PHP - Option Retaining Values In Text Box Help!!
Hi all,
i need some help in retaining the values in an <option> textbox.. I have found a code to loop the date online, and use it for Javascript jump menu, but i have no ideas how to retain the option.. here are my codes(in red), $year1 = date('Y'); $month1 = date('m'); $day1 = date('j'); $day1 = $day1 + 1; $year2 = date('Y') + 2; $month2 = date('m'); $day2 = date('j'); $start_date = "$year1-$month1-$day1"; $end_date = "$year2-$month2-$day2"; $date = mktime(0,0,0,$month1,$day1,$year1); //Gets Unix timestamp START DATE $date1 = mktime(0,0,0,$month2,$day2,$year2); //Gets Unix timestamp END DATE $difference = $date1-$date; //Calcuates Difference $daysago = floor($difference /60/60/24); //Calculates Days Old $i = 0; while ($i <= $daysago +1) { if ($i != 0) { $date = $date + 86400; } else { $date = $date - 86400; } $yy1 = date('Y',$date); $mm1 = date('m',$date); $dd1 = date('d',$date); $date3 = "$yy1-$mm1-$dd1" ; echo "<option value=\"test1234.php?m=$mm1&y=$yy1&d=$dd1\" "; echo ">".$date3 ."</option>"; $i++; } i know that i have to something selected, but i dun understand how this set of date code loop which left me helpless. Similar TutorialsHi I have a list of states using the array method in a form. The drop down menu works fine. I want to save the user choice,if the form is re-displayed due to a blank field or pattern mismatch. I know I can use the selected=selected, but don't know wher to put the statement: My array is: state_province = array ("list of states", "provinces") Var in my labels array is "state"=>"state" Here is my code for the select/option statement: { if($field == "state") { echo "<div class='province_state'><label for='state' size='10'>* Province/State</label><select>"; foreach($state_province as $state) { echo "\n<option value='$state_province' /> "; echo $state ; echo "</option>"; } echo "</select></div>\n"; } ?Is this the correct code to add and where would I add it? if(@$_POST['state'] == $value) { echo "selected='selected' "; } I have a form and on each input I set the value as the post variable of it's self. I do this so that if the user submit the form and it has errors they haven't lost the data they have inputted. For example <input name="input1" type="text" value="<?php echo($_POST['input1']); ?>" /> This works fine for text field and textarea's but how do I retain the value of radio boxes and checkboxes? Hi guys, I had a problem before which is similar to the problem im about to describe, however Pikachu helped me solve the last problem so a big thanks to him! However this problem is slightly different and im struggling to find a solution. I will first describe the problem I had which pikachu helped me solve as it would be easier for me to describe this past problem. I have an update form, so a user can edit existing records which are pulled from a database. When the user changes the values in a text field then submits the form, the values in the text fields updates fine. However if they change the values in the text box and then tries to submit the form but there was an error, the values in the text fields go back to the values from the database and the user has to change the text fields again. Well this problem was solved with the help of pikachu and the this is what I did... <?php if(isset($_POST['title'])){echo htmlspecialchars($_POST['title']);} else if (isset($title)) {echo htmlspecialchars($title);}?> Again, the above works fine however the problem I have now is the same problem as described above however it relates to drop down menus and not text fields. I can't seem to figure out how to retain the values in a drop down menu the same way I have for text fields. Below is the code that im using that I thought would work however it isnt working, the values of the drop down menu (if there was an error) are going back to the values from the database. Heres the code... <option value="Psychopathic" <?php if ((isset($_POST['category']))&&($_POST['category'] == 'Psychopathic')) { echo ' selected=selected'; } else if ((isset($category))&&($category == 'Psychopathic')) { echo ' selected=selected'; } ?