I'm building a drop-down list with options containing data from a database and want to hide options which have zero data.
I have tried an if statement using continue but failing to catch the live data values from the database.
<select name='Database' title="Choose from database">
<option value="">All</option>
<?php foreach($database as $row):
if ($row['topic'] == 0) {
continue;
}
else {
?>
<option value="<?= $row['topic']; ?>"
<?php if ($row['topic'] == $_SESSION['prosess']){echo "
selected";}?>>
<?= $row['topic']; ?>
<?php }?>
</option>
<?php endforeach; ?>
</select>
Is there any clever javascript-, php-, etc. code that can deactivate/hide options from a database which are empty.
Add this in your css:
select option:empty {
display:none
}
I think you should try this.
<select name='Database' title="Choose from database">
<option value="">All</option>
<?php
if(count($database) > 0)
{
foreach($database as $row)
{
?>
<option value="<?= $row['topic']; ?>"
<?php if ($row['topic'] == $_SESSION['prosess']){echo "
selected";}?>>
<?= $row['topic']; ?>
<?php }?>
</option>
<?php
}
}
?>
</select>
Hope it helpful for you.
Related
I want to update "profile of a user" in php. There is a repetition of one value for two times in dropdown list. for example i take language value='Punjabi' from database but there is also a value placed in dropdown with name of 'Punjabi'.
The issue is simply that there is a repetition of value which i don't want.
<?php $result=mysqli_query($conn, "select * from profile where id=$firstPerson");
while($queryArray=mysqli_fetch_array($result)){ ?>
<select name="language" id="language" >
<option value='<?php echo $queryArray["language"];?> '> <?php echo $queryArray["language"]; ?></option>
//for example, the value from database is "Punjabi"
<option value="Hindi">Hindi</option>
<option value="Punjabi">Punjabi</option>
<option value="Urdu">Urdu</option>
</select>
<?php } ?>
when a value='Punjabi' from database is selected in dropdown list, the dropdown should not show the value='Punjabi' that is already placed in dropdown.
Remember: i have more than 1000 values in my dropdown(html) list.
screenshot
Instead of creating a new option according to the user data, Check if existing options are equal to user data:
<select name="language" id="language" >
<option value="Punjabi" <?php if ($queryArray["language"]=="Punjabi"){echo 'selected="selected"'} ?>>Punjabi</option>
<option value="Hindi" <?php if ($queryArray["language"]=="Hindi"){echo 'selected="selected"'} ?>>Hindi</option>
<option value="Urdu" <?php if ($queryArray["language"]=="Urdu"){echo 'selected="selected"'} ?>>Urdu</option>
</select>
If there are large number of options and you don't want to hard code these conditions, you can remove the second option using javascript on DOM ready:
$(document).ready(function(){
$('option[value="<?php echo $queryArray["language"] ?>"]').eq(1).remove();
})
skip the loop when value is equal to Punjabi, Urdu and Hindi.
<?php $result=mysqli_query($conn, "select * from profile where id=$firstPerson");
while($queryArray=mysqli_fetch_array($result)){ ?>
<select name="language" id="language" >
<?php if($queryArray["language"]!="Punjabi" && $queryArray["language"]!="Urdu" &&
$queryArray["language"]!="Hindi") { ?>
<option value="Hindi">Hindi</option>
<option value="Punjabi">Punjabi</option>
<option value="Urdu">Urdu</option>
<?php } ?>
I think you are doing it wrong way the correct way would be having a table which stored all the languages along with values
using selected attribute to achieve your objective
<?php
$result=mysqli_query($conn, "select * from profile where id=$firstPerson");
$queryArray1=mysqli_fetch_array($result);
$langOfUser=$queryArray1["language"];
?>
<select name="language" id="language" >
<?php $result=mysqli_query($conn, "select * from langtab");
while($queryArray=mysqli_fetch_array($result)){ ?>
<option value='<?php echo $queryArray["languageValue"];?> ' <?php if($langOfUser== $queryArray["languageValue"]){ echo 'selected';}?>> <?php echo $queryArray["languageName"]; ?></option>
<?php } ?>
</select>
You have to use if condition to display values in select option.
