Question

HTML：saucydares.freehostia.com/saucy.php我应该补充一点，虽然没有图片，但该网站包含成人主题，因此不安全。

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();
    }
*/
    ?>

它旨在读取两个下拉列表，并根据组合显示mySQL数据库中的随机字段。回应变量不是永久性的，只是一个测试。两个问题：

如果选择，使用var_dump（$ _ POST）经典系列它看到了所有右侧框中的选项为“For 他“如果你选择长游戏它认为它们是“为了她”。
因此if语句不工作。

感谢您的回复！

Answer 1

价值观是'他'，'她'不是'为了他'，'为了她'

Answer 2

您的代码可能会使用一些严重的冷凝。 if语句应该是嵌套的，并且函数可以使用参数压缩到一个函数。一个可能的函数签名：

function game($player, $mode)

然后根据参数在函数内部建立业务逻辑和查询。

PHP无法识别Javascript中的表单元素

2 个答案: