Question

我一直在努力解决这个问题一个小时左右，看起来很基本，但我似乎无法深究它。我在这里要做的是用我的mysql数据库中的内容创建一个块5次。

工作正常，但我真正的问题是当我尝试回显错误消息时。如果$ _POST ['champ5v5name [1-5]']处于默认值，我会在FOR检查后立即获得IF。如果是，它应该回显出一条错误消息，说“选择你的冠军！”，但它没有（它只是进入下一页，好像它是成功的）。我尝试了各种不同的方法，但没有一种方法有效，有人可以帮忙吗？

for($i=1;$i<=5;$i++) {
    $result = mysql_query("SELECT name,health,damage,armour,aspeed FROM champions");
    $htmltext .= '<label>Champion '.$i.'</label><br/>';
    $htmltext .=  '<select name="champ5v5name'.$i.'">';
    $htmltext .= '<option value="select'.$i.'">Select champion:</option>';
}
while($rowschamp = mysql_fetch_array($result,MYSQL_NUM)) {
    $htmltext .= '<option value="'.$rowschamp[0].'">'.$rowschamp[0].' (HP: '.$rowschamp[1].' DMG: '.$rowschamp[2].' ARMOUR: '.$rowschamp[3].' ASPEED: '.$rowschamp[4].')</option>';
}
$htmltext .= '</select><br/><br/>';

if($_POST['champ5v5name1'] != 'Select champion:' || $_POST['champ5v5name2'] != 'Select champion:' || $_POST['champ5v5name3'] != 'Select champion:' || $_POST['champ5v5name4'] != 'Select champion:' || $_POST['champ5v5name5'] != 'Select champion:') {
    if($_POST['champ5v5name1'] == $_POST['champ5v5name2'] || $_POST['champ5v5name1'] == $_POST['champ5v5name3'] || $_POST['champ5v5name1'] == $_POST['champ5v5name4'] || $_POST['champ5v5name1'] == $_POST['champ5v5name5']) $error = 'A champion is repeated.';
    if($_POST['champ5v5name2'] == $_POST['champ5v5name3'] || $_POST['champ5v5name2'] == $_POST['champ5v5name4'] || $_POST['champ5v5name2'] == $_POST['champ5v5name5'] || $_POST['champ5v5name2'] == $_POST['champ5v5name1']) $error = 'A champion is repeated.';
    if($_POST['champ5v5name3'] == $_POST['champ5v5name2'] || $_POST['champ5v5name3'] == $_POST['champ5v5name4'] || $_POST['champ5v5name3'] == $_POST['champ5v5name5'] || $_POST['champ5v5name3'] == $_POST['champ5v5name1']) $error = 'A champion is repeated.';
    if($_POST['champ5v5name4'] == $_POST['champ5v5name2'] || $_POST['champ5v5name4'] == $_POST['champ5v5name3'] || $_POST['champ5v5name4'] == $_POST['champ5v5name5'] || $_POST['champ5v5name4'] == $_POST['champ5v5name1']) $error = 'A champion is repeated.';
    if($_POST['champ5v5name5'] == $_POST['champ5v5name2'] || $_POST['champ5v5name5'] == $_POST['champ5v5name3'] || $_POST['champ5v5name5'] == $_POST['champ5v5name4'] || $_POST['champ5v5name5'] == $_POST['champ5v5name1']) $error = 'A champion is repeated.';
}
else {
    $error = 'Choose your champions!';
}

Answer 1

您正在做的就是为变量分配一个字符串，无论条件是true还是false，您都会设置错误。如果出现错误，我认为没有逻辑可以执行任何特定任务。

话虽如此，你真的需要简化一些事情：

try {
    $champions = array();
    $error     = null;
    for ($i = 1; $i < 6; $i++){
        $champions[] = $_POST['champ5v5name' . $i];
    }
    $dups      = array_count_values($champions);
    rsort($dups);

    // Check for missed assignments, assuming no champions have the word 'select' in their name
    if(in_array('select',$champions)){
        $error = 'Choose your champions!';
    }

    // Check for duplicate champs
    if($dups[0] != 1){
        $error = 'Champion Repeated!';
    }

    if($error){
        throw new Exception($error);
    }
}
catch (Exception $e) {
    echo $e->getMessage();
    exit(); // Probably don't exit, just show the form again.
}

// Do other stuff, everything is ok if the code gets here...

Answer 2

您是否正在检查所有冠军是否为默认值，或任何是否为默认值？

你的条件是：

$_POST['champ5v5name1'] != 'Select champion:' ||
$_POST['champ5v5name2'] != 'Select champion:' ||
$_POST['champ5v5name3'] != 'Select champion:' ||
$_POST['champ5v5name4'] != 'Select champion:' ||
$_POST['champ5v5name5'] != 'Select champion:'

如果您要参加任何活动，那应该是

$_POST['champ5v5name1'] != 'Select champion:' &&
$_POST['champ5v5name2'] != 'Select champion:' &&
$_POST['champ5v5name3'] != 'Select champion:' &&
$_POST['champ5v5name4'] != 'Select champion:' &&
$_POST['champ5v5name5'] != 'Select champion:'

编辑：

问题是表单提交的值为select$i，而不是Select champion:

编辑2：

您应该使用PHP的表单数组功能：

for($i = 1; $i <= 5; $i++) {
    $result = mysql_query("SELECT name,health,damage,armour,aspeed FROM champions");
    $htmltext .= '<label>Champion '.$i.'</label><br/>';

    $htmltext .= '<select name="champ5v5name[]">';
    #Field names ending in `[]` turn into arrays!

    $htmltext .= '<option value="">Select champion:</option>';
    while($rowschamp = mysql_fetch_array($result, MYSQL_NUM)) {
        $htmltext .= '<option value="'.$rowschamp[0].'">'.$rowschamp[0].' (HP: '.$rowschamp[1].' DMG: '.$rowschamp[2].' ARMOUR: '.$rowschamp[3].' ASPEED: '.$rowschamp[4].')</option>';
    }
    $htmltext .= '</select><br /><br />';
}

$champions = $_POST['champ5v5name'];

$error = "No champions chosen!";
for($champions as $champion) {
    if($champion) {
        $error = "";
        break;
    }
}
if(!$error && count($array) != count(array_unique($champions))) {
    $error = "A champion is repeated"
}

PHP $ _POST问题

2 个答案:

编辑：

编辑2：