i am trying below code to display list of enum values in select dropdown box.
but its displaying only dropdown box, but values are not displaying....
tablename = tbl_users, column name = userStatus
<select>
<?
$stmt = $user_home->runQuery('SHOW COLUMNS FROM '.tbl_users.' WHERE field="'.userStatus.'"');
while($data = $stmt->fetch()) {
foreach(explode("','",substr($row[1],6,-2)) as $option) {
print("<option>$option</option>");
}
}
?>
<select>
Note : I really tried lot before posting question here & i am new to php coding, still learning....
To display list of enum values in select dropdown:
<select name="select">
<?php
$sql = 'SHOW COLUMNS FROM table_name WHERE field="field_name"';
$row = $dbh->query($sql)->fetch(PDO::FETCH_ASSOC);
foreach(explode("','",substr($row['Type'],6,-2)) as $option) {
print("<option value='$option'>$option</option>");
}
?>
</select>
For display enum value as dropdown you can do something like this.
<?php $status = array('Y'=>'Approve','N'=>'unapprove'); ?>
<select>
<?php foreach($status as $key=>$state) { ?>
<option value="<?php echo $key;?>"><?php echo $state;?></option>
<?php } ?>
</select>
Related
I'm building a system that tracks contact lenses. I'm storing the contact lens info in a database as sometimes prices/availabilities change and i access this info from multiple points in the program. I'm trying to interface with this list using a dropdown by doing "SELECT * FROM contacts" as a query. my code looks like this :
$contact_list = mysqli_query($link, "SELECT brand FROM contacts ORDER BY brand");
Then I echo that list out in a while loop using PHP to populate the options in the dropdown.
My question is this: I have these dropdowns for each eye on the same form. So it's "Brand Right Eye"....other miscellaneous info about the right eye....then "Brand Left Eye". But ONLY the right eye is populating with the brand info because it appears first in the code. What i'm having to do is copy/paste the exact same query and do
$contact_list2 = mysqli_query($link, "SELECT brand FROM contacts ORDER BY brand");
then later if I need the dropdown again, I need to do $contact_list3..and so on. Why can i not generate a drop down using the same variable? Why does it stop responding to calling the variable after the first execution of it and is there any work around that I can implement that would allow me to not have to copy/paste the same query with a different variable association each time?
just for refernce, my php while code is this:
<select class="form-control" name = "brandOS">
<option value="0">Please Select</option>
<?php
while($row = mysqli_fetch_array($contact_list))
{
?>
<option value = "<?php echo($row['brand'])?>" name = "brandOS">
<?php echo($row['brand']) ?>
</option>
<?php
}
?>
</select>
I have this loop copy/pasted for right eye and left eye. But it only works on which ever drop down appears first in the code.
A possible solution will be more efficient in term of performance could be :
<?php
$left_eye = '<option value="0">Please Select</option>';
$rigth_eye = '<option value="0">Please Select</option>';
while($row = mysqli_fetch_array($contact_list))
{
//logic for left eye
$left_eye .= <<<HTML
<option value ="{$row['brand']}" name = "brandOS">
{$row['brand']}
</option>
HTML;
//logic for rigth eye
$rigth_eye .= <<<HTML
<option value ="{$row['brand']}" name = "brandOS">
{$row['brand']}
</option>
HTML;
}
?>
<select class="form-control" name = "brandOS">
<?php echo $left_eye ; ?>
</select>
<select class="form-control" name = "brandOS">
<?php echo $rigth_eye ; ?>
</select>
With this solution you get your result in the same while loop. If the left and right select are the same you can use the same variable.
Store the brands in an array, then you can just loop through the array.
<?php
$contact_list = mysqli_query($link, "SELECT brand FROM contacts ORDER BY brand");
$brands = array();
while($row = mysqli_fetch_array($contact_list))
{
array_push($brands, $row['brand']);
}
?>
<select class="form-control" name = "brandOS">
<option value="0">Please Select</option>
<?php
foreach($brands as $brand){
?>
<option value = "<?php echo($brand[0])?>" name = "brandOS">
<?php echo($brand[0]) ?>
</option>
<?php
}
?>
</select>
You can use a PHP array, like the SESSION one, to store values and use them across your site. Be sure you call "session_start()" method on each page you use that array, though.
//Initialize sessions
session_start();
...
//Right after getting result from query
$_SESSION['contact_list'] = $contact_list;
To use it, just be sure to call the method I told you above, and call the variable with the same syntax:
<?php
while($row = mysqli_fetch_array($_SESSION['contact_list'])) { ?>
Hope this helps.
i am using Php/Mysql , i have the client table and trying to display data in a drop down list. Unfortunately, only one client is displayed in drop down which i have the total of 3 clients. Why only one ? For example : Michael King, Michael Jordan , Michael John when i select all the data from table and make an output to display in dropdown, Michael John is only in the dropdown.
Here my Mysql code :
//All data is selected from client_tb
<?php
$sql = "SELECT * FROM client_tb";
$result = $conn->query($sql);
while($row=mysqli_fetch_array($result))
{
$id = $row['id'];
$lname = $row['lname'];
$fname = $row['fname'];
}
?>
//my dropdown which will show the clients from client_tb but only one will appear.