<select name="language" id="language" >
<?php $result=mysqli_query($conn, "select * from profile where id=$firstPerson");
while($queryArray=mysqli_fetch_array($result)){
if($queryArray["language"]!="Punjabi") {
$opval = "<option value=" . $queryArray["language"] . ">". $queryArray["language"]. " </option> "
echo $opval;
}
?>
<option value="Punjabi">Punjabi</option>
<option value="Hindi">Hindi</option>
<option value="Urdu">Urdu</option>
</select>
So your problem is that you have html hardcoded options and database options. You need to merge them into one on that website.
So you can use some javascript
elements = [1, 2, 9, 15].join(',')
$.post('post.php', {elements: elements})
But you can fill your elements like this is you don´t want to write it by hand
$("#id select").each(function()
{
allOptionsInSelect.push($(this).val());
});
Than on php side you can do
$elements = $_POST['elements'];
$elements = explode(',', $elements);
And now you have html hardcoded select on server side. Now you need to check if it doesn´t already exist when you are printing from database
You can do that like this
if(in_array(value_from_database, $elements) {
// It is so skip
} else {
// It is not, so print it
}
You can use if elseif this way.
<select name="language" id="language" >
<option value='<?php echo $queryArray["language"];?>'><?php echo $queryArray["language"]; ?></option>
<?php if ($queryArray["language"] == "Hindi") { ?>
<option value="Punjabi">Punjabi</option>
<option value="Urdu">Urdu</option>
<?php } elseif ($queryArray["language"] == "Urdu") { ?>
<option value="Punjabi">Punjabi</option>
<option value="Hindi">Hindi</option>
<?php } elseif ($queryArray["language"] == "Punjabi") { ?>
<option value="Urdu">Urdu</option>
<option value="Hindi">Hindi</option>
<?php } ?>
Good morning, I have a problem that is: I can not keep several options selected after submitting the form and I would like someone to help me.
<select name="utilizadores[]" id="utilizadores" multiple="multiple" class="selectpicker" data-live-search="true" data-actions-box="true" title="Utilizadores">
<?php while ($reg_sql=mysqli_fetch_array($res_sql)){?>
<option value="<?php echo $reg_sql['ID_USER']; ?>"><?php echo $reg_sql['NOMEUSER']; ?></option>
<?php } ?>
</select>
<script type="text/javascript">
document.getElementById('utilizadores').value = "<?php echo $_POST['utilizadores[]'];?>";
</script>
this is my code to have the various options in the select box
You have to check if $_POST['utilizadores'] or $_GET['utilizadores'] it depends on your request type. I will use $_POST in here for explain my answer.
your select is multiple, you can use in_array function for checking that if result from db record is in array of $_POST['utilizadores']
<select name="utilizadores[]" id="utilizadores" multiple="multiple" class="selectpicker" data-live-search="true" data-actions-box="true" title="Utilizadores">
<?php while ($reg_sql=mysqli_fetch_array($res_sql)){?>
**<option value="<?php echo $reg_sql['ID_USER']; ?>"
<?php
if(isset($_POST['utilizadores'])){
if(in_array($reg_sql['ID_USER'], $_POST['utilizadores'])){
echo 'selected';
}else{
echo '';
}
}
?>
>**<?php echo
$reg_sql['NOMEUSER']; ?></option>
<?php } ?>
</select>
You may be able to do it more efficiently if your database result also contains which rows are selected, but when you loop through, just add the selected="selected" attribute to the <option> tag.
Assuming your $_POST array exists in this scope, you can use the in_array function in PHP to determine if the option has been selected (docs).
The ternary based operation is as follows:
in_array($reg_sql['ID_USER'],$_POST['utilizadores']) ? 'selected="selected"' : ''
Which says "if the ID_USER is in the post array, then print the selected attribute, otherwise, print a blank string"
Putting it all together:
<select name="utilizadores[]" id="utilizadores" multiple="multiple" class="selectpicker" data-live-search="true" data-actions-box="true" title="Utilizadores">
<?php while ($reg_sql=mysqli_fetch_array($res_sql)){?>
<option value="<?php echo $reg_sql['ID_USER']; ?>" <?= in_array($reg_sql['ID_USER'],$_POST['utilizadores']) ? 'selected="selected"' : '' $?>>
<?php echo $reg_sql['NOMEUSER']; ?>
</option>
<?php } ?>
</select>
An example of how you can do this. Either add the "selected" string if yes or leave blank if no. You can also write selected="selected". You can do the same thing to set disabled or readonly.
