Currently setting up a form for users to select aircraft information from. Currently, the form allows users to select an airplane and in return when the form is submitted it passes the value"aircraft type" to an API. Due to this, I cannot set the values to be the registration of the airplanes so have put this into the ID for now (open to suggestions).
<select name="type" id="plane-list">
<option selected="selected">Choose one</option>
<?php
foreach($allaircraft as $item){
?>
<option id="<?php echo $item->registration; ?>" value="<?php echo $item->icao; ?>"><?php echo $item->name; ?> - <?php echo $item->registration; ?></option>
<?php
}
?>
</select>
As you can from above these are all dynamic values which change over time so hard coding the options are not possible.
So summed up I need to pull the ID or $item->registration; instead of ICAO code. This then needs to be set as a variable so that I can call a function with this code.
Let me know if you need me to clear up anything. Don't normally questions on here so be nice.
Related
I have a create.php which creates a user, and edit.php for editing information of the student.
I'm a beginner in PHP so if this might be a question that has a very easy method, please do tell.
I've been searching for retrieving a value that has already been set on a dropdown, but the only things I've found are:
a.) They only print the selected value, not retrieve the already-set option.
sample code:
<p>Schedule</p>
<select value="<?= $data->schedule; ?>" name = "schedule">
<?php
$schedule = array ('Morning', 'Afternoon');
foreach ($schedule as $sched) { ?>
<option value="<?php echo $sched; ?>"><?php echo $sched; ?></option>
</select>
<?php } ?>
b.) I have no knowledge of javascript, I tried it, but as I have no knowledge of it, I can't seem to make it work.
sample codes I tried:
<p>Schedule</p>
<select id = "my_select" name = "schedule">
<option id = "o1" id="id1">Morning</option>
<option id = "o2" id="id2">Afternoon</option>
</select>
<script type="text/javascript">
$("#my_select").change(function() {
var id = $(this).find('option:selected').attr('id');
});
</script>
I also tried
document.getElementById('schedule').selectedOptions[0].text
but still does not output "afternoon"
This is my last resort as I have no other resource to find.
I added a picture of what I wanted to do since my question might be a little confusing.
Selects use selected attribute not the value to define the selected state.
So in your code check the value and then apply selected.
<p>Schedule</p>
<select name="schedule">
<?php foreach (['Morning', 'Afternoon'] as $sched): ?>
<option value="<?= $sched ?>"<?= ($data->schedule == $sched ? ' selected' : null) ?>><?= $sched ?></option>
<?php endforeach ?>
</select>
i want to get the id of selected option in dropdown.
I have a dropdow that displays company names from database. Now I want get the id of that company. How do i do that?
<select required name="org-list" id="org-list" class="form-control">
<option value="">Select</option>
<?php foreach($org as $value) { ?>
<option id="org" value="<?php echo $value['org_name'];?>"><?php echo $value['org_name']; ?></option>
<?php } ?>
</select>
The Model
public function get_organisation()
{
$q = $this->db->select('*')
->from('company')
->get();
return $q->result_array();
}
And in my controller i want the id of selected option from database.
$data = $this->key_m->array_from_post(array('id','org-list','keys'));
$data['keys'] = $license;
var_dump($data);
You should put the organization id in option tag's value. like this
<option value="<?php echo $value['org_id'];?>"><?php echo $value['org_name']; ?></option>
No need to provide id attribute to each option tag. When you will submit the form value of "org-list" will be selected organization id.
This is the ajax way
$.post(url,{ id : "id value from drop down here" },function(data){
//This is the call back for success
})
For that one, you need jquery. The is the easiest way to use. Url is where you are posting back to server.
In your server side catch like this
$this->input->post('id')
Because you used "id" in the ajax request. So you have to retrieve it using the field name 'id' . Hope you get it.
There are a couple things which you "should" / "could" change. At first I noticed that your ID attribute is "org", Id's are supposed to be unique and yours is not. Though, back to the actual question. You should change your code as follow:
<select required name="org-list" id="org-list" class="form-control">
<option value="">Select</option>
<?php foreach($org as $value): ?>
<option id="<?php echo $value['org_id'];?>" value="<?php echo $value['org_id'];?>"><?php echo $value['org_name']; ?></option>
<?php endforeach; ?>
</select>
Now in your controller you grab the selected ID like: $this->input->post('org-list').
