selected value get from db into dropdown select box option using php mysql error

Question

I need to get selected value from db into select box. please, tell me how to do it. Here is the code. Note: 'options' value depends on the category.

<?php 
  $sql = "select * from mine where username = '$user' ";
  $res = mysql_query($sql);
  while($list = mysql_fetch_assoc($res)){
    $category = $list['category'];
    $username = $list['username'];
    $options = $list['options'];
?>

<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
   <option value="0">Please Select Option</option>
   <option value="PHP">PHP</option>
   <option value="ASP">ASP</option>
</select>

<?php 
  }
?>

Here i need to print Category (which category user submitted along with option select box which option user submitted when posting form). this form is edit form user can see which category and option he selected . if user wants to change option he can do in edit form. — Php Gemini
– Php Gemini, Commented Sep 11, 2013 at 6:01
i have tried as <option value="<?php echo $options ?>>"<?php echo $options ?> </option> but it only prints selected value instead of this i need as i want to show user option value in select box first and second remain option also. but i have no idea how to do. — Php Gemini
– Php Gemini, Commented Sep 11, 2013 at 6:04

rakeshjain · Accepted Answer · 2013-09-11 06:02:33Z

41

I think you are looking for below code changes:

<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP" <?php if($options=="PHP") echo 'selected="selected"'; ?> >PHP</option>
<option value="ASP" <?php if($options=="ASP") echo 'selected="selected"'; ?> >ASP</option>
</select>

answered Sep 11, 2013 at 6:02

rakeshjain

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4 Comments

rakeshjain Over a year ago

@BolemVeeru If $options from your code contains one of the courses from the drop down for courses, it should work. Can you just print and check the value of $options before the dropdown

Bhargav Chudasama Over a year ago

how its done when select tag pass into php variable lik

$role= '<select class="form-control" name="role" id="role">                             <option value="">Select</option>                             <option value="a" <?php if($admin[0]->role=="Admin") echo selected="selected"; ?>>Admin</option>                             </select>';

Robusto Over a year ago

Just echo " selected" (note space at beginning) and you'll be fine.

Dhruv Thakkar Over a year ago

how to achieve this in for loop? if your data is coming from database

Dharman · Accepted Answer · 2021-02-01 14:17:47Z

9

The easiest way I can think of is the following:

<?php

$selection = array('PHP', 'ASP');
echo '<select>
      <option value="0">Please Select Option</option>';

foreach ($selection as $selection) {
    $selected = ($options == $selection) ? "selected" : "";
    echo '<option '.$selected.' value="'.$selection.'">'.$selection.'</option>';
}

echo '</select>';

The code basically places all of your options in an array which are called upon in the foreach loop. The loop checks to see if your $options variable matches the current selection it's on, if it's a match then $selected will = selected, if not then it is set as blank. Finally the option tag is returned containing the selection from the array and if that particular selection is equal to your $options variable, it's set as the selected option.

edited Feb 1, 2021 at 14:17

Dharman♦

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answered Aug 22, 2016 at 0:38

independent.guru

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Comments

chirag ode · Accepted Answer · 2013-09-11 06:06:16Z

5

for example ..and please use mysqli() next time because mysql() is deprecated.

<?php
$select="select * from tbl_assign where id='".$_GET['uid']."'"; 
$q=mysql_query($select) or die($select);
$row=mysql_fetch_array($q);
?>

<select name="sclient" id="sclient" class="reginput"/>
<option value="">Select Client</option>
<?php $s="select * from tbl_new_user where type='client'";
$q=mysql_query($s) or die($s);
while($rw=mysql_fetch_array($q))
{ ?>
<option value="<?php echo $rw['login_name']; ?>"<?php if($row['clientname']==$rw['login_name']) echo 'selected="selected"'; ?>><?php echo $rw['login_name']; ?></option>
<?php } ?>
</select>

edited Sep 11, 2013 at 6:06

answered Sep 11, 2013 at 6:00

chirag ode

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Comments

Nasser Hajlawi · Accepted Answer · 2019-08-07 02:58:27Z

2

Just Add an extra hidden option and print selected value from database

<option value="<?php echo $options;?>" hidden><?php echo $options;?></option>
<option value="PHP">PHP</option>
<option value="ASP">ASP</option>

answered Aug 7, 2019 at 2:58

Nasser Hajlawi

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Comments

Dharman · Accepted Answer · 2021-03-30 20:55:59Z

2

Select value from drop down.

