Javascript Function to populate a select menu from MySQL

Question

I have this PHP Code which populates a select menu from a MySQL Database...

<select name="input" id="input">
<?php
$sql="SELECT * from table ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
while($result=mysql_fetch_array($rs))
{
    echo '<option value="'.$result["db_field"].'">'.$result["db_field"].'</option>';
}
?>
</select>

which works perfectly fine but i need to somehow get it into a javascript function.

I have the javascript code that when you click a button it adds more text boxes and another select menu but it does not populate the data from the database in any new (added on) select menus

i think thats what i need but im not sure how, any chance you could give me a big hand on this please — user2135867
– user2135867, Commented Mar 20, 2013 at 23:26
This post can give you an idea, hope that help you! =) stackoverflow.com/questions/9665372/… — Superlandero
– Superlandero, Commented Mar 20, 2013 at 23:28

jsshah · Accepted Answer · 2013-03-20 23:32:41Z

0

You can probably convert the options into JSON using json_encode (I am not a PHP programmer and dont know exact semantics of using it)

In PHP do something like:

echo '<script>var optionsJSON = '.json_encode(mysql_fetch_array($rs)).'</script>'

In javascriptn do something like (I am using jquery):

var select = $('select.classOfThisSelect');
var options = JSON.parse(optionsJSON);
for(var i = 0; i < options.length; i++)
  $('option').attr({value: options[i]}).append(options[i]).appendTo(select);

optionsJSON will be the JSON string which will be globally available You can freely use it in your Javascript function

Note: You may need to surround the json_encode with quotes

answered Mar 20, 2013 at 23:32

jsshah

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Comments

RobG · Accepted Answer · 2013-03-20 23:47:22Z

0

Start by converting the string of values into an array, so you have something like:

var values = ['value0','value1','value2'];

Then you can convert them to options and add them to a select element like:

function addOptions(select, values) {
  for (var i=0, iLen=values.length; i<iLen; i++) {
    select.appendChild(new Option(values[i],values[i]));
  }
}

And call it like:

addOptions(document.getElementById('input'), values);

after the select is added to the page.

Incidentally, you don't need to add both an id and name to form controls. You must have a name for them to be submitted, the ID is unnecessary. If you get a reference to the form you can access controls as named properties of the form, so you might reference the select using:

document.forms[0]['input'];

or

document.forms[0].input;

and so on. Note that "input" isn't a good choice of control name.

edited Mar 20, 2013 at 23:47

answered Mar 20, 2013 at 23:42

RobG

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3 Comments

user2135867 Over a year ago

aha okay - how would i get the values from the database into the "var values" ?

user2135867 Over a year ago

i cannot work out how to get the mysql values into the values var in javascript?

RobG Over a year ago

Are you getting them from an XMLHttpRequest? If so, you can have them returned as JSON and convert that to an array, or just as a delimited string and use split on the delimiter.

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