Here I have html that has been generated through PHP and my issue is that I want the dropdown's option to stay selected after form submission. What am I missing?
$core_areas = array("Category 1", "Category 2", "Category 3");
if (isset($_POST["core"]) && $_POST["core"] != "")
{
echo "<select name=\"core\" id=\"core\">
<option value=\"\"> -Any- </option>";
foreach($core_areas as $value)
{
echo "<option value=\"$value\">$value</option>";
}
echo "</select>";
}
else
{
echo "<select name=\"core\" id=\"core\">
<option value=\"\"> -Any- </option>";
foreach($core_areas as $value)
{
echo "<option value=\"$value\">$value</option>";
}
echo "</select>";
}
Add the selected attribute to the selected option element.
echo "<option value=\"$value\" " .
($value == $_POST["core"] ? "selected" : "") .
">$value</option>";
Related
I am attempting to create a single PHP page that opens a salesman or customer window based off a hidden input field in a form being posted. The tricky part for me is I also need two select option dropdowns that appear based on the same hidden input field being posted but ALSO retain their value after submit.
I know how to retain the values of a form like so..
<form action="" method="post">
<select name="newyear">
<option <?php if ($sqlyear == '2015cust') { ?>selected="true" <?php }; ?>value="2015cust">2015</option>
<option <?php if ($sqlyear == '2016cust') { ?>selected="true" <?php }; ?>value="2016cust">2016</option>
</select>
</form>
But when I try and implement this method inside my PHP if statements it does not retain the selected option.
if ($open == 'salesman_window'){
echo "This is salesman dropdown menu";
echo "$sqlyear";
echo "<form action='' method='post'>";
echo "<input type='hidden' name='newyear_trigger' value=''>";
echo "<input type='hidden' name='window' value='salesman_window'>";
echo "<select name='newyear'>";
echo "<option if ($sqlyear == '2015users') { selected='true' }; value='2015users'> 2015</option>";
echo "<option if ($sqlyear == '2016users') { selected='true' }; value='2016users'>2016</option>";
echo "</select>";
echo "<button class='documentation_button'>Submit</button>";
echo "</form>";
}
if ($open == 'customer_window'){
echo "this is customer dropdown menu";
echo "$sqlyear";
echo "<form action='' method='post'>";
echo "<input type='hidden' name='newyear_trigger' value=''>";
echo "<input type='hidden' name='window' value='customer_window'>";
echo "<select name='newyear'>";
echo "<option if ($sqlyear == '2015cust') { selected='true' }; value='2015cust'>2015</option>";
echo "<option if ($sqlyear == '2016cust') { selected='true' }; value='2016cust'>2016</option>";
echo "</select>";
echo "<button class='documentation_button'>Submit</button>";
echo "</form>";
}
?>
I have tried breaking up the option with multiple echo ""; and tried a few ".." of these in there as well but either get syntax errors or it just doesn't retain the selection. Everything works in regards to showing the correct dropdown menu based on what is posted but I just cannot get the select option to retain. The $sqlyear is for sure getting the correct value each time the dropdown option is selected so I know it isn't that either. Can anyone help?
A couple of options:
Use a ternary operator in your echo:
echo "<option" . (($sqlyear == '2015users') ? " selected='true'" : '')
. " value='2015users'>2015</option>";
echo "<option" . (($sqlyear == '2016users') ? " selected='true'" : '')
. " value='2016users'>2016</option>";
Keep using the if, but without opening/closing PHP tags as much:
<option <?php if ($sqlyear == '2015users') echo " selected='true'"; ?> value='2015users'>
2015</option>
<option <?php if ($sqlyear == '2016users') echo " selected='true'"; ?> value='2016users'>
2016</option>
in my dropdown list I insert a value from a mySQL query. This Value is my selected Option.
The problem is, that he duplicate my selected Value (it is logical because I create "6 Options" but the user only should have "5 Options").
How can I prevent this?
This is my Code:
echo "<select id='qty' name='qty' onchange='this.form.submit()'>";
echo "<option value='".$row['a_qty']."' selected>".$row['a_qty']."</option> ";
echo "<option value='1'>1</option>";
echo "<option value='2'>2</option>";
echo "<option value='3'>3</option>";
echo "<option value='4'>4</option>";
echo "<option value='5'>5</option>";
echo "</select>";
echo "<script>
And my Output:
Try as below :
echo "<select id='qty' name='qty' onchange='this.form.submit()'>";
for($i=1;$i<=5;$i++)
{
$selected = '';
if($i==$row['a_qty'])
$selected = 'selected="selected"';
echo "<option value='$i' $selected>$i</option>";
}
echo "</select>";
I have a php select populated and given a matching value by using php to loop through the results of a sql query.
$result = mysqli_query($con, "select * from course")
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['Title'] ." '>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
That gives me a drop down list populated with all course titles,
upon submission I can access the selected value using $_POST['CourseSelect'];
However the drop down (select) resets itself to the default value when the page reloads.
How can I keep that option selected using php?
I know that I can use the selected keyword inside of an a select option to make that option the default selected option.
for example the second option would be selected when loading the page:
<select>
<option>One</option>
<option selected>Two</option>
<option>Three</option>
</select>
You can make it simple as
while($row = mysqli_fetch_array($result))
{
$select = '';
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] == $row['Title'] ) $select = 'SELECTED';
echo "<option value='".$row['Title']."' ".$select.">" . $row['Title'] . "</option>";
}
you can use like below..
