-1024568 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.