[SOLVED] Form update without submit button

zang8027 · December 30, 2008

I have a php file that has 2 drop down boxes. Each box is filled with values using a php and a database. I want to make it so that when the user selects a value out of the top drop down box, it automatically updates the values that are in the lower one, without hitting a submit button. I have it working with 2 submit buttons but I dont want users to have to keep hitting submit.

Here is the code:

<FORM action="dropdown.php" method="get">
<SELECT NAME="cityLargeList">
<?php
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM cityLarge WHERE state_ID = 39";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$cityLargeID=$row['cityLarge_ID'];
//store into a variable the Name for the current row
$cityLargeName=$row['cityLarge_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$cityLargeID'>$cityLargeName";
}
//close the connection
mysql_close($link);

?>
</SELECT>
<input type="submit" value="Find City" />
</FORM>

<FORM action="test.php" method="get">
<SELECT NAME="citySmallList">
<?php
//Get the value from City Large
$userCityLarge = $_GET['cityLargeList'];
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM citySmall WHERE cityLarge_ID = $userCityLarge";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$citySmallID=$row['citySmall_ID'];
//store into a variable the Name for the current row
$citySmallName=$row['citySmall_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$citySmallID'>$citySmallName";
}
//close the connection
mysql_close($link);

?>
</SELECT>
<input type="submit" value="Find hotels" />
</FORM>

That code has 2 submit buttons. Any ideas?

ILMV · December 30, 2008

Could you send us the generated HTML code.

ILMV

zang8027 · December 30, 2008

What code?

dMilesFox · December 30, 2008

What code?

The one you usually get by right clicking with your mouse on the page ("View Source") That way u can get the rendered html source and not the php server source.

revraz · December 30, 2008

You'll need to use javascript if you want it to auto update.

zang8027 · December 30, 2008

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Untitled Document</title>

</head>

<body>

To find the restaurants in your area, please select where you are staying.

<br />

<OPTION VALUE='1'>Pittsburgh<OPTION VALUE='2'>Philadelphia<OPTION VALUE='3'>Scranton<OPTION VALUE='4'>Harrisburg<OPTION VALUE='5'>Erie<OPTION VALUE='6'>State College</SELECT>

</FORM>

<br />

<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>C:\xampp\htdocs\Click-N-Dine\dropdown.php</b> on line <b>59</b><br />

</SELECT>

</FORM>

</body>

</html>

Ok, i'll look on line for some javascript.

premiso · December 30, 2008

Not a php issue.

Look into the onChange event in HTML/Javascript.

You will have to use an ajax call to make it purely dynamic. I would look into jQuery for the Ajax call. The other option is when onchange is activated (meaning they selected a city in the list) it automatically submits the form and re-generates the page. I think that is what you are after.

onChange myfriend =)

<FORM action="dropdown.php" name="dropdown" method="get">
<select name=myselect onChange="this.dropdown.submit()">

Should submit the form onchange.

dMilesFox · December 30, 2008

Use the "onchange" attribute to call the proper function, Then you can evaluate the selected value on the first drop down box and depending on its value, something needs to be set on the second drop down box.

zang8027 · December 30, 2008

ok, thats exactly what i need. I added the Javascript there but nothing happened. Sorry, I have no idea about java so i must be missing something.

<FORM action="dropdown.php" name="dropdown" method="get">
<SELECT NAME="cityLargeList" onChange="this.dropdown.submit()">

zang8027 · December 30, 2008

Here i attached the 3 files. The file i am working on is dropdown.php. There is also the database and another test file.

[attachment deleted by admin]

dennismonsewicz · December 30, 2008

change

this.dropdown.submit()

to

this.form.submit()

zang8027 · December 30, 2008

Hey that works! Thanks a bunch!

dMilesFox · December 30, 2008

ok, thats exactly what i need. I added the Javascript there but nothing happened. Sorry, I have no idea about java so i must be missing something.
<FORM action="dropdown.php" name="dropdown" method="get">
<SELECT NAME="cityLargeList" onChange="this.dropdown.submit()">

this "this.dropdown.submit()" it's a bad call

Note: great to know its working now!

premiso · December 30, 2008

change

this.dropdown.submit()

to

this.form.submit()

lol, well at least this is the php section and not javascript or else I might be embarrassed.

dennismonsewicz · December 30, 2008

hey it works...

zang8027 · December 30, 2008

One other problem i keep getting tho..

Since my values are set using the drop down from above, my default values are not being set on the lower drop downs until the top is selected so i keep getting errors like this in the lower ones until I select the top drop menu:

Please Select a City Warning: mysql_fetch_array():supplied arguent is not a a valid mySQL result resource:...

I tried adding an if statement and moving the default value down that says

if($stateID == NULL) {
   print"<OPTION>Please Select a state";
} else {
   print "<OPTION VALUE='$cityLarge_ID'>$cityLarge_name";
}

But that does nothing.

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[SOLVED] Form update without submit button

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