<option value ="<?=$lname?><?=$fname?>"><?=$lname?> , <?=$fname?> </option> </select><br><br>
You can also achieve dropdown outside the while loop. Try this:
$sql = "SELECT * FROM client_tb";
$result = $conn->query($sql);
$options =array();
while($row=mysqli_fetch_array($result))
{
$options[] =$row;
}
Your dropdown:
<select name="">
<?php
foreach($options as $option):
echo '<option value ="'.$option['lname'].''.$option['fname'].'">'.
$option['lname'].','.$option['fname'].'</option>';
endforeach;
?>
</select>
You could also add your db query into a function , then call it.
function myFunction() {
$sql = "SELECT * FROM client_tb";
$result = $conn->query($sql);
while($row=mysqli_fetch_array($result))
{
$myvalues[] =$row;
}
return $myvalues;
}
Now the dropdown,
Note the options are inside the loop
<select name="">
<?php foreach($myvalues as $myvalue) {
echo '<option value="'.$myvalue['lname'].''.$myvalue['fname'].'">'.
$myvalue['lname'].','.$myvalue['fname'].'</option>';
}
?>
</select>
i have a list box that contain name of persons. In the column "Requestor" there are many names that repeat with the same name. So when I make the list box, the name shows all the name and some name shows more than one with the same name. How can i make the name just come out one only. Below are my codes.
<select name="requestor" id="requestor">
<option value="0">-- Select requestor --</option>
<?php
$getallCapex_transaction = mysql_query("SELECT * FROM capex_transaction ");
while($viewallCapex_transaction = mysql_fetch_array($getallCapex_transaction))
{
?>
<option id="<?php echo $viewallCapex_transaction ['Project_id'];?>"><?php
echo $viewallCapex_transaction['Requestor']; ?></option>
<?php } ?>
</select>
And one more question how can I show list of month and year that connect with mySql? can someone show me some codes. Thanks
Well, if there are multiple rows with the same "Requestor" but different "Project_id", that's just how it's gonna turn out. You can select distinct by Requestor in your SQL, but that will leave you unaware of what record you're pointing at.
Put your values in an array and some something like array_unique to create an unique array. Try something like this:
<?php
$requestorArray = array();
?>
<select name="requestor" id="requestor">
<option value="0">-- Select requestor --</option>
<?php
$getallCapex_transaction = mysql_query("SELECT * FROM capex_transaction ");
while($viewallCapex_transaction = mysql_fetch_array($getallCapex_transaction))
{
$requestorArray[$viewallCapex_transaction ['Project_id']] = $viewallCapex_transaction['Requestor'];
}
$requestorArray = array_unique($requestorArray);
foreach($requestorArray as $projectId => $requestor)
?>
<option id="<?php echo $projectId;?>"><?php
echo $requestor; ?></option>
<?php } ?>
</select>
SO I have about 60 fields of data queried from a database. They look like this in page source:
<option>Genesis</option>
<option>Exodus</option>
<option>Leviticus</option>
For example I wanna have it so that 1-20 are a certain optgroup and then 20-60 is another. Could I do it using my format of options of would they have to be numbered like this:
<option value="1">Genesis</option>
<option value="2">Exodus</option>
<option value="3">Leviticus</option>
This is for my php class, but i dont think php is involved in making optgroups here, or is it? Thank you hope you understand my question. Hoping for help.
I pull the data from mysql using this :
//Query the database for the results we want
$query = $mysqli->query("select distinct bname as Name from kjv"); ?>
And then output it in the select dropdown box using this:
<select>
<?php while($option = $query->fetch_object()){ ?>
<option><?php echo $option->Name; ?></option>
<?php } ?>
</select>
Use:
<select>
<?php
$i=1;
while($option = $query->fetch_object()){
if($i%10==1) echo "<optgroup label='Option Group'>";
echo "<option value='$i'>".$option->Name."</option>";
$i++;
if($i%10==1) echo "</optgroup>";
}
?>
</select>
Using while loop :
$query = "SELECT * FROM example";
$result = mysql_query($query) or die(mysql_error());
foreach ($row = mysql_fetch_array($result)) {
echo "'<option value={$row['id']}>{$row['value']}</option>'";
}
I have dropdown list that gets values form array based on a mySQL SELECT query. Everything is working fine except that I would like to add the option to select ALL values in the list. Here is my code...
$dataArray = array();
$result = mysql_query("SELECT id, user_name FROM apsc_customers");
while($row = mysql_fetch_assoc($result)) {
$dataArray[$row['id']] = $row['user_name'];
}
AND
if($this->customer_id == ""){
$this->arrFilteringFields[_CUSTOMER] = array("table"=>DB_PREFIX."customers", "field"=>"id", "type"=>"dropdownlist", "source"=>$dataArray, "sign"=>"like%", "width"=>"");
}
Looking forward to any replies.
Thx
Where is your dropdown list? I can't see it in your code. Do you mean html element select created by PHP and based on data from MySQL table? Then you need to use javascript code to select all options in such dropdown list. What is $this in your code? Provide more information, more code and more clarificatios.
Do you mean:
<select>
<?php
while($row = mysql_fetch_assoc($result)) {
?>
<option value="<?php echo $row['id']; ?>"><?php echo $row['user_name']; ?></option>
<?php } ?>
</select>