How to get the value which user has selected in php? - php

I have a little problem here. I have a list, which is:
<form action="display.php" method="post">
<select name="holiday">
<option value="month">Kiekvieno mėnesio skaičiavimas</option>
<option value="day">Kiekvienos dienos skaičiavimas</option>
</select>
<br> <br>
<input type="submit" name="select" value="Pasirinkti" />
</form>
What I need to do is that when the user selects value ="month", php code would do one action, and when the user selects value ="day", php code would do another action?
My php code looks like this:
<?php
if($_POST['select']){
// Storing selected value in a variable
$selectedValue= $_POST['holiday'];
}
else if ($selectedValue == $_POST['month']) {
$todaysDate = new DateTime();
while ($employee = $select->fetch()){
$employmentDateValue = new DateTime($employee['employment_date']);
// Comparing employment date with today's date
$differenceBetweenDates = date_diff($employmentDateValue, $todaysDate);
$workMonths = $differenceBetweenDates->y * 12 + $differenceBetweenDates->m;
$holidayDays = $workMonths *4;
echo "<tr>";
echo "<td>".$employee['name']."</td>";
echo "<td>".$employee['surname']."</td>";
echo "<td>".$employee['employment_date']."</td>";
echo "<td>".$workMonths."</td>";
echo "<td>".$holidayDays."</td>";
echo "</tr>";
}
}
else {
echo "Lalalala";
}
?>
I've tried to do so with $_POST['select'], but it doesn't work. Thanks for any help guys

<?php
if($_POST['select']){
// Storing selected value in a variable
$selectedValue= $_POST['holiday'];
if ($selectedValue == 'month') {
}
else if ($selectedValue == 'day') {
}
else{
echo "Lalalala";
}
}
?>

You need to do $_POST['holiday'], so change:
if($_POST['select']){
// Storing selected value in a variable
$selectedValue= $_POST['holiday'];
}
to
if($_POST['holiday']){
// Storing selected value in a variable
$selectedValue = $_POST['holiday'];
}
You also need to change the line:
else if ($selectedValue == $_POST['month']) {
So it's not part of the original if statement:
if ($selectedValue == 'month') {
// your code
}
else {
echo "Lalalala";
}

Related

If condition not checking properly in php?

I am working on submission system. I have a drop down list when user want to edit the section drop-down need to be matched select option to visible in drop down list based on before saved in database value. I write some code for this option but it's not working properly. This same code I have used for the antoher project it's working well. An error also not coming I have checked syntax no problem.
<?php
$journalid = $_GET['jiid'];
$sqlold = "SELECT `jtype`, `jtitle` FROM `journal` WHERE jid=? LIMIT 1";
$db->query($sqlold);
$db->bind(1, $journalid);
$db->execute();
$resjid = $db->resultset();
foreach($resjid as $resjtype){
$resjournaltype = $resjtype['jtype'];
}
echo 'Res Ournal Type: '.$resjournaltype;
$db->query("SELECT article_name, article_value FROM article_type");
$db->execute();
$resultrow = $db->resultset();
foreach($resultrow as $vals){
if($resjournaltype == $vals['article_value'] ){
echo "yes";
}
?>
<option <?php //if ( $jtype == $res['article_value'] ) echo 'selected = "selected"'; ?> value="<?php echo $res["article_value"];?>"><?php echo $res['article_name']; ?> </option>
<?php
else {?>
<option <?php //if ( $jtype == $res['article_value'] ) echo 'selected = "selected"'; ?> value="<?php echo $res["article_value"];?>"><?php echo $res['article_name']; ?> </option>
<?php
}
}*/
else{
?>
<option <?php //if ( $resjid[0]['jtype'] == $vals['article_value'] ) echo 'selected = "selected"'; ?> value="<?php echo $vals["article_value"];?>"><?php echo $vals['article_name']; ?> </option>
<?php
}
}
?>
I am searching for a solution every where on the Internet but I forgot some syntax problems. These problems not giving an error. If here I am checking if condition inside for-each loop here is I made a mistake. My actual code is this. The problem was I am checking two string, not numbers. In PHP string values compare with strcmp() function.
if($resjournaltype == $vals['article_value'] )
{ echo "yes"; }
I changed condition like this finally my problem has been solved.
if(strcmp($resjournaltype, $vals['article_value']) == true){
echo $resjournaltype .'-'. $vals['article_value'] ; }