>>Psychopathic</option> Does anybody know what i am doing wrong and how i can fix this problem? I understand what i need to do but i cant seem to get the code to do what i want to do and thats to retain values of a drop down menu when the page is returned with an error. Any help would be much appreciated. AJay Hey, Im needing some help with an update form that I have created using PHP/MySQL. Basically the form is working great (updating records, retrieving records and showing errors) but there is a problem with the errors. If there are no errors and I edit the existing text input values, the database updates fine. However if I edit the existing text input values and there is an error, the edited input values dont save, they revert back to the values of the database. I understand why this is happening, its because the text inputs are set to show the values of the database so if i edit the existing values and an error occurs or the page is refreshed, the values revert back to the database values. However id like it so that if any text in the text field have been edited and an error occurs, the text stays to how it is until the user by passes any errors and then the database is updated. Hopefully this makes sense. I have attached the php page just incase anybody wants to look into it but any help would be much appreciated! Also, im not looking for someone to just to fix this for me but if someone knows my problem and can guide me in the right direction, that would be great. Hi guys, I am having a real problem with this and I can't figure it out. I am not that much of a techie so take it easy on me. I am building a survey form for a website to get feedback on the service, and naturally it can fail validation if the email address is misspelled or the security question is answered wrong. Retaining text values was easy and I managed to dig around for some solutions to retain radio button values. The problem is IT ONLY WORKS ON ONE OPTION! I tried everything including coming up with completely separate value names, yet the page still loads with all radio options set to the same shared value (usually of the last radio button question). Here is my code: <?PHP $Excellent_status = 'unchecked'; $VeryGood_status = 'unchecked'; $Satisfactory_status = 'unchecked'; $Weak_status = 'unchecked'; $Horrible_status = 'unchecked'; if (isset($_POST['send'])) { $selected_tk = $_POST['tk']; if ($selected_tk == 'Excellent') { $Excellent_status = 'checked'; } else if ($selected_tk == 'VeryGood') { $VeryGood_status = 'checked'; } else if ($selected_tk == 'Satisfactory') { $Satisfactory_status = 'checked'; } else if ($selected_tk == 'Weak') { $Weak_status = 'checked'; } else if ($selected_tk == 'Horrible') { $Horrible_status = 'checked'; } } $Excellent2_status = 'unchecked'; $VeryGood2_status = 'unchecked'; $Satisfactory2_status = 'unchecked'; $Weak2_status = 'unchecked'; $Horrible2_status = 'unchecked'; if (isset($_POST['send'])) { $selected_ts = $_POST['ts']; if ($selected_ts == 'Excellent') { $Excellent2_status = 'checked'; } else if ($selected_ts == 'VeryGood') { $VeryGood2_status = 'checked'; } else if ($selected_ts == 'Satisfactory') { $Satisfactory2_status = 'checked'; } else if ($selected_ts == 'Weak') { $Weak2_status = 'checked'; } else if ($selected_ts == 'Horrible') { $Horrible2_status = 'checked'; } } ?> <h1 align="center">Feedback Form:</h1> <p align="center">We highly value your input. Your feedback can help us serve you better and provide you with a holistic experience that meets or exceeds all your expectations. <br /> Please take the time to fill out the form below.</p> <br /> <br /> <p align="center"><strong>(Fields marked with a <span class="required">*</span> are necessary. The course information is to validate which course and batch the survey is regarding, and the name and email allows our admin to verify the details and gain more information from you if needed)</strong></p> <p align="center"><strong>The identity in this form is strictly confidential, and it will not be revealed to course trainer, co-ordinator, or your sponsor.</strong></p> <!--Error Message--> <?php echo $error;?> <form method="post" name="contFrm" id="contFrm" action=""> <label><span class="required">*</span>Name:</label> <input name="name" type="text" class="box" id="name" size="40" value="<?php echo $name;?>" /><br /> <label><span class="required">*</span>Email: </label> <input name="email" type="text" class="box" id="email" size="40" value="<?php echo $email;?>" /><br /> <label><span class="required">*</span>Course attended: </label> <input name="course" type="text" class="box" id="course" size="40" value="<?php echo $course;?>" /><br /> <label><span class="required">*</span>Course Date and City: </label> <input name="dateplace" type="text" class="box" id="dateplace" size="40" value="<?php echo $dateplace;?