<select class="form-control" name="category" id="sel1">
    <?php foreach($data as $key =>$value) { ?>                         
        <option value="<?php echo $data[$key]->name; ?>"<?php if($id_name[0]->p_name==$data[$key]->name) echo 'selected="selected"'; ?>><?php echo $data[$key]->name; ?></option>
    <?php } ?>
</select>

edited Mar 30, 2021 at 20:55

Dharman♦

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answered Jun 11, 2017 at 14:55

Siddharth Shukla

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

THE EASIEST SOLUTION

It will add an extra in your options but your problem will be solved.

<?php 
    if ($editing == Yes) {
        echo "<option value=\".$MyValue.\" SELECTED>".$MyValue."</option>";
    }
?>

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 13, 2020 at 18:08

K. Raza

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Comments

Haroon Hayat · Accepted Answer · 2021-06-12 10:30:26Z

$option = $result['semester'];
<option >Select</option>
                    <option value="1st" <?php if($option == "1st") echo 'selected = "selected"'; ?>>1st</option>
                    <option value="2nd" <?php if($option == "2nd") echo 'selected = "selected"'; ?>>2nd</option>
                    <option value="3rd" <?php if($option == "3rd") echo 'selected = "selected"'; ?>>3rd</option>
                    <option value="4th" <?php if($option == "4th") echo 'selected = "selected"'; ?>>4th</option>
                    <option value="5th" <?php if($option == "5th") echo 'selected = "selected"'; ?>>5th</option>
                    <option value="6th" <?php if($option == "6th") echo 'selected = "selected"'; ?>>6th</option>
                    <option value="7th" <?php if($option == "7th") echo 'selected = "selected"'; ?>>7th</option>
                    <option value="8th" <?php if($option == "8th") echo 'selected = "selected"'; ?>>8th</option>
                </select>

Tempo · Accepted Answer · 2016-04-14 15:53:31Z

0

BEST code and simple

<select id="example-getting-started" multiple="multiple" name="category">

    <?php
    $query = "select * from mine";
    $results = mysql_query($query);

    while ($rows = mysql_fetch_assoc(@$results)){ 
    ?>
    <option value="<?php echo $rows['category'];?>"><?php echo $rows['category'];?></option>

    <?php
    } 
    ?>
</select>

edited Apr 14, 2016 at 15:53

Tempo

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answered Aug 22, 2015 at 13:42

razvan

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Comments

Draken · Accepted Answer · 2016-05-04 12:09:06Z

0

You can also do like this ....

<?php  $countryname = $all_meta_for_user['country']; ?>

<select id="mycountry"  name="country" class="user">

    <?php $myrows = $wpdb->get_results( "SELECT * FROM wp_countries order by country_name" );
    foreach($myrows as $rows){
        if( $countryname == $rows->id ){ 
            echo "<option selected = 'selected' value='".$rows->id."'>".$rows->country_name."</option>";
        } else{ 
            echo "<option value='".$rows->id."'>".$rows->country_name."</option>";
        }
    }
    ?>
</select>

edited May 4, 2016 at 12:09

Draken

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answered May 4, 2016 at 11:12

narinder kumar

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Comments

double-beep · Accepted Answer · 2020-08-23 15:40:27Z

0

Answer is simple. when u pass value from dropdown.

