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>
Related
Not experienced with creating forms in PHP.
I can get my form to produce a dropdown list that has one of my rows listed as an option, but as soon as I try to concatenate 2 rows together (from the same table) for option output...
a) It just doesn't work and I get errors
b) I get the first row as a single option, then my next row as a separate option.
I know there is a simple solution to this, but I am an online student just learning, and I can't seem to find a good example of the code to write it. I'm pretty sure it's an issue of quotes not being placed correctly.
MySQLTable Data:
Table Name: courses
Table Rows: course_id, course_name, max_enrolment
Sample Data: LO-COMP-8001, Intro to HTML, 20
function select_course(){
global $open;
$select = "SELECT * FROM courses";
$result = mysqli_query($open, $select);
return $result;
}
<form action="insert.php" method="post">
<dl>
<dt>Select Course</dt>
<dd><select name="course_id">
<?php // CREATE dropdown menu
$result = select_course();
while ($row = mysqli_fetch_assoc($result)) {
foreach ($row as $selection) {
echo "<option value=\"$selection\">$selection</option>";
}}
?>
</select>
</dd>
</dl>
Then there are a few more form fields such as student name and student id afterwards...
Goal Output:
course_id course-name
"LO-COMP-8001 Intro to HTML" ... as a single connected dropdown option and other remaining courses in a dropdown menu
Current Output:
LO-COMP-8001 (as an option)
Intro to HTML (as another option! ... No good)
20 (must be hidden, I need to check if course is full in another function and either allow or deny a student to enrolled etc.)
I have tried:
// output is the one mentioned above..
echo "<option value=\"$selection\">$selection</option>";
// or alternatively...
echo '<option value="'.$row['course_id'].'">'.$row['course_id'].'</option>';
But the second option creates all kinds of weird results.
This is what I am experimenting with right now...
echo '<option value="'.$row['course_id'] $row['course_name']'">'.$row['course_id'] $row['course_name'].'</option>';
But there is a bunch of issues with quotes and square brackets, and I just don't know how to format it correctly for the output.
Any assistance is appreciated.
$row holds the entire row as an associative array therefore no need for the 'foreach' loop.
function select_course(){
global $open;
$select = "SELECT * FROM courses";
$result = mysqli_query($open, $select);
return $result;
}
<form action="insert.php" method="post">
<dl>
<dt>Select Course</dt>
<dd><select name="course_id">
<?php // CREATE dropdown menu
$result = select_course();
while ($row = mysqli_fetch_assoc($result)) {?>
<option value="<?php echo $row["course_id"]; ?>"><?php echo $row["course_name"]; ?></option>
<?php }
?>
</select>
</dd>
</dl>
</form>
I was able to come up with another solution as well:
Once the foreach loop was removed, I tried cleaning up the code some... I'm not sure if this is uncommon or 'bad' style, but it does work.
$result = select_course();
while ($row = mysqli_fetch_assoc($result)) {
$course_id = $row['course_id'];
$course_name = $row['course_name'];
echo "<option value=\"$course_id\">$course_id $course_name</option>";
Results in: LO-COMP-8001 Intro to HTML as a single option, plus all my other courses in the database.
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>
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 am using a PHP script within an HTML page to create a dropdown menu with results populated from a database. I am using the PHP script in the HTML page because I'd like the dropdown menu to remain on the same page as a number of options.
When creating an option in the dropdown menu via an echo, it won't allow me to use the value of a variable (i.e. the value of a field in the database that has been retrieved) in the option tag. Here is the code:
<select name="comics">
<OPTION>Select an option</OPTION>;
<?php
include_once('includes/conn.inc.php');
$query = ("SELECT comicID, comicName FROM comic");
$result = mysqli_query($conn, $query);
while ($row = $result->fetch_assoc())
{
echo "<option> .$row['comicName']. </option>";
}
?>
</select>
As is, the code creates the drop down menu and adds the "Select an option" line. The second option is created, but the value reads as " .$row['comicName']. " instead of, for example, "Superman".
Thanks in advance.
Have you named the file with a php extension ? try doing a print_r with $result before the while loop
Your solution tries performing concatenation without closing the double-quotes first, so it treats the variable as a string. Try this:
echo "<option>" . $row['comicName'] . "</option>";
Try:
<select name="comics">
<OPTION>Select an option</OPTION>;
<?php
include_once('includes/conn.inc.php');
$query = ("SELECT comicID, comicName FROM comic");
$result = mysqli_query($conn, $query);
while ($row = $result->fetch_assoc())
{
echo "<option value='".$row['comicName']."'>".$row['comicName']."</option>";
}
?>
</select>
Let me know if works.
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>'";
}