>" /><br /><br /><br /> <table width="980" border="1" cellpadding="5" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td width="280" align="left" bgcolor="#285A80"><strong><font color="#FFFFFF" size="+1">Attribute</font></strong></td> <td width="115"> <div align="center">EXCELLENT! </div></td> <td width="115"> <div align="center">Really Good. </div></td> <td width="115"> <div align="center">Satisfactory. </div></td> <td width="115"> <div align="center">Weak, needs attention. </div></td> <td width="115"> <div align="center">Horrible! Unacceptable. </div></td> </tr> <tr> <td width="280" align="left" bgcolor="#71C4CD"><strong><font color="#333333">Trainer Knowledge</font></strong></td> <td> <div align="center"> <input type="radio" name="tk" id="Excellent" value="Excellent" <?PHP print $Excellent_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="tk" id="VeryGood" value="VeryGood" <?PHP print $VeryGood_status; ?>/> </div> </label></td> <td> <div align="center"> <input type="radio" name="tk" id="Satisfactory" value="Satisfactory" <?PHP print $Satisfactory_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="tk" id="Weak" value="Weak" <?PHP print $Weak_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="tk" id="Horrible" value="Horrible" <?PHP print $Horrible_status; ?>/> </div> </td> </tr> <tr> <td width="280" align="left" bgcolor="#71C4CD"><strong><font color="#333333">Trainer Style and Presentation</font></strong></td> <td> <div align="center"> <input type="radio" name="ts" id="Excellent2" value="Excellent2" <?PHP print $Excellent2_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="ts" id="VeryGood2" value="VeryGood2" <?PHP print $VeryGood2_status; ?>/> </div> </label></td> <td> <div align="center"> <input type="radio" name="ts" id="Satisfactory2" value="Satisfactory2" <?PHP print $Satisfactory2_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="ts" id="Weak2" value="Weak2" <?PHP print $Weak2_status; ?>/> </div> </td> <td> <div align="center"> <input type="radio" name="ts" id="Horrible2" value="Horrible2" <?PHP print $Horrible2_status; ?>/> </div> </td> </tr> </table> <br /><br /> <label>Particularly positive points you appreciated: </label> <textarea name="message1" cols="80" rows="3" id="message1"><?php echo $message1;?> </textarea><br /> <label>Particularly negative points you found disappointing: </label> <textarea name="message2" cols="80" rows="3" id="message2"><?php echo $message2;?> </textarea><br /> <label>Any specific message you would like to provide us: </label> <textarea name="message3" cols="80" rows="3" id="message3"><?php echo $message3;?> </textarea><br /> <label><span class="required">* Security Question</span></label><? require_once ("ClassMathGuard.php"); MathGuard::insertQuestion(); ?> <br /> <br /> <!-- Submit Button--> <input name="send" type="submit" class="button" id="send" value="" /> </form> Fact is there are 8 questions (all multiple choice with radio button answers) but I decided to save you a long code read. Basically what I get is that despite choosing different answers for each question, the retention is only of the last option or question, and all the above question choices are set to that choice button. What is wrong? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=307346.0 I cannot figure out how to retain the inputted information once submit is pressed. I want the user to enter their first name, last name, hours of work, and hourly pay rate. Then, I want them to press submit. Their pay will be calculated and displayed, but I still want their inputted information to be visible. How can I do that?? Here is my code: Code: [Select] <html> <head> <?php $pay = $_POST['pay']; $hours = $_POST['hours']; $extra = $_POST['extra']; $rate = $_POST['rate']; if ($hours > 40){ $extra=$hours-40; $hours=40; $pay=($extra*1.5*$rate)+($hours*$rate); } else { $pay=$hours*$rate; } echo "Your pay is: <br /> $ $pay <br />" ; ?> </head> <body> <form method='post' action='pay2.php'> First Name: <input type='text' name='first'><br/> Last Name: <input type='text' name='last'><br/> Hours of Work: <input type='text' name='hours'><br/> Hourly Pay Rate: <input type='text' name='rate'><br/> <input type='submit' value='submit'> </form> </body> </html> Hello all. I'm new here... and new to php... so please be gentle if I'm overlooking something that's blindingly obvious. I have a 'view blog entries' page and an 'edit blog entries' page. When I go to the 'edit blog entries' page, I want the form fields to display the preexisting blog content for whatever entry I'm about to edit. The code below does display the 'title' but it does not display the 