Just use as if else.

for eg:

foreach($result as $row) {                      
   $GLOBALS['output'] .='<option value="'.$row["dropdownid"].'"'. 
   ($GLOBALS['passselectedvalueid']==$row["dropwdownid"] ? ' Selected' : '').' 
   >'.$row['valueetc'].'</option>';
}

edited Aug 23, 2020 at 15:40

double-beep

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answered Aug 23, 2020 at 13:58

siraj

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Comments

Nuzhat Shirgaonkar · Accepted Answer · 2021-03-22 09:02:48Z

0

<?php 
  $sql = "select * from mine where username = '$user' ";
  $res = mysql_query($sql);
  while($list = mysql_fetch_assoc($res)){
    $category = $list['category'];
    $username = $list['username'];
    $options = $list['options'];
?>

<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP" <?php echo $options == 'PHP' ? 'selected' : ''; ?> >PHP</option>
<option value="ASP" <?php echo $options == 'ASP' ? 'selected' : ''; ?> >ASP</option>
</select>

<?php 
  }
?>

edited Mar 22, 2021 at 9:02

answered Mar 22, 2021 at 8:49

Nuzhat Shirgaonkar

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1 Comment

jasie Over a year ago

Thank you for this code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you’ve made.

Dharman · Accepted Answer · 2021-03-30 20:58:28Z

0

USING PDO

<?php
$username = "root";
$password = "";
$db = "db_name";

$dns = "mysql:host=localhost;dbname=$db;charset=utf8mb4";
$conn = new PDO($dns,$username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$sql = "select * from mine where username = ? ";

$stmt1 = $conn->prepare($sql);
$stmt1->execute(array($_POST['user']));
$all = $stmt1->fetchAll(); ?>

<div class="controls">
    <select  data-rel="chosen"  name="degree_id" id="selectError">
        <?php 
        foreach($all as $nt) { 
            echo "<option value =$nt[id]>$nt[name]</option>";
        }
        ?>
    </select>
</div>

edited Mar 30, 2021 at 20:58

Dharman♦

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answered May 20, 2018 at 8:37

7ani9

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Comments

cheinan · Accepted Answer · 2021-05-19 09:17:59Z

0

I'm using eval() PHP function like this:

My PHP code:

$selOps1 = $selOps2 = $selOps3 = '';
eval('$selOps'. $dbRow["DBitem"] . ' = "selected";');

Then in my select box I use it like this:

<select>
   <option <?=$selOps1?> value="1">big</option>
   <option <?=$selOps2?> value="2">Middle</option>
   <option <?=$selOps3?> value="3">Small</option>
</select>

answered May 19, 2021 at 9:17

cheinan

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Comments

Hassan Qasim · Accepted Answer · 2021-10-08 07:55:12Z

0

Put value from db into a variable and check like following code example

<select class="form-control" name="currency_selling" required >
     <option value="">Select Currency</option>
     <option value="pkr" <?=$selected_currency == 'pkr' ? ' selected="selected"' : '';?> >PKR</option>
  <option value="dollar"  <?=$selected_currency == 'dollar' ? ' selected="selected"' : '';?> >USD</option>
  <option value="pounds"  <?=$selected_currency == 'pounds' ? ' selected="selected"' : '';?> >POUNDS</option>
  <option value="dirham"  <?=$selected_currency == 'dirham' ? ' selected="selected"' : '';?> >DRHM</option>
  </select>

answered Oct 8, 2021 at 7:55

Hassan Qasim

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Comments

Pradeeshnarayan · Accepted Answer · 2013-09-11 06:04:05Z

-1

This may help you.

?php 
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res))
{
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
// Assuming $list['options'] is a coma seperated options string 
$arr=explode(",",$list['options']);
<?php foreach ($arr as $value) { ?>
   <option value="<?php echo $value; ?>"><?php echo $value; ?></option>
<?php } >
</select>

<?php 
}
?>

answered Sep 11, 2013 at 6:04

Pradeeshnarayan

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