CODEWITHSUNDEEP
phphtmlmysqldatabaseselect
Yasmine Ra'fat
Yasmine Ra'fat

Reputation: 97

Fill drop down list via mysql select query

I have this php script that's supposed to fill a drop down list; I actually embedded it into the <select> element but it didn't work. Here's my script to clearly explain my problem :

<select name="cats">
   <?php
   require_once("connection.php");
   $rs = mysql_query("select cat_name from category");
   $count = 0;
   while($array = mysql_fetch_array($rs)){
    echo "<option>".$array[$count]."</option>";
   }
   mysql_close($con);
   ?>
 
  </select>

Can help me to define the error and if this is the wrong way to do it? What is the better approach?

Upvotes: 0

Views: 912

Answers (3)

Devang Rathod
Devang Rathod

Reputation: 6736

i think there is not any issues with your code , check your database connection again and try.because i try your code in my localhost and fetch cat_name in dropdown list.

Upvotes: 0

Sudhir Bastakoti
Sudhir Bastakoti

Reputation: 100205

change:

while($array = mysql_fetch_array($rs)){
   echo "<option>".$array[$count]."</option>";
}

to

while($array = mysql_fetch_array($rs)){
    echo "<option value='".$array["cat_id"]."'>".$array["cat_name"]."</option>";
}

where cat_id is id of your category table

Upvotes: 1

Ynhockey
Ynhockey

Reputation: 3932

In general terms this is the right way to do it. I assume the problem is here:

echo "<option>".$array[$count]."</option>";

Your $count is necessarily 0, so you will always be outputting $array[0], which is the first field that's being selected (usually some kind of ID, but it depends on the table structure). This may or may not be the right field in the table.

What I recommend is first to specify the exact fields you're selecting. For example:

select `id`, `text` from `category`

And then use the associative array $array to get what you want, for example: $array['id']

Upvotes: 0

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