You have to provide value for each id attr. for corresponding <option>
<select required name="org-list" id="org-list" class="form-control">
<option value="">Select</option>
<?php foreach($org as $value): ?>
<option id="<?php echo $value['org_id'];?>" value="<?php echo $value['org_name'];?>"><?php echo $value['org_name']; ?></option>
<?php endforech; ?>
</select>
i am developing a website for meeting schedule.the user which was currently logged in need to send request to the guests.the names of the guests are provided in the user table in database along with the host.so when i try to fetch the names of all the user expect that of the host only the host's name is displayed in the list menu.my code is as follows
<?php
include("db.php");
$q=mysql_query("select * from user where name='$name'");
while($r=mysql_fetch_array($q))
{
$uid=$r['uid'];
?>
<select name="">
<option name="">geust name</option>
<?php
echo'<option name="">'.$name.'</option>';
?>
</select>
<?php
}?>
i want to list out the guest names in the option other than that of the host.but i dont know how to display it
Try this:
<?php
include("db.php");
$q=mysql_query("select * from user where name='$name'");
?>
<select name="sample">
<option value="<?php echo $r['guestname']; ?>">geust name</option>
<?php
while($r=mysql_fetch_array($q))
?>
<option value="<?php echo $r['name']; ?>"><?php echo $name; ?></option>;
<?php
}
?>
</select>
In your above example you are creating a new select element each Time you loop through results. you are using $name as option value. $name is the value of your search criteria. You want to use $r['name']. Witherwind's answer will get you what you asked. The only thing you need to change is the first option outside loop should be blank.
currently i am working on a system. here i have a drop down list and its values are populating from database. now what i need is when ever some one selects a record from the list the selected value should be get stored in a session variable which ll get displayed in an another page. can we do so. i am coding in php.
my code is:
<td align="center">
<?php
$sql_dep=" Select * from places_tbl";
$row_dep = $db->Execute($sql_dep);
$total_dep = $row_dep->RecordCount();
?>
<select name="place" class="txtbx08" id="place">
<option value="">--Please Select City--</option>
<?php if($total_dep>0) {
while(!$row_dep->EOF)
{
?>
<option value="<?php echo $row_dep->fields["place_id"];?>">
<?php echo ucfirst($row_dep->fields["place_name"]); ?>
</option>
<?php
$row_dep->MoveNext();
}}
?>
</select></td>
I'm not sure how you're retrieving the selected value (assumingly through POST), but the normal procedure would be:
session_start();
$_SESSION['place'] = $_POST['place'];
Disclaimer: It's been a while since I last wrote any code. The quality of my code is likely to be sub-par. You've been warned.
Greetings.
I am coding a basic form that uses a SELECT list I populate from my database.
However, my user needs the form to be dynamic and wants to be able to select either MasterTable1 or MasterTable2 or MasterTable3...
Instead of hardcoding the table name for the database query that populates the SELECT list, I attempted to implement a basic Ajax action (used example from w3schools)...and that's when I lost my sanity...
I can output <div id='txtHint'></div> in my page and it shows the correct table name that was picked.
But how do I pass the correct table name to my query that will populate my SELECT list???
I tried
<select name="DBFilename" id="DBFilename" size="0">
<option value="">Select Filename</option>
<?php
$sql="select distinct filename from "."<div id='txtHint'></div>";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)){ ?>
<option value="<?php echo $row['filename']; ?>"><?php echo $row['filename']; ?></option>
<?php } ?>
</select>
But to no avail. This is confusing since I can do this...
...
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","gettable.php?q="+str,true);
xmlhttp.send();
}
</script></head>
<body><form><select name="SrcTbl" id="SrcTbl" size="0" onchange="showTable(this.value)">
<option value="">Select Data Table</option>
<option value=""> </option>
<option value="MasterTable1">up to 31 days old</option>
<option value="MasterTable2">62 days old</option>
</select>
</form><br /><div id="txtHint"><select name="tabList"><option></option></select> </div>
</body></html>
And the name of my table will be displayed in the SELECT list 'tablist'.
How do I pass the correct table name to my query that will populate my SELECT list? Thanks!!
(Pastebin =>form code)
m8, ajax is mainly used for user experience and populating a select list is so easy mode that it shouldnt be bothered with ajax in the first place!
You should use ajax if you want to use some methods on user submitted data and create an illusion of seamless data exchange between the server and the client, and based on the return results render a corresponding view or an element of the view.
unless you load every element of the view with ajax, you should populate your html with php from the start!
<select name="DBFilename" id="DBFilename" size="whatever since style belongs to css">
<option selected="selected">Select Filename</option>
<?php
$sql="SELECT DISTINCT filename from 'wherever'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)){ ?>
<option value="<?php echo $row['filename']; ?>"><?php echo $row['filename'];?>
</option>
<?php } ?>
</select>
Create a separate php script which returns a list of select options -> values depending on the table name given to it. You must remember to protect against sql injection. Then use ajax to retrieve the list and insert it into the select.