Option Menu with Last Record as Default

I have an option menu as follows:
<select name="selCycle" id="selCycle" onChange="formFilter.submit()">
<option value="%">all cycles</option>
<?php
do {
?>
<option value="<?php echo $row_Recordset2['Cycle'] ?>"
<?php
if ($varCycle_DetailRS4 == $row_Recordset2['Cycle']) {
echo 'selected';
} elseif ($varCycle2_DetailRS4 == $row_Recordset2['Cycle']) {
echo 'selected';
} else {
echo '';
}
?>
>
<?php echo $row_Recordset2['Cycle'] ?>
</option>
<?php
} while ($row_Recordset2 = mysql_fetch_assoc($Recordset2));
$rows = mysql_num_rows($Recordset2);
if ($rows > 0) {
mysql_data_seek($Recordset2, 0);
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
}
?>
</select>
Currently, the default selection is showing all records. I would like for the default to be the latest set of data equal to:
<?php echo $row_RecordsetCycle['Cycle']; ?>
So option menu would list 1,2,3,4,5 with 5 being the default when the page loads. User can then pick any option available. is set to last record in table with a limit of 1 so it will always echo the last record, which composes the option menu.
Help please. What should I edit so that the one record in
<?php echo $row_RecordsetCycle['Cycle']; ?>
is the default or selected option menu when the page loads? Currently, the default just shows all records and is extremely slow to load, hence why I want the latest record to be the default.
I took a stab at rewriting this. (code at the end)
the first thing I did was turn your do_while loop into a while loop as I think it was adding unnecessary confusion to the code
after doing that I moved some code
$rows = mysql_num_rows($Recordset2);
if ($rows > 0) {
mysql_data_seek($Recordset2, 0);
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
}
above the while loop as I thought it was needed to "prime the pump" of the while loop
I also pulled the
if ($varCycle_DetailRS4 == $row_Recordset2['Cycle']) {
return ' selected';
} elseif ($varCycle2_DetailRS4 == $row_Recordset2['Cycle']) {
return ' selected';
} else {
return '';
}
out into a function so as to simplify the code further
now here come my assumptions I assumed that $varCycle_DetailRS4 was the last value and that $varCycle2_DetailRS4 was the currentlySelected that was set befor the code provided;
if that is the case and that $varCycle2_DetailRS4 would be unset if it was the first then all that would need to be done is to slightly modify the isSelected to only set one option as selected.
<?php
function isSelect($value)
{
$lastValue = $varCycle_DetailRS4;
$currentlySelected = $varCycle2_DetailRS4;
if(isset($currentlySelected))
{
$selectValue = $currentlySelected;
}
else
{
$selectValue = $lastValue;
}
if ($selectValue == $value) {
return ' selected';
} else {
return '';
}
}
?>
<select name="selCycle" id="selCycle" onChange="formFilter.submit()">
<option value="%">all cycles</option>
<?php
$rows = mysql_num_rows($Recordset2);
if ($rows > 0) {
mysql_data_seek($Recordset2, 0);
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
}
while ($row_Recordset2 = mysql_fetch_assoc($Recordset2))
{
$value = $row_Recordset2['Cycle']
echo " <option value=\"".$value."\"".isSelect($value).">".$value."</option>/n";
} ;
?>
</select>
Try using end() to get the last array:
<?php
$arr =array(1,2,3,4,5);
$last = end($arr);
?>
<select name="selCycle" id="selCycle" onChange="formFilter.submit()">
<option value="%">all cycles</option>
<?php foreach($arr as $key=>$val):
if(in_array("$last", array($val))==true){
echo '<option value="" selected="selected">'.$val.'</option>';
continue;
}
echo '<option>'.$val.'</option>';
endforeach; ?>
</select>
Stop using mysql extension, use pdo or mysqli. Using your question code, it should be:
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
$last = end($row_Recordset2);
foreach($row_Recordset2 as $key=>$val){
//other code
if(in_array("$last", array($val))==true){
echo '<option value="" selected="selected">'.$val.'</option>';
continue;
}
echo '<option>'.$val.'</option>'
}
Working Demo

