PHP doesn't recognise form elements from Javascript

HTML: saucydares.freehostia.com/saucy.php I should just add that while there are no images, the site contains adult themes so isn't work safe.

PHP:

    <?php

    mysql_connect("mysql4.freehostia.com", sebsal2_db, "");

    function him()
    {
    $HIMquery = "SELECT dares FROM sebsal2_db.him UNION SELECT dares FROM sebsal2_db.other ORDER BY Rand() LIMIT 1";
    $HIMresult = mysql_query($HIMquery);
    while ($HIMrow = mysql_fetch_array($HIMresult, MYSQL_NUM)) {echo "$HIMrow[0]";}
    }

    function her()
    {
    $HERquery = "SELECT dares FROM sebsal2_db.her UNION SELECT dares FROM sebsal2_db.other ORDER BY Rand() LIMIT 1";
    $HERresult = mysql_query($HERquery);
    while ($HERrow = mysql_fetch_array($HERresult, MYSQL_NUM)) {echo "$HERrow[0]";}
    }

    function double()
    {
    $DOUBLEquery = "SELECT dares FROM sebsal2_db.double ORDER BY Rand() LIMIT 1";
    $DOUBLEresult = mysql_query($DOUBLEquery);
    while ($DOUBLErow = mysql_fetch_array($DOUBLEresult, MYSQL_NUM)) {echo "$DOUBLErow[0]";}
    }

    function himlong()
    {
    $HIMLONGquery = "SELECT dares FROM sebsal2_db.him2 UNION SELECT dares FROM sebsal2_db.other2 ORDER BY Rand() LIMIT 1";
    $HIMLONGresult = mysql_query($HIMLONGquery);
    while ($HIMLONGrow = mysql_fetch_array($HIMLONGresult, MYSQL_NUM)) {echo "$HIMLONGrow[0]";}
    }

    function herlong()
    {
    $HERLONGquery = "SELECT dares FROM sebsal2_db.her2 UNION SELECT dares FROM sebsal2_db.other2 ORDER BY Rand() LIMIT 1";
    $HERLONGresult = mysql_query($HERLONGquery);
    while ($HERLONGrow = mysql_fetch_array($HERLONGresult, MYSQL_NUM)) {echo "$HERLONGrow[0]";}
    }

    function doublelong()
    {
    $DOUBLELONGquery = "SELECT dares FROM sebsal2_db.double2 ORDER BY Rand() LIMIT 1";
    $DOUBLELONGresult = mysql_query($DOUBLELONGquery);
    while ($DOUBLELONGrow = mysql_fetch_array($DOUBLELONGresult, MYSQL_NUM)) {echo "$DOUBLELONGrow[0]";}
    }

var_dump($_POST)

/*    $mode = mysql_real_escape_string($_POST['mode']);
    $player = mysql_real_escape_string($_POST['player']);
    echo $player;
    echo $mode;

    if ($player=="For him" && $mode=="Classic Collection") {
    him();
    }

    if ($player=="For her" && $mode=="Classic Collection") {
    her();
    }

    if ($player=="Double dare" && $mode=="Classic Collection") {
    double();
    }

    if ($player=="For him" && $mode=="The Long Game") {
    himlong();
    }

    if ($player=="For her" && $mode=="The Long Game") {
    herlong();
    }

    if ($player=="Double dare" && $mode=="The Long Game") {
    doublelong();
    }
*/
    ?>

It is designed to read two drop down lists and display a random field from a mySQL database depending on the combination. Echoing 开发者_StackOverflow中文版the variables is not permanent, was just a test. Two problems:

• Using var_dump($_POST), if you select Classic Collection it sees all the options in the right hand box as "For him" and if you select The Long Game it sees them as "For her".

• For this reason the if statements don't work.

Thanks for your replies!

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Values are 'him','her' not 'For him','For her'

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Your code could use some serious condensing. The if statements should be nested, and the functions could be condensed to one function using parameters. A possible function signature:

function game($player, $mode)

Then base your business logic and queries inside the function on the parameters.