<select name="utilizadores[]" id="utilizadores" multiple="multiple" class="selectpicker" data-live-search="true" data-actions-box="true" title="Utilizadores">
<?php while ($reg_sql=mysqli_fetch_array($res_sql)){?>
<?php $selected = isset($reg_sql["mi_variable"]) ? "selected" : ""; ?>
<option value="<?php echo $reg_sql['ID_USER'];?>" <?php echo $selected; ?> >
<?php echo $reg_sql['NOMEUSER']; ?>
</option>
<?php } ?>
</select>
<script type="text/javascript">
document.getElementById('utilizadores').value = "<?php echo $_POST['utilizadores[]'];?>";
</script>
Can someone help me on how to link up 2 dropdown list which data came from the same table in the database, is it possible to do this way because I can only find a solution which there must be 2 table in the database to link up those two. this is how should it work, 1st, the user choose the department and the 2nd dropdown will show only the people that assigned from that particular department.
database table has id,name,department.
the department works already but the 2nd dropdown didnt work to filter from the department. Can someone help me to know what is wrong with this?
index.php
<?php
$query ="SELECT DISTINCT company from hrar";
$results = $db_handle->runQuery($query);
?>
<script>
function getName(val) {
$.ajax({
type: "POST",
url: "getName.php",
data:'department='+val,
success: function(data){
$("#nameList").html(data);
}
});
}
function selectDepartment(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
</script>
<label>Department:</label><br/>
<select name="country" id="country-list" class="demoInputBox" onChange="getName(this.value);">
<option value="">Select Department</option>
<?php
foreach($results as $country) {
?>
<option value="<?php echo $country["id"]; ?>"><?php echo $country["company"]; ?></option>
<?php
}
?>
</select>
</div>
<div class="row">
<label>Name:</label><br/>
<select name="state" id="nameList" class="demoInputBox">
<option value="">Select Name</option>
</select>
//below is the getName.php
if(!empty($_POST["company"])) {
$query ="SELECT DISTINCT name from hrar = '" . $_POST["company"] . "'";
$results = $db_handle->runQuery($query);
?>
<option value="">Select Name</option>
<?php
foreach($results as $name) {
?>
<option value="<?php echo $name["id"]; ?>"><?php echo $name["name"]; ?></option>
<?php
}
}
?>
Try this.
<select name="filter" id="filter" onchange="this.form.submit();">
<option value="name">Select Name</option>
<?php
$sql = "SELECT DISTINCT name FROM hr";
$result = mysqli_query($conn,$sql);
while ($row = mysqli_fetch_array($result))
{
echo "<option value='". $row['name'] ."'>" .$row['name'] ."</option>";
}
?>
</select>
I have 10 dropdown menus that are like:
Please Select
Item 1
Item 2
Item 3
---Appetizers---
Item 4
Item 5
---Main Courses---
Item 6
Item 7
---Lunch Specials---
Item 8
I want to grab the value if an Item is selected and print it on a confirmation page. Can I do that with a for loop and javascript? Like if I do this, how would I call it in HTML?
function getItems() {
var items = [
document.getElementById("item1").value,
document.getElementById("item2").value,
document.getElementById("item3").value,
document.getElementById("item4").value,
document.getElementById("item5").value,
document.getElementById("item6").value,
document.getElementById("item7").value,
document.getElementById("item8").value,
document.getElementById("item9").value,
document.getElementById("item10").value
];
for (int i = 0; i < items.length; i++) {
var count = 0;
if (
(item[i] != "Please Select") ||
(item[i] != "---Appetizers---") ||
(item[i] != "---Main Courses---") ||
(item[i] != "---Lunch Specials---")
) {
document.getElementById(i).innerHTML = item[i];
count++;
}
}
}
The dropdowns are populated from a MySQL database using php. My HTML/php:
Item 4
<select id="item4">
<option value"">Please Select</option>
<?php if ($resultApp4->num_rows > 0) { ?>
<option value"">---Appetizers---</option>
<?php while($row = $resultApp4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>"><?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<?php } } ?>
<?php if ($resultMain4->num_rows > 0) { ?>
<option value"">---Main Courses---</option>
<?php while($row = $resultMain4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>"><?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<br>
<?php } } ?>
<?php if ($result4->num_rows > 0) { ?>
<option value"">---Lunch Specials---</option>
<?php while($row = $result4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>"><?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<br>
<?php } } ?>
</select>
<br>
You seem to generate a single drop down with all options. If that is intended, then make it one in which multiple selections can be made, which is not the default behaviour. For that to happen, add the multiple attribute to the select.