'article' content. (However, the 'title' and 'article' content both show when I view the source code.) Can anyone tell me what I could be doing wrong? Thanks in advance. Code: [Select] // Retrieve the blog post's information: $q = "SELECT title, article FROM entries WHERE article_id=$article_id"; $r = @mysqli_query ($dbc, $q); if (mysqli_num_rows($r) == 1) { // Valid blog entry ID, show the form. // Get the blog entry's information: $row = mysqli_fetch_array ($r, MYSQLI_NUM); echo '<form action="edit_entry.php" method="post"> <p>Title:<br /> <input type="text" name="title" size="45" maxlength="80" value="' . $row[0] . '" /></p> <p>Blog entry<br /> <textarea name="article" id="article" cols="45" rows="5" ' . $row[1] . '></textarea> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> <input type="hidden" name="article_id" value="' . $article_id . '" /> </form>'; } else { // Not a valid entry ID. echo '<p class="error">This page has been accessed in error.</p>'; } mysqli_close($dbc); ?> okay all i need is a way to make 4 <OPTION> values into a string which i can compare to a row in my db. This is my form. Code: [Select] <select name='n1'> <option value='0'>0</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> </select> <select name='n2'> <option value='0'>0</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> </select> <select name='n3'> <option value='0'>0</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> </select> <select name='n4'> <option value='0'>0</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> </select> here is the php code which i tried but doesnt seem to work. Code: [Select] $n1 = $_POST['n1']; $n2 = $_POST['n2']; $n3 = $_POST['n3']; $n4 = $_POST['n4']; $string = ("" . $n1 . "" . $n2 . "" . $n3 . "" . $n4 . ""); echoing $string does give to correct numbers.. but when i try and compare it to my database it doesnt match. all help would be great. Hi, I am developing a web application using PHP and MySQL. I am filling my combobox with the value fetched from the MySQL database. I want to retain the selected value of the combo box in postback. I am able to do it with the statiic values. But in this case I am binding the combbox dynamically with the values from the database. Can anybody please help me? Thanks, Regards, Viji how do I populate the textbox values in form.html? Thanks in advance for any help you might offer. Code: [Select] //formfill.php <?php $hours="akj"; //split string characters to array $hours_array = str_split($hours); //iterate through the characters in the array foreach($hours_array as $char) { echo($char."<br />"); //somehow populate text boxes in form.html } ?> //form.html <html> <body> <form> <table> <tr> <td><input type="text" name="hour[0]" value="" /></td> <td><input type="text" name="hour[1]" value="" /></td> <td><input type="text" name="hour[2]" value="" /></td> </tr> </table> </form> </body> </html> I'm stuck trying to figure out why when I make a selection in drop down list of a php form, the selection does not stay when form is refreshed or re-displayed due to an pattern error of another field. My array is state_province = array(list of the states, and provinces) field name in the label array is: "state" => "State", Here is the code that I'm using: echo "<form action='$_SERVER[PHP_SELF]' method='POST'>"; foreach($labels as $field => $label) { if($field == "state") { echo "<div class='province_state'><label for='state' size='15'>* Province/State</label><select>"; foreach($state_province as $state) { echo "<option value=$state"; if(@$_POST['state'] == $state) { echo "selected='selected'"; } echo ">"; echo $state; echo "</option>\n"; } echo "</select></div>\n"; } Hi everyone, I need to retain the checkbox selections after the form is submitted, but I'm stuck and wondering if anyone can offer some assistance? Thank you. <?php $categories = mysqli_query($db,'SELECT id, category FROM category'); if (!$categories) { exit('<p>Unable to obtain category list from the database.</p>'); } // Check if there are any categories and if so, list them. if (mysqli_num_rows($categories) != 0){ echo '<p>Select the related categories:<br />'; while ($category = mysqli_fetch_array($categories)) { $categoryid = $category['id']; $categoryname = htmlspecialchars($category['category']); $checked= ($category['id']==$categories) ? ' checked="checked"' : ''; echo '<input type="checkbox" name="categories[]" value="'.$categoryid.'" '.$checked.' > '.$categoryname.' <br/>'; } } ?>
my name is fairooj and ama new to php and jquery. i have a proplam. i want your help.
