POST multiple columns of data to one row

[lukep11a]

Hi, I am trying to post 2 selections from radio buttons into 1 row of a mySQL table, this is my code:

 

<form name="form1" method="post" action="new1.php">
<h2>Section 1</h2>
<input type="radio" name="section1" value="Man Utd">Man Utd<br>
<input type="radio" name="section1" value="Chelsea">Chelsea<br>
<input type="radio" name="section1" value="Man City">Man City<br>
<input type="radio" name="section1" value="Arsenal">Arsenal<br>
<h2>Section 2</h2>
<input type="radio" name="section2" value="Liverpool">Liverpool<br>
<input type="radio" name="section2" value="Tottenham">Tottenham<br>
<input type="radio" name="section2" value="Everton">Everton<br>
<input type="radio" name="section2" value="Fulham">Fulham<br>
    <br>
    <input type="submit" value="Submit Team Selections">
</form>

 

on submit, the following code is run:

 

$section1 = $_POST['section1'];
$section2 = $_POST['section2'];
$query="INSERT INTO selections (section1)VALUES ('".$section1."')";
$query="INSERT INTO selections (section2)VALUES ('".$section2."')";
mysql_query($query) or die ('Error updating database');

 

it worked fine when i tested with only one selection, but after trying with 2 selections i found that only the second selection was posted to the mySql table. Does anybody know why this is? Any help would be very much appreciated.

 

Thanks

Luke

[Pikachu2000]

You can't run two queries with one call to mysql_query(). In this case there isn't even a reason to run two queries. Validate that both values are present and valid and run one query to insert them.

 

$query = "INSERT INTO selections ( section1, section2 ) VALUES ( '$section1', '$section2' )";

[fugix]

when you specify your $query value twice, you are actually overriding the first value of $query. I recommend doing something like this.

$section1 = $_POST['section1'];
$section2 = $_POST['section2'];
$query="INSERT INTO selections (section1, section2) VALUES ('$section1', '$section2')";
mysql_query($query) or die ("Error updating database: " . mysql_error());

[lukep11a]

Thanks for your help, knew it would be something simple!