Then you need to add a listener to the change event, so that your code executes whenever the user selects or unselects item(s).
Note that you have forgotten some = after the value attributes. Also, it is not allowed to put a <br> tag after an option element. A select element should only have option children elements.
Here is how the corrected PHP could look:
Item 4
<select id="item4" multiple size="8">
<option value="">Please Select</option>
<?php if ($resultApp4->num_rows > 0) { ?>
<option value="">---Appetizers---</option>
<?php while($row = $resultApp4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>">
<?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<?php } } ?>
<?php if ($resultMain4->num_rows > 0) { ?>
<option value="">---Main Courses---</option>
<?php while($row = $resultMain4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>">
<?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<?php } } ?>
<?php if ($result4->num_rows > 0) { ?>
<option value="">---Lunch Specials---</option>
<?php while($row = $result4->fetch_assoc()) { ?>
<option value="<?php echo $row['price']; ?>">
<?php echo $row['name']; ?> - $<?php echo $row['price']?></option>
<?php } } ?>
</select>
The JavaScript could be much simpler, as in this working snippet:
// Listen to the selection changes made:
document.querySelector('#item4').addEventListener('change', function () {
// For this demo, the text is stored in a DIV
document.getElementById('output').textContent =
[...this.selectedOptions] // convert selected options list to array
.filter ( option => option.value.length ) // option must have a value
.map( option => option.textContent ) // get text of option
.join('\n'); // add line breaks
});
Item 4 (hold Ctrl key to add selections)<br>
<select id="item4" multiple size=8>
<option value="">Please Select</option>
<option value="">---Appetizers---</option>
<option value="1">drink 1 - $1</option>
<option value="1.10">drink 2 - $1.10</option>
<option value="">---Main Courses---</option>
<option value="15">dish 1 - $15</option>
<option value="16">dish 2 - $16</option>
<option value="">---Lunch Secials---</option>
<option value="10">lunch 1 - $10</option>
<option value="11">lunch 2 - $11</option>
</select>
<div id="output" style="white-space:pre"></div>
I have texbox and dropdown list which is populated from mysql database. I want to retrieve data from database and wants to display in textbox using dropdown selected list, without refreshing the page. Here is my code and Thanks in Advance.
<select name="select1" class="form-control" id="dropdownlist1">
<option id="0">-- Select the Company --</option>
<?php
require("dbcon.php");
$getallcompanies = mysql_query("SELECT * FROM ifcandetails6");
while($viewallcompanies = mysql_fetch_array($getallcompanies)){
?>
<option id="<?php echo $viewallcompanies['tcuid']; ?>"><?php echo $viewallcompanies['tcname'] ?></option>
<?php
}
?>
</select>
Input Textbox:
<input type="text" id="field1" value="<?php echo $viewallcompanies['tccontact']?>" disabled/>
Separate things out like this:
<?php
require("dbcon.php");
$query = mysql_query("SELECT tcuid, tcname, tccontact FROM ifcandetails6");
while($row = mysql_fetch_array($query)){
$contacts[] = $row['tccontact'];
$companies[] = $row;
}
?>
<select name="select1" class="form-control" id="dropdownlist1">
<option id="0">-- Select the Company --</option>
<?php foreach($companies as $company) { ?>
<option id="<?= $company['tcuid']; ?>"><?= $company['tcname'] ?></option>
<?php } ?>
</select>
<?php foreach($contacts as $contact) { ?>
<input type="text" id="field1" value="<?= $contact['tccontact']?>"/>
<?php } ?>