I am attempting to create a drop-down menu that links to user profile pages and includes user images next to it. I am a bit new to PHP so bear with me if this is a simple question. Here is the code I've created:
function dropdown() {
$query = "SELECT * FROM usersbadges";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$name=$row['name'];
$image=$row['image'];
}
mysql_close();
}
$options.=<option value=\$name\>.$image </option>;
<select name=name>
<option name=0>
print $options
</select>
It tells me that the < brackets are unexpected but I'm at a loss as to how to put them. If those were fixed, would this code create me a drop down menu with user names accompanied by their images? I'd just like to know if I'm on the right track/what the next step is. All help is greatly appreciated
use this
$query = "SELECT * FROM usersbadges";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$name=$row['name'];
$image=$row['image'];
}
mysql_close();
}
$options.="<option value='$name'>$image </option> <select name=name> <option name=0>";
echo $options;
echo '</select>';
Related
Newbie to php/mysql. Would like to have the option of inserting a new location in my form that is then added to the db or choose from a pulldown of locations currently registered in db. Below is my code (that works) to present the pulldown options. Not sure how to add the ability to include new location. I can do either/or (input/text or select/option) but not both in same form. Thanks in advance for any help.
......
<p>Location: <select name="e_location">
<?php if (isset($_POST['e_location'])) {echo $_POST['e_location'];
}
?>" />
<?php $q = "SELECT DISTINCT e_location FROM events";
echo '<option selected value="">Choose Previous City/State</option>';
$r = mysqli_query($dbc, $q);
if (mysqli_num_rows($r) >= 1) {
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
echo '<option
value="'.$row['e_location'].'">'.$row['e_location'].'</option>';
}
.....
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 drop down menu with two options in html, I also created a PHP script that checks what option from the drop down menu has been selected and based on the selection executes a mysql query to fetch data from database.
But I am also trying to echo out a new drop down menu with the results obtained from database and that is where the I am struggling because no errors are diaplayed but also no drop down menu is 'echoed' out onto the page.
HTML:
<?php require "course.php" ?>
<select id="workshop" name="workshop" onchange="return test();">
<option value="">Please select a Workshop</option>
<option value="Forex">Forex</option>
<option value="BinaryOptions">Binary Options</option>
</select>
PHP code:
$form['workshop'] = $_POST['workshop'];
$form['forex'] = $_POST['Forex'];
$form['binary'] = $_POST['Binary'];
//Retrieve Binary Workshops
if($form['workshop'] == 'Forex'){
$sql2 = "SELECT id, course, location FROM courses WHERE course LIKE '%Forex%' OR course LIKE '&forex%'";
$query2 = mysqli_query($link, $sql2);
echo "<select id='Forex' name='Forex' style='display: none'>";
while($result2 = mysqli_fetch_assoc($query2)){
echo "<option value=''>".$result2['course']."</option>";
}
echo "</select>";
echo '</br>';
}
Could someone point out a mistake I am doing or perhaps suggest where I could look for answers
you said $query2 is displaying value in print_r.so the only mistake i find in your code is display:none .
remove display:none
echo "<select id='Forex' name='Forex' style='display: none'>";
----------------------------------------------------------------------^
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>