Passing information using post method without session variables

I will admit immediately that this is homework. I am only here as a last resort after I cannot find a suitable answer elsewhere. My assignment is having me pass information between posts without using a session variable or cookies in php. Essentially as the user continues to guess a hidden variable carries over all the past guesses up to that point. I am trying to build a string variable that holds them all and then assign it to the post variable but I cannot get anything to read off of the guessCounter variable i either get an undefined index error at the line of code that should be adding to my string variable or im just not getting anything passed over at all. here is my code any help would be greatly appreciated as I have been at this for awhile now.
<?php
if(isset($_POST['playerGuess'])) {
echo "<pre>"; print_r($_POST) ; echo "</pre>";
}
?>
<?php
$wordChoices = array("grape", "apple", "orange", "banana", "plum", "grapefruit");
$textToPlayer = "<font color = 'red'>It's time to play the guessing game!(1)</font>";
$theRightAnswer= array_rand($wordChoices, 1);
$passItOn = " ";
$_POST['guessCounter']=$passItOn;
$guessTestTracker = $_POST['guessCounter'];
$_POST['theAnswer'] = $theRightAnswer;
if(isset($_POST['playerGuess'])) {
$passItOn = $_POST['playerGuess'];
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$guessTestTracker = $_GET['guessCounter'];
$theRightAnswer = $_GET['theAnswer'];
}
else if ($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_POST['playerGuess'])) {
if(empty($_POST['playerGuess'])) {
$textToPlayer = "<font color = 'red'>Come on, enter something(2)</font>";
}
else if(in_array($_POST['playerGuess'],$wordChoices)==false) {
$textToPlayer = "<font color = 'red'>Hey, that's not even a valid guess. Try again (5)</font>";
$passItOn = $_POST['guessCounter'].$passItOn;
}
if(in_array($_POST['playerGuess'],$wordChoices)&&$_POST['playerGuess']!=$wordChoices[$theRightAnswer]) {
$textToPlayer = "<font color = 'red'>Sorry ".$_POST['playerGuess']." is wrong. Try again(4)</font>";
$passItOn = $_POST['guessCounter'].$passItOn;
}
if($_POST['playerGuess']==$wordChoices[$theRightAnswer]) {
$textToPlayer = "<font color = 'red'>You guessed ".$_POST['playerGuess']." and that's CORRECT!!!(3)</font>";
$passItOn = $_POST['guessCounter'].$passItOn;
}
}
}
}
$_POST['guessCounter'] = $passItOn;
$theRightAnswer=$_POST['theAnswer'];
for($i=0;$i<count($wordChoices);$i++){
if($i==$theRightAnswer) {
echo "<font color = 'green'>$wordChoices[$i]</font>";
}
else {
echo $wordChoices[$i];
}
if($i != count($wordChoices) - 1) {
echo " | ";
}
}
?>
<h1>Word Guess</h1>
Refresh this page
<h3>Guess the word I'm thinking</h3>
<form action ="<?php echo $_SERVER['PHP_SELF']; ?>" method = "post">
<input type = "text" name = "playerGuess" size = 20>
<input type = "hidden" name = "guessCounter" value = "<?php echo $guessTestTracker; ?>">
<input type = "hidden" name = "theAnswer" value = "<?php echo $theRightAnswer; ?>">
<input type = "submit" value="GUESS" name = "submitButton">
</form>
<?php
echo $textToPlayer;
echo $theRightAnswer;
echo $guessTestTracker;
?>
This is a minimal functional example of what you need to do. There are still a couple of minor bugs (like duplicate entries in the history), but I've left these as an exercise for you. Treat this as a starting point and build up what you need from it.
I've added comments to explain what's happening, so hopefully it is clear to you.
$answer = null;
$history = [];
$choices = ['apple', 'grape', 'banana'];
$message = '';
// check if a guess has been made.
if (!empty($_POST) && !empty($_POST['guess'])) {
// check if previous guesses have been made.
if (!empty($_POST['history'])) {
$history = explode(',', $_POST['history']);
}
// check guess.
if (!empty($_POST['answer']) && !empty($_POST['guess'])) {
// check guess and answer are both valid.
if (in_array($_POST['guess'], $choices) && isset($choices[$_POST['answer']])) {
if ($_POST['guess'] == $choices[$_POST['answer']]) {
// correct; clear history.
$history = [];
$message = 'correct!';
} else {
// incorrect; add to history and set previous answer to current.
$history[] = $_POST['guess'];
$answer = $_POST['answer'];
$message = 'incorrect!';
}
} else {
// invalid choice or answer value.
}
}
}
if (empty($answer)) {
// no answer set yet (new page load or correct guess); create new answer.
$answer = rand(0, count($choices) - 1);
}
?>
<p>Guess the word I'm thinking:</p>
<p><?php echo implode(' | ', $choices) ?></p>
<form method="POST">
<input type="hidden" name="answer" value="<?php echo $answer; ?>">
<input type="hidden" name="history" value="<?php echo implode(',', $history); ?>">
<input type="text" name="guess">
<input type="submit" name="submit" value="Guess">
</form>
<p><?php echo $message; ?></p>