$result = mysqli_query($con, "select * from course");
$selected = "";
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
$selected = $row['Title'] == $_REQUEST['CourseSelect'] ? "Selected" : "";
echo "<option value='" . $row['Title'] ." ' $selected>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
try this:
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
$selected = $_POST['CourseSelect'] == $row['Title'] ? 'selected' : '';
echo "<option value='{$row['Title']}' {$selected}>{$row['Title']}</option>";
}
you can do like this
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] != "0")
{
echo "<option >".$_POST['CourseSelect']."</option>";
}
else
{
echo "<option value='0'> - Select Course - </option>";
}
while($row = mysqli_fetch_array($result))
{
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] != $row['Title'])
echo "<option value='" . $row['Title'] ." '>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
PHP Function
# select box
/*
Example:
Parameter 1:
$options[1] = 'Course 1';
$options[2] = 'Course 2';
$options[3] = 'Course 3';
Parameter 2:
$selectedOption = 2; The dropdown need to be selected
*/
function buildOptions($options, $selectedOption)
{
foreach ($options as $value => $text)
{
if ($value == $selectedOption)
{
$Return .='<option value="'.$value.'" selected="selected">'.stripslashes($text).'</option>';
}
else
{
$Return .='<option value="'.$value.'">'.stripslashes($text).'</option>';
}
}
return $Return;
}
Function Call
$result = mysqli_query($con, "select * from course");
while ($row = mysqli_fetch_array($result))
{
$UniqueId = $row['Title'];
$Value = $row['Title'];
$optionsArray[$UniqueId] = $Value; // Store the values into an array
}
$CourseSelect = isset($_POST['CourseSelect']) ? $_POST['CourseSelect'] : '0';
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
echo buildOptions($optionsArray, $CourseSelect);
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
Notes
You can use buildOptions() for all of your projects to display select box. I have been using this for years.
How can I make the filtered selection still appear after the request.
So if I have options 1,2,and 3. When I select 2 and data shows I still want 2 to display signifying that the data is filtered through option 2.
echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >";
echo "<select name='Country' tabindex='1' >";
while($row = mysql_fetch_array($result))
{
echo " <option value='". $row['Fund_Manager_Company_Code'] ."'>". $row['Fund_Manager_Company_Name'] ."</option>";
}
echo "</select>";
echo "<input type='submit' value='Filter' />";
echo "</form>";
You can do something like this :
$country = isset($_POST['Country']) ? $_POST['Country'] : '';
while($row = mysql_fetch_array($result))
{
echo " <option value='". $row['Fund_Manager_Company_Code'] ."' ".(($row['Fund_Manager_Company_Code'] == $country) ? 'selected="selected"' : '').">". $row['Fund_Manager_Company_Name'] ."</option>";
}
What you need is to add selected attribute to option:
$Country = $_POST['Country'];
$sected = 'selected = "selected" ';
while($row = mysql_fetch_array($result))
{
echo " <option ".($row['Fund_Manager_Company_Code'] == $Country? $selected : '')."value='". $row['Fund_Manager_Company_Code'] ."'>". $row['Fund_Manager_Company_Name'] ."</option>";
}
The one that the value selected and posted then will be selected...
something like
echo" <option value='" . $row['Fund_Manager_Company_Code'] . "' " . ((isset($_POST['Country']) && $_POST['Country'] == $row['Fund_Manager_Company_Code'])
? 'selected="selected"' : '') . ">" . $row['Fund_Manager_Company_Name'] . "</option>";
After your submit, you need to catch the selection in your PHP code:
$selection = $_POST['Country'];
echo "<form name='country_list' method='POST' action='http://opben.com/colombia/familias-de-carteras' >";
echo "<select name='Country' tabindex='1' >";
while($row = mysql_fetch_array($result))
{
$selected = "";
if ($row['Fund_Manager_Company_Code'] == $selection) {
$selected = "selected";
}
echo " <option value='". $row['Fund_Manager_Company_Code'] ."' ".$selected.">". $row['Fund_Manager_Company_Name'] ."</option>";
}
echo "</select>";
echo "<input type='submit' value='Filter' />";
echo "</form>";
can someone tell me is there a possibility to add a "blank option" once the dropdown is populated?
echo '<select name="dropdown" style="width:150px">';
while($rec=mysql_fetch_array($run1))
{
$value = $rec['route'];
echo "<option value=\"$value\">$value</option>";
}
echo '</select>';
Why can't you do it beforehand?
echo '<select name="dropdown" style="width:150px">';
echo '<option value=""></option>';
while($rec=mysql_fetch_array($run1))
{
$value = $rec['route'];
echo "<option value=\"$value\">$value</option>";
}
echo '</select>';
Or if you really need it after (don't know why), just add a selected option to the <option>
echo '<select name="dropdown" style="width:150px">';
while($rec=mysql_fetch_array($run1))
{
$value = $rec['route'];
echo "<option value=\"$value\">$value</option>";
}
echo '<option value="" selected="selected"></option>';
echo '</select>';
After your while block, just add an additional echo:
echo "<option value=\"$value\"></option>";
As already mentioned, it's generally best to do this before your while loop, so
the default is blank. That way you can catch when/if a user forgets to select an
option.
simple trick, add extra line after the while loop.
while($rec=mysql_fetch_array($run1))
{
$value = $rec['route'];
echo "<option value=\"$value\">$value</option>";
}
echo "<option value=\"$value\"></option>";