<script type="text/javascript"> var count = 0; $(function(){ $('p#add_field').click(function(){ count += 1; $('#container').append( '<strong>Link #' + count + '</strong><br />' + '<input id="field_' + count + '" name="fields[]' + '" type="text" />' + '<input id="code_' + count + '" name="code[]' + '" type="text" /><br />' ); }); }); </script> Ok from one brick wall to the next. After lots of help with my last query I moved onto the next task an I'm completely stuck again. I have a form thats created with a loop. Here is that form <form action="" method="post" id=""> <?php for ($i = 1; $i <= $totalRows_rs_cacheNum; $i++) { echo "<label>Cache ".$i."</label> <span id='sprytextfield".$i."'> <input name='cache[$i]' type='text' value='".$_POST['cache[$i]']."'/> <span class='textfieldRequiredMsg'>Required!</span></span> <div class='clear'></div>"; } ?> <input name="nextbtn" type="submit" value="Next" /> When a user enters data in all the fields, If one piece of information is incorrect an error pops up saying one of the fields you entered does not match the fields in the database. At the same time however i wish to keep the data they have entered in each of the fields. Normally you'd just write value="<?php echo $_POST['fieldName'];?>" but for some reason in this case it does not work. can someone point me in the right direction. I've tried Google many different terms but I can not find the correct term to find my answer. Hence me asking you guys again. Thanks Am new to php... I have been battling on my dynamic checkboxes in such a way that if none is checked the form is return, also I need to retain what was checked when the form postback due to other invalid inputs. Code: [Select] $result = mysql_query("SELECT * FROM course") or die(mysql_error()); if ($result) { while ($row = mysql_fetch_array($result)){ if (isset($_POST['courses']) and $_POST['courses'] == $row['cid']) {echo $row['cid'];} print "<input type=\"checkbox\" name=\"courses[]\" value=\"$row[cid]\">$row[cname]\n"; } } Help needed purely on php codes. Thanks in advance Hi there,
I am getting this in a PHP generated email:
ACWA Member/Subscriber: Melody’s Children’s Service
The line of code in the php is:
$subject = "ACWA Member/Subscriber: $membername";
I have tried htmlentitiies, html_entity_decode, htmlspecialchars but nothing seems to work.
I haven't used htmlentities anywhere else in the code for this variable.
Any suggestions?
All advice gratefully received.
I have a form where I have inserted 7 pre-populated relational lists. All of the information is pulling correctly from the databases, but when it posts, it's posting the value "ids" instead of the chosen text. The files a www.kcwell.com/gcc_form.php and www.kcwell.com/gccsuccess_form.php How do I set up a query to obtain the data that I need? Help! Code what i made so far.
$inputText = 'This is testing http://www.youtube.com'; $allowedDomains = 'www.google.com youtube.com/ http://www.test.org'; $array = preg_split('/[\s]+/', $allowedDomains); $regex = '';//Need this line if(preg_match($regex, $inputText)){ print 'Domain match!'; }else{ print 'Domain not match!'; } Edited by jacob21, 16 October 2014 - 08:31 AM. the title sounds quite ridiculous but i dont know how else to put it... i want to return results, but my site has got quite popular, quicker than i would have thought. i want to paginate the results. but i want the results to appear at random. but the results should not appear twice... can this even be done? any questions...? thanks! |