How to Set a Dropdowns Default Selection Based on the Logged in User

I have a dropdown control which and I would like it to default to a specific option based on the access of the logged in user. For example, the dropdown has 10 options but only administrators have access to view all 10 options. The majority of users only have access to 1 of the options though. The page contents are hidden/displayed based on whether or not the value of the dropdown is null.
Question: Using the example above, if an admin is logged in I need the dropdown to default to "Select an option". This way the page content is hidden. On the other hand, if a user with access to only 1 is logged in, I need it to default to that 1. This way they don't have to select anything and, by default the page content is displayed. How do I go about doing this?
Below is my current code which handles what the dropdown displays based on when a selection is made.
PHP/HTML
// Hide/Show main content div
<?php
if (isset($_GET['src'])) {
$src = $_GET['src'];
} else {
?>
<style>
#divmain { display: none; }
</style>
}
<?php } ?>
// Start of form, header, etc.
<select name="select1" id="select1">
<?php
$sql = getOptions();
$data = makeConnection($sql);
if ($src == null) { // If value is null, default to 'Select an option'
echo "<option selected value=\"\" disabled=\"disabled\">--Select an option--</option>";
while ($row = odbc_fetch_array($db)) {
echo "<option value=\"".$row['content']."\">".$row['content']."</option>";
}
} else { // If value not null keep the selected value selected
while ($row = odbc_fetch_array($db)) {
if ($row['content'] == $src) { $selected = " selected "; }
else { $selected = " "; }
echo "<option value=\"".$row['content']."\" ".$selected.">".$row['content']."</option>";
}
}
?>
</select>
JS
// Pass selected value on change
$('#select1').change(function() {
var sel = $(this).val();
location.href = "page1.php?src=" + sel;
}
SQL
// Hardcoding user for testing purposes, THIS WILL BE CHANGED
function getOptions() {
$results = "SELECT content, userid FROM table WHERE userid = 'username'";
return $results;
}
Any help is much appreciated and please let me know if I'm not clear about anything.
Got some help and have it figured out now. Here is the revised code:
PHP/HTML
// Hide/Show main content div
<?php
$src = null;
if (isset($_GET['src'])) {
$src = $_GET['src'];
} else {
?>
<style>
#divmain { display: none; }
</style>
}
<?php } ?>
// Start of form, header, etc.
<select name="select1" id="select1">
<?php
$sql = getOptions();
$data = makeConnection($sql);
if ($src == null) {
$i = 0;
$content = "";
while ($row = odbc_fetch_array($db)) {
$content .= "<option value=\"".$row['content']."\">".$row['content']."</option>";
$i++;
}
if ($i > 1) {
echo "<option selected value=\"\" disabled=\"disabled\">--Select an option--</option>";
}
echo $content;
if ($i > 1) { $oneopt = 1; }
else { $oneopt = 0; }
} else {
while ($row = odbc_fetch_array($db)) {
if ($row['content'] == $src) { $selected = " selected "; }
else { $selected = " "; }
echo "<option value=\"".$row['content']."\" ".$selected.">".$row['content']."</option>";
}
}
?>
</select>
JS
$('#select1').change(function() {
var sel = $(this).val();
location.href = "page1.php?src=" + sel;
}
<?php
global $optone;
if ($optone == 1) {
echo "$('#select1').trigger('change');";
}
?>
SQL -- Stays the same
#chenasraf really appreciate the help! Hope this can be of some help to someone in the future!
The preselected option is always the first to have a "selected" attribute. Your first disabled one has it first on all cases, so it's always chosen. Remove that and work from there.
On a side note, I think you can manage to make this options part work better. I've taken the liberty of rewriting it for you:
<select name="select1" id="select1">
<?php
$selected = '';
while ($row = obdc_fetch_array($db)) {
$options[] = '<option value="'.$row['content'].'">'.$row['content'].'</option>';
if ($row['content'] == $src)
$selected = count($options) - 1;
}
array_unshift($options, '<option disabled="disabled"', $selected == '' ? ' selected="selected"' : '','>--Select an option--</option>');
foreach ($options as $option) {
echo $option;
}
?>
</select>

PHP - pre select/save option on dynamically populated select dropdowns

I have a dropdown box that holds the days 1-31 and i want to store/save what the user has previously selected if they return to the page.
My function to generate the box is:
public function fetchDDMMYYYYDropdown($select_d,$session_d) {
$days = range (1, 31);
$dropdown .= '<select name="'.$select_d.'">';
foreach($days as $key=>$name){
if($session_d==$name){
$session = 'selected';
}
$dropdown .= '<option value="'.sprintf("%02d", $name).'" selected="'.$session.'">'.sprintf("%02d", $name).'</option>';
}
$dropdown .= '</select>';
return $dropdown;
}
And my form is on this page:
<?php
session_start();
include("includes/func.class.php");
$dob = $func->fetchDDMMYYYYDropdown('dob_d', $_SESSION['dob_d']);
?>
<form action="t35t_send.php" method="get">
<?php echo $dob;?>
<input type="submit" value="send">
</form>
And it goes to this to save the SESSION variable:
session_start();
$_SESSION['dob_d'] = $_GET['dob_d'];
$dob = $_SESSION['dob_d'];
echo $dob;
I can tell that the $_SESSION['dob_d'] is correct and saved as i can output it inside both the function and the initial form page - so it's just the following which must not be right but at the moment the dropdown box just resets back to the first value, not the saved session:
if($session_d==$name){
$session = 'selected';
}
$dropdown .= '<option value="'.sprintf("%02d", $name).'" selected="'.$session.'">'.sprintf("%02d", $name).'</option>';
try this
function fetchDDMMYYYYDropdown($select_d,$session_d) {
$days = range (1, 31);
$dropdown .= '<select name="'.$select_d.'">';
foreach($days as $key=>$name){
if($session_d==$name){
$session = 'selected = selected';
}
else
{
$session = '';
}
$dropdown .= '<option value="'.sprintf("%02d", $name).'"'.$session.'">'.sprintf("%02d", $name).'</option>';
}
$dropdown .= '</select>';
return $dropdown;
}
The Problem was if even 'selected' is there in "option" even then value gets selected and in your previous code .. 'selected will be there for every date .. so it eas showing '31'.
I have changed the code so that 'selected = selected ' gets echo for the saved value.
Hope it helps you
A rough guess your if condition is not returning true,
Try checking the value of both variables.
Just to confirm can you replace the code as below and try.
if($session_d == sprintf("%02d", $name)){
$session = 'selected';
}
$dropdown .= '<option value="'.sprintf("%02d", $name).'" selected="'.$session.'">'.sprintf("%02d